id stringlengths 11 15 | title stringlengths 7 171 | problem stringlengths 9 4.33k | solution stringlengths 6 19k | problem_wikitext stringlengths 9 4.42k | solution_wikitext stringlengths 7 19.1k | proof_title stringlengths 9 171 | theorem_url stringlengths 34 198 | proof_url stringlengths 36 198 | categories listlengths 0 9 | theorem_references listlengths 0 36 | proof_references listlengths 0 253 |
|---|---|---|---|---|---|---|---|---|---|---|---|
proofwiki-5300 | Characterization of Measurable Functions | Let $\struct {X, \Sigma}$ be a measurable space.
Let $f: X \to \overline \R$ be an extended real-valued function.
Then the following are all equivalent:
{{begin-eqn}}
{{eqn | n = 1
| o =
| r = f\) is measurable \(
}}
{{eqn | n = 2
| o =
| r = \forall \alpha \in \R: \set {x \in X: \map f x \le ... | Each of $(2)$ up to $(5')$ is equivalent to $(1)$ by combining Mapping Measurable iff Measurable on Generator and Generators for Extended Real Sigma-Algebra.
{{qed}} | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $f: X \to \overline \R$ be an [[Definition:Extended Real-Valued Function|extended real-valued function]].
Then the following are all equivalent:
{{begin-eqn}}
{{eqn | n = 1
| o =
| r = f\) is [[Definition:Measurable Fu... | Each of $(2)$ up to $(5')$ is equivalent to $(1)$ by combining [[Mapping Measurable iff Measurable on Generator]] and [[Generators for Extended Real Sigma-Algebra]].
{{qed}} | Characterization of Measurable Functions | https://proofwiki.org/wiki/Characterization_of_Measurable_Functions | https://proofwiki.org/wiki/Characterization_of_Measurable_Functions | [
"Measurable Functions"
] | [
"Definition:Measurable Space",
"Definition:Extended Real-Valued Function",
"Definition:Measurable Function"
] | [
"Mapping Measurable iff Measurable on Generator",
"Generators for Extended Real Sigma-Algebra"
] |
proofwiki-5301 | Characterization of Extended Real Sigma-Algebra | Let $\map \BB \R$ be the Borel $\sigma$-algebra on $\R$.
Let $\overline \BB$ be the extended real $\sigma$-algebra.
Define $\SS := \powerset {\set {-\infty, +\infty} }$, where $\PP$ denotes power set.
Then:
:$\overline \BB = \set {B \cup S: B \in \map \BB \R, S \in \SS}$ | Let $\overline B \in \overline \BB$.
Then by Extended Real Sigma-Algebra Induces Borel Sigma-Algebra on Reals, we have:
:$\overline B \cap \R \in \map \BB \R$
We also have, by definition of the extended real numbers $\overline \R$, that:
:$\overline \R \setminus \R = \set {-\infty, +\infty}$
and therefore, $\overline B... | Let $\map \BB \R$ be the [[Definition:Borel Sigma-Algebra|Borel $\sigma$-algebra]] on $\R$.
Let $\overline \BB$ be the [[Definition:Extended Real Sigma-Algebra|extended real $\sigma$-algebra]].
Define $\SS := \powerset {\set {-\infty, +\infty} }$, where $\PP$ denotes [[Definition:Power Set|power set]].
Then:
:$\ov... | Let $\overline B \in \overline \BB$.
Then by [[Extended Real Sigma-Algebra Induces Borel Sigma-Algebra on Reals]], we have:
:$\overline B \cap \R \in \map \BB \R$
We also have, by definition of the [[Definition:Extended Real Number Line|extended real numbers]] $\overline \R$, that:
:$\overline \R \setminus \R = \s... | Characterization of Extended Real Sigma-Algebra | https://proofwiki.org/wiki/Characterization_of_Extended_Real_Sigma-Algebra | https://proofwiki.org/wiki/Characterization_of_Extended_Real_Sigma-Algebra | [
"Extended Real Numbers",
"Sigma-Algebras"
] | [
"Definition:Borel Sigma-Algebra",
"Definition:Extended Real Sigma-Algebra",
"Definition:Power Set"
] | [
"Extended Real Sigma-Algebra Induces Borel Sigma-Algebra on Reals",
"Definition:Extended Real Number Line",
"Definition:Set Difference",
"Set Difference Union Intersection",
"Sigma-Algebra Closed under Union",
"Closed Set Measurable in Borel Sigma-Algebra",
"Definition:Closed Set/Topology",
"Extended ... |
proofwiki-5302 | Extended Real Sigma-Algebra Induces Borel Sigma-Algebra on Reals | Let $\overline \BB$ be the extended real $\sigma$-algebra.
Let $\map \BB \R$ be the Borel $\sigma$-algebra on $\R$.
Then:
:$\overline \BB_\R = \map \BB \R$
where $\overline \BB_\R$ denotes a trace $\sigma$-algebra. | We have Euclidean Space is Subspace of Extended Real Number Space.
The result follows from Borel Sigma-Algebra of Subset is Trace Sigma-Algebra.
{{qed}} | Let $\overline \BB$ be the [[Definition:Extended Real Sigma-Algebra|extended real $\sigma$-algebra]].
Let $\map \BB \R$ be the [[Definition:Borel Sigma-Algebra|Borel $\sigma$-algebra]] on $\R$.
Then:
:$\overline \BB_\R = \map \BB \R$
where $\overline \BB_\R$ denotes a [[Definition:Trace Sigma-Algebra|trace $\sigma... | We have [[Euclidean Space is Subspace of Extended Real Number Space]].
The result follows from [[Borel Sigma-Algebra of Subset is Trace Sigma-Algebra]].
{{qed}} | Extended Real Sigma-Algebra Induces Borel Sigma-Algebra on Reals | https://proofwiki.org/wiki/Extended_Real_Sigma-Algebra_Induces_Borel_Sigma-Algebra_on_Reals | https://proofwiki.org/wiki/Extended_Real_Sigma-Algebra_Induces_Borel_Sigma-Algebra_on_Reals | [
"Extended Real Numbers",
"Sigma-Algebras"
] | [
"Definition:Extended Real Sigma-Algebra",
"Definition:Borel Sigma-Algebra",
"Definition:Trace Sigma-Algebra"
] | [
"Euclidean Space is Subspace of Extended Real Number Space",
"Borel Sigma-Algebra of Subset is Trace Sigma-Algebra"
] |
proofwiki-5303 | Generators for Extended Real Sigma-Algebra | Let $\overline \BB$ be the extended real $\sigma$-algebra.
Then $\overline \BB$ is generated by each of the following collections of extended real intervals:
{{begin-eqn}}
{{eqn | n = 1
| o =
| r = \set {\ \closedint a \to: a \in \R}
}}
{{eqn | n = 1'
| o =
| r = \set {\ \closedint a \to: a \i... | Let us first establish that $(1)$ up to $(4')$ all generate the same $\sigma$-algebra.
Denote $\GG_i$ for the collection at point $(i)$, and $\GG'_i$ for that at $(i')$, where $i = 1, 2, 3, 4$.
Furthermore, write $\Sigma_i$ for $\map \sigma {\GG_i}$ and $\Sigma'_i$ for $\map \sigma {\GG'_i}$.
Here $\sigma$ denotes gene... | Let $\overline \BB$ be the [[Definition:Extended Real Sigma-Algebra|extended real $\sigma$-algebra]].
Then $\overline \BB$ is [[Definition:Sigma-Algebra Generated by Collection of Subsets|generated]] by each of the following collections of [[Definition:Extended Real Interval|extended real intervals]]:
{{begin-eqn}}
... | Let us first establish that $(1)$ up to $(4')$ all [[Definition:Sigma-Algebra Generated by Collection of Subsets|generate]] the same [[Definition:Sigma-Algebra|$\sigma$-algebra]].
Denote $\GG_i$ for the collection at point $(i)$, and $\GG'_i$ for that at $(i')$, where $i = 1, 2, 3, 4$.
Furthermore, write $\Sigma_i$ f... | Generators for Extended Real Sigma-Algebra | https://proofwiki.org/wiki/Generators_for_Extended_Real_Sigma-Algebra | https://proofwiki.org/wiki/Generators_for_Extended_Real_Sigma-Algebra | [
"Extended Real Numbers",
"Sigma-Algebras"
] | [
"Definition:Extended Real Sigma-Algebra",
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Definition:Extended Real Interval"
] | [
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Definition:Sigma-Algebra",
"Definition:Sigma-Algebra Generated by Collection of Subsets",
"Generated Sigma-Algebra Preserves Subset",
"Definition:Subset",
"Definition:Set Complement",
"Definition:Sigma-Algebra Generated by Collection of Sub... |
proofwiki-5304 | Characteristic Function Measurable iff Set Measurable | Let $\struct {X, \Sigma}$ be a measurable space.
Let $E \subseteq X$.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$E \in \Sigma$; that is, $E$ is a $\Sigma$-measurable set}}
{{item|(2):|$\chi_E: X \to \set {0, 1}$, the characteristic function of $E$, is $\Sigma$-measurable}}
{{end-itemize}} | === $(1)$ implies $(2)$ ===
Assume that $E \in \Sigma$.
It is clear that $x \notin \set {0, 1}$ implies $\map {\chi_E^{-1} } x = \O$.
Hence Preimage of Union under Mapping and Characteristic Function Determined by 1-Fiber yield, for any $\alpha \in \R$:
:<nowiki>$\set {x \in X: \map {\chi_E} x \ge \alpha} = \begin{case... | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $E \subseteq X$.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$E \in \Sigma$; that is, $E$ is a [[Definition:Measurable Set|$\Sigma$-measurable set]]}}
{{item|(2):|$\chi_E: X \to \set {0, 1}$, the [[Definition:Characteristic Function of S... | === $(1)$ implies $(2)$ ===
Assume that $E \in \Sigma$.
It is clear that $x \notin \set {0, 1}$ implies $\map {\chi_E^{-1} } x = \O$.
Hence [[Preimage of Union under Mapping]] and [[Characteristic Function Determined by 1-Fiber]] yield, for any $\alpha \in \R$:
:<nowiki>$\set {x \in X: \map {\chi_E} x \ge \alpha} ... | Characteristic Function Measurable iff Set Measurable | https://proofwiki.org/wiki/Characteristic_Function_Measurable_iff_Set_Measurable | https://proofwiki.org/wiki/Characteristic_Function_Measurable_iff_Set_Measurable | [
"Characteristic Functions",
"Measurable Sets",
"Measurable Functions"
] | [
"Definition:Measurable Space",
"Definition:Measurable Set",
"Definition:Characteristic Function (Set Theory)/Set",
"Definition:Measurable Function"
] | [
"Preimage of Union under Mapping",
"Characteristic Function Determined by 1-Fiber",
"Definition:Sigma-Algebra",
"Sigma-Algebra Contains Empty Set",
"Characterization of Measurable Functions",
"Definition:Measurable Function",
"Definition:Measurable Function",
"Definition:Measurable Function"
] |
proofwiki-5305 | Simple Function is Measurable | Let $\struct {X, \Sigma}$ be a measurable space.
Let $f: X \to \R$ be a simple function.
Then $f$ is $\Sigma$-measurable. | Let $f$ be written in the following form:
:$f = \ds \sum_{i \mathop = 1}^n a_i \chi_{S_i}$
where $a_i \in \R$ and the $S_i$ are $\Sigma$-measurable.
Next, for each ordered $n$-tuple $b$ of zeroes and ones define:
:<nowiki>$\map {T_b} i := \begin{cases}
S_i & : \text {if $\map b i = 0$}\\
X \setminus S_i & : \text {if $... | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $f: X \to \R$ be a [[Definition:Simple Function|simple function]].
Then $f$ is [[Definition:Measurable Function|$\Sigma$-measurable]]. | Let $f$ be written in the following form:
:$f = \ds \sum_{i \mathop = 1}^n a_i \chi_{S_i}$
where $a_i \in \R$ and the $S_i$ are [[Definition:Measurable Set|$\Sigma$-measurable]].
Next, for each [[Definition:Ordered Tuple|ordered $n$-tuple]] $b$ of zeroes and ones define:
:<nowiki>$\map {T_b} i := \begin{cases}
S_i... | Simple Function is Measurable | https://proofwiki.org/wiki/Simple_Function_is_Measurable | https://proofwiki.org/wiki/Simple_Function_is_Measurable | [
"Measurable Functions",
"Simple Functions"
] | [
"Definition:Measurable Space",
"Definition:Simple Function",
"Definition:Measurable Function"
] | [
"Definition:Measurable Set",
"Definition:Ordered Tuple",
"Sigma-Algebra Closed under Finite Intersection",
"Definition:Pairwise Disjoint",
"Sigma-Algebra Closed under Union",
"Definition:Measurable Set",
"Characterization of Measurable Functions",
"Definition:Measurable Function"
] |
proofwiki-5306 | Identity Mapping is Relation Isomorphism | Let $\struct {S, \RR}$ be a relational structure.
Then the identity mapping $I_S: S \to S$ is a relation isomorphism from $\struct {S, \RR}$ to itself. | By definition of identity mapping:
:$\forall x \in S: \map {I_S} x = x$
So:
:$x \mathrel \RR y \implies \map {I_S} x \mathrel \RR \map {I_S} y$
From Identity Mapping is Bijection, $I_S$ is a bijection.
Hence:
:$\map {I_S^{-1} } x = x$
So:
:$x \mathrel \RR y \implies \map {I_S^{-1} } x \mathrel \RR \map {I_S^{-1} } y$
{... | Let $\struct {S, \RR}$ be a [[Definition:Relational Structure|relational structure]].
Then the [[Definition:Identity Mapping|identity mapping]] $I_S: S \to S$ is a [[Definition:Relation Isomorphism|relation isomorphism]] from $\struct {S, \RR}$ to itself. | By definition of [[Definition:Identity Mapping|identity mapping]]:
:$\forall x \in S: \map {I_S} x = x$
So:
:$x \mathrel \RR y \implies \map {I_S} x \mathrel \RR \map {I_S} y$
From [[Identity Mapping is Bijection]], $I_S$ is a [[Definition:Bijection|bijection]].
Hence:
:$\map {I_S^{-1} } x = x$
So:
:$x \mathrel \R... | Identity Mapping is Relation Isomorphism | https://proofwiki.org/wiki/Identity_Mapping_is_Relation_Isomorphism | https://proofwiki.org/wiki/Identity_Mapping_is_Relation_Isomorphism | [
"Relation Isomorphisms",
"Identity Mappings"
] | [
"Definition:Relational Structure",
"Definition:Identity Mapping",
"Definition:Relation Isomorphism"
] | [
"Definition:Identity Mapping",
"Identity Mapping is Bijection",
"Definition:Bijection"
] |
proofwiki-5307 | Inverse of Relation Isomorphism is Relation Isomorphism | Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be relational structures.
Let $\phi: \struct {S, \RR_1} \to \struct {T, \RR_2}$ be a bijection.
Then:
:$\phi: \struct {S, \RR_1} \to \struct {T, \RR_2}$
is a relation isomorphism
{{iff}}:
:$\phi^{-1}: \struct {T, \RR_2} \to \struct {S, \RR_1}$
is also a relation isomorp... | Follows directly from the definition of relation isomorphism.
{{Qed}} | Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be [[Definition:Relational Structure|relational structures]].
Let $\phi: \struct {S, \RR_1} \to \struct {T, \RR_2}$ be a [[Definition:Bijection|bijection]].
Then:
:$\phi: \struct {S, \RR_1} \to \struct {T, \RR_2}$
is a [[Definition:Relation Isomorphism|relation isomo... | Follows directly from the definition of [[Definition:Relation Isomorphism|relation isomorphism]].
{{Qed}} | Inverse of Relation Isomorphism is Relation Isomorphism | https://proofwiki.org/wiki/Inverse_of_Relation_Isomorphism_is_Relation_Isomorphism | https://proofwiki.org/wiki/Inverse_of_Relation_Isomorphism_is_Relation_Isomorphism | [
"Relation Isomorphisms"
] | [
"Definition:Relational Structure",
"Definition:Bijection",
"Definition:Relation Isomorphism",
"Definition:Relation Isomorphism"
] | [
"Definition:Relation Isomorphism"
] |
proofwiki-5308 | Measurable Function is Simple Function iff Finite Image Set | Let $\struct {X, \Sigma}$ be a measurable space.
Let $f: X \to \R$ be a measurable function.
Then $f$ is a simple function {{iff}} its image is finite:
:$\card {\Img f} < \infty$ | === Necessary Condition ===
Suppose that $f$ is a simple function, and that:
:$\ds \forall x \in X: \map f x = \sum_{i \mathop = 1}^n a_i \map {\chi_{S_i} } x$
Since each of the $\chi_{S_i}$ is a characteristic function, it can take only the values $0$ and $1$.
Thus each summand can take two values.
It follows immediat... | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $f: X \to \R$ be a [[Definition:Measurable Function|measurable function]].
Then $f$ is a [[Definition:Simple Function|simple function]] {{iff}} its [[Definition:Image of Mapping|image]] is [[Definition:Finite Set|finite]]:
:$\card ... | === Necessary Condition ===
Suppose that $f$ is a [[Definition:Simple Function|simple function]], and that:
:$\ds \forall x \in X: \map f x = \sum_{i \mathop = 1}^n a_i \map {\chi_{S_i} } x$
Since each of the $\chi_{S_i}$ is a [[Definition:Characteristic Function of Set|characteristic function]], it can take only th... | Measurable Function is Simple Function iff Finite Image Set | https://proofwiki.org/wiki/Measurable_Function_is_Simple_Function_iff_Finite_Image_Set | https://proofwiki.org/wiki/Measurable_Function_is_Simple_Function_iff_Finite_Image_Set | [
"Measurable Functions",
"Simple Functions"
] | [
"Definition:Measurable Space",
"Definition:Measurable Function",
"Definition:Simple Function",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Finite Set"
] | [
"Definition:Simple Function",
"Definition:Characteristic Function (Set Theory)/Set",
"Definition:Addition/Summand",
"Simple Function is Measurable",
"Definition:Simple Function"
] |
proofwiki-5309 | Composite of Relation Isomorphisms is Relation Isomorphism | Let $\struct {S_1, \RR_1}$, $\struct {S_2, \RR_2}$ and $\struct {S_3, \RR_3}$ be relational structures.
Let:
:$\phi: \struct {S_1, \RR_1} \to \struct {S_2, \RR_2}$
and:
:$\psi: \struct {S_2, \RR_2} \to \struct {S_3, \RR_3}$
be relation isomorphisms.
Then $\psi \circ \phi: \struct {S_1, \RR_1} \to \struct {S_3, \RR_3}$ ... | From Composite of Bijections is Bijection, $\psi \circ \phi$ is a bijection, as, by definition, an relation isomorphism is also a bijection.
By definition of composition of mappings:
:$\map {\psi \circ \phi} x = \map \psi {\map \phi x}$
As $\phi$ is a relation isomorphism, we have:
:$\forall x_1, y_1 \in S_1: x_1 \math... | Let $\struct {S_1, \RR_1}$, $\struct {S_2, \RR_2}$ and $\struct {S_3, \RR_3}$ be [[Definition:Relational Structure|relational structures]].
Let:
:$\phi: \struct {S_1, \RR_1} \to \struct {S_2, \RR_2}$
and:
:$\psi: \struct {S_2, \RR_2} \to \struct {S_3, \RR_3}$
be [[Definition:Relation Isomorphism|relation isomorphisms]... | From [[Composite of Bijections is Bijection]], $\psi \circ \phi$ is a [[Definition:Bijection|bijection]], as, by definition, an [[Definition:Relation Isomorphism|relation isomorphism]] is also a [[Definition:Bijection|bijection]].
By definition of [[Definition:Composition of Mappings|composition of mappings]]:
:$\map... | Composite of Relation Isomorphisms is Relation Isomorphism | https://proofwiki.org/wiki/Composite_of_Relation_Isomorphisms_is_Relation_Isomorphism | https://proofwiki.org/wiki/Composite_of_Relation_Isomorphisms_is_Relation_Isomorphism | [
"Relation Isomorphisms"
] | [
"Definition:Relational Structure",
"Definition:Relation Isomorphism",
"Definition:Relation Isomorphism"
] | [
"Composite of Bijections is Bijection",
"Definition:Bijection",
"Definition:Relation Isomorphism",
"Definition:Bijection",
"Definition:Composition of Mappings",
"Definition:Relation Isomorphism",
"Definition:Relation Isomorphism",
"Definition:Relation Isomorphism"
] |
proofwiki-5310 | Pointwise Sum of Simple Functions is Simple Function | Let $\struct {X, \Sigma}$ be a measurable space.
Let $f, g : X \to \R$ be simple functions.
Then the pointwise sum $f + g: X \to \R$ of $f$ and $g$:
:$\forall x, y \in X: \map {\paren {f + g} } x := \map f x + \map g x$
is also a simple function. | We have $f + g = + \circ \innerprod f g \circ \Delta_X$, where:
:$\Delta_X: X \to X \times X$ is the diagonal mapping on $X$
:$\innerprod f g: X \times X \to \R \times \R, \map {\innerprod f g} {x, y} := \tuple {\map f x, \map g y}$
:$+: \R \times \R \to \R$ is real addition.
{{explain|What is the meaning of the notati... | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $f, g : X \to \R$ be [[Definition:Simple Function|simple functions]].
Then the [[Definition:Pointwise Addition|pointwise sum]] $f + g: X \to \R$ of $f$ and $g$:
:$\forall x, y \in X: \map {\paren {f + g} } x := \map f x + \map g x$
... | We have $f + g = + \circ \innerprod f g \circ \Delta_X$, where:
:$\Delta_X: X \to X \times X$ is the [[Definition:Diagonal Mapping|diagonal mapping]] on $X$
:$\innerprod f g: X \times X \to \R \times \R, \map {\innerprod f g} {x, y} := \tuple {\map f x, \map g y}$
:$+: \R \times \R \to \R$ is [[Definition:Real Additio... | Pointwise Sum of Simple Functions is Simple Function | https://proofwiki.org/wiki/Pointwise_Sum_of_Simple_Functions_is_Simple_Function | https://proofwiki.org/wiki/Pointwise_Sum_of_Simple_Functions_is_Simple_Function | [
"Simple Functions",
"Pointwise Operations"
] | [
"Definition:Measurable Space",
"Definition:Simple Function",
"Definition:Pointwise Addition",
"Definition:Simple Function"
] | [
"Definition:Diagonal Mapping",
"Definition:Addition/Real Numbers",
"Definition:Restriction/Mapping",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Measurable Function is Simple Function iff Finite Image Set",
"Definition:Finite Set",
"Cardinality of... |
proofwiki-5311 | Pointwise Product of Simple Functions is Simple Function | Let $\struct {X, \Sigma}$ be a measurable space.
Let $f, g : X \to \R$ be simple functions.
Then $f \cdot g: X \to \R, \map {\paren {f \cdot g} } x := \map f x \cdot \map g x$ is also a simple function. | From Measurable Function is Simple Function iff Finite Image Set, it follows that there exist $x_1, \ldots, x_n$ and $y_1, \ldots y_m$ comprising the image of $f$ and $g$, respectively.
But then it immediately follows that any value attained by $f \cdot g$ is of the form $x_i \cdot y_j$.
Hence, there are at most $n \ti... | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $f, g : X \to \R$ be [[Definition:Simple Function|simple functions]].
Then $f \cdot g: X \to \R, \map {\paren {f \cdot g} } x := \map f x \cdot \map g x$ is also a [[Definition:Simple Function|simple function]]. | From [[Measurable Function is Simple Function iff Finite Image Set]], it follows that there exist $x_1, \ldots, x_n$ and $y_1, \ldots y_m$ comprising the [[Definition:Image of Mapping|image]] of $f$ and $g$, respectively.
But then it immediately follows that any value attained by $f \cdot g$ is of the form $x_i \cdot ... | Pointwise Product of Simple Functions is Simple Function | https://proofwiki.org/wiki/Pointwise_Product_of_Simple_Functions_is_Simple_Function | https://proofwiki.org/wiki/Pointwise_Product_of_Simple_Functions_is_Simple_Function | [
"Simple Functions",
"Pointwise Operations",
"Pointwise Product of Simple Functions is Simple Function"
] | [
"Definition:Measurable Space",
"Definition:Simple Function",
"Definition:Simple Function"
] | [
"Measurable Function is Simple Function iff Finite Image Set",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Pointwise Product of Measurable Functions is Measurable",
"Definition:Measurable Function",
"Measurable Function is Simple Function iff Finite Image Set",
"Definition:Simple Function"
] |
proofwiki-5312 | Positive Part of Simple Function is Simple Function | Let $\struct {X, \Sigma}$ be a measurable space.
Let $f: X \to \R$ be a simple function.
Then $f^+: X \to \R$, the positive part of $f$, is also a simple function. | Let $f$ have the following standard representation:
:$f = \ds \sum_{i \mathop = 0}^n a_i \chi_{E_i}$
Then we see that $f^+$ must satisfy:
:$f^+ = \ds \sum_{i \mathop = 0}^n \max \set {a_i, 0} \chi_{E_i}$
as the $E_i$ are disjoint, and $\chi_{E_i} \ge 0$ pointwise.
Since all of the $E_i$ are measurable, it follows that ... | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $f: X \to \R$ be a [[Definition:Simple Function|simple function]].
Then $f^+: X \to \R$, the [[Definition:Positive Part|positive part]] of $f$, is also a [[Definition:Simple Function|simple function]]. | Let $f$ have the following [[Definition:Standard Representation of Simple Function|standard representation]]:
:$f = \ds \sum_{i \mathop = 0}^n a_i \chi_{E_i}$
Then we see that $f^+$ must satisfy:
:$f^+ = \ds \sum_{i \mathop = 0}^n \max \set {a_i, 0} \chi_{E_i}$
as the $E_i$ are [[Definition:Disjoint Sets|disjoint]... | Positive Part of Simple Function is Simple Function | https://proofwiki.org/wiki/Positive_Part_of_Simple_Function_is_Simple_Function | https://proofwiki.org/wiki/Positive_Part_of_Simple_Function_is_Simple_Function | [
"Positive Parts",
"Simple Functions",
"Positive Parts"
] | [
"Definition:Measurable Space",
"Definition:Simple Function",
"Definition:Positive Part",
"Definition:Simple Function"
] | [
"Definition:Standard Representation of Simple Function",
"Definition:Disjoint Sets",
"Definition:Pointwise Inequality",
"Definition:Measurable Set",
"Definition:Simple Function"
] |
proofwiki-5313 | Negative Part of Simple Function is Simple Function | Let $\struct {X, \Sigma}$ be a measurable space.
Let $f: X \to \R$ be a simple function.
Then $f^-: X \to \R$, the negative part of $f$ is also a simple function. | Let $f$ have the following standard representation:
:$f = \ds \sum_{i \mathop = 0}^n a_i \chi_{E_i}$
Then we see that $f^-$ must satisfy:
:$f^- = \ds \sum_{i \mathop = 0}^n \min \set {a_i, 0} \chi_{E_i}$
as the $E_i$ are disjoint, and $\chi_{E_i} \ge 0$ pointwise.
Since all of the $E_i$ are measurable, it follows that ... | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $f: X \to \R$ be a [[Definition:Simple Function|simple function]].
Then $f^-: X \to \R$, the [[Definition:Negative Part|negative part]] of $f$ is also a [[Definition:Simple Function|simple function]]. | Let $f$ have the following [[Definition:Standard Representation of Simple Function|standard representation]]:
:$f = \ds \sum_{i \mathop = 0}^n a_i \chi_{E_i}$
Then we see that $f^-$ must satisfy:
:$f^- = \ds \sum_{i \mathop = 0}^n \min \set {a_i, 0} \chi_{E_i}$
as the $E_i$ are [[Definition:Disjoint Sets|disjoint]... | Negative Part of Simple Function is Simple Function | https://proofwiki.org/wiki/Negative_Part_of_Simple_Function_is_Simple_Function | https://proofwiki.org/wiki/Negative_Part_of_Simple_Function_is_Simple_Function | [
"Negative Parts",
"Simple Functions",
"Negative Parts"
] | [
"Definition:Measurable Space",
"Definition:Simple Function",
"Definition:Negative Part",
"Definition:Simple Function"
] | [
"Definition:Standard Representation of Simple Function",
"Definition:Disjoint Sets",
"Definition:Pointwise Inequality",
"Definition:Measurable Set",
"Definition:Simple Function"
] |
proofwiki-5314 | Difference of Positive and Negative Parts | Let $X$ be a set, and let $f: X \to \overline{\R}$ be an extended real-valued function.
Let $f^+$, $f^-: X \to \overline{\R}$ be the positive and negative parts of $f$, respectively.
Then $f = f^+ - f^-$. | Let $x \in X$.
Consider the following four cases for the value of $\map f x$ in $\overline{\R}$:
$(1): \quad \map f x = -\infty$
: By ordering on extended reals:
::$\map {f^+} x = \map \max {0, \map f x} = \map \max {0, -\infty} = 0$
::$\map {f^-} x = - \map \min {0, \map f x} = - \map \min {0, -\infty} = +\infty$
: B... | Let $X$ be a [[Definition:Set|set]], and let $f: X \to \overline{\R}$ be an [[Definition:Extended Real-Valued Function|extended real-valued function]].
Let $f^+$, $f^-: X \to \overline{\R}$ be the [[Definition:Positive Part|positive]] and [[Definition:Negative Part|negative parts]] of $f$, respectively.
Then $f = f^... | Let $x \in X$.
Consider the following four cases for the value of $\map f x$ in $\overline{\R}$:
$(1): \quad \map f x = -\infty$
: By [[Definition:Ordering on Extended Real Numbers|ordering on extended reals]]:
::$\map {f^+} x = \map \max {0, \map f x} = \map \max {0, -\infty} = 0$
::$\map {f^-} x = - \map \min {0,... | Difference of Positive and Negative Parts | https://proofwiki.org/wiki/Difference_of_Positive_and_Negative_Parts | https://proofwiki.org/wiki/Difference_of_Positive_and_Negative_Parts | [
"Positive Parts",
"Negative Parts",
"Mapping Theory",
"Positive Parts",
"Negative Parts"
] | [
"Definition:Set",
"Definition:Extended Real-Valued Function",
"Definition:Positive Part",
"Definition:Negative Part"
] | [
"Definition:Ordering on Extended Real Numbers",
"Definition:Extended Real Subtraction",
"Definition:Ordering on Extended Real Numbers",
"Definition:Extended Real Subtraction"
] |
proofwiki-5315 | Sum of Positive and Negative Parts | Let $X$ be a set, and let $f: X \to \overline \R$ be an extended real-valued function.
Let $f^+, f^-: X \to \overline \R$ be the positive and negative parts of $f$, respectively.
Then $\size {f} = f^+ + f^-$, where $\size {f}$ is the absolute value of $f$. | Let $x \in X$.
Suppose that $\map f x \ge 0$, where $\ge$ signifies the extended real ordering.
Then $\size {\map f x} = \map f x$, and:
:$\map {f^+} x = \map \max {\map f x, 0} = \map f x$
:$\map {f^-} x = - \map \min {\map f x, 0} = 0$
Hence $\map {f^+} x + \map {f^-} x = \map f x = \size {\map f x}$.
Next, suppose t... | Let $X$ be a [[Definition:Set|set]], and let $f: X \to \overline \R$ be an [[Definition:Extended Real-Valued Function|extended real-valued function]].
Let $f^+, f^-: X \to \overline \R$ be the [[Definition:Positive Part|positive]] and [[Definition:Negative Part|negative parts]] of $f$, respectively.
Then $\size {f} ... | Let $x \in X$.
Suppose that $\map f x \ge 0$, where $\ge$ signifies the [[Definition:Extended Real Ordering|extended real ordering]].
Then $\size {\map f x} = \map f x$, and:
:$\map {f^+} x = \map \max {\map f x, 0} = \map f x$
:$\map {f^-} x = - \map \min {\map f x, 0} = 0$
Hence $\map {f^+} x + \map {f^-} x = \m... | Sum of Positive and Negative Parts | https://proofwiki.org/wiki/Sum_of_Positive_and_Negative_Parts | https://proofwiki.org/wiki/Sum_of_Positive_and_Negative_Parts | [
"Positive Parts",
"Negative Parts",
"Mapping Theory",
"Positive Parts",
"Negative Parts"
] | [
"Definition:Set",
"Definition:Extended Real-Valued Function",
"Definition:Positive Part",
"Definition:Negative Part",
"Definition:Absolute Value of Mapping/Extended Real-Valued Function"
] | [
"Definition:Ordering on Extended Real Numbers",
"Definition:Ordering on Extended Real Numbers"
] |
proofwiki-5316 | Absolute Value of Simple Function is Simple Function | Let $\struct {X, \Sigma}$ be a measurable space.
Let $f: X \to \R$ be a simple function.
Then $\size f: X \to \R$, the absolute value of $f$, is also a simple function. | By Sum of Positive and Negative Parts, we have:
:$\size f = f^+ + f^-$
We also have that Positive Part of Simple Function is Simple Function and Negative Part of Simple Function is Simple Function.
Hence $\size f$ is a pointwise sum of simple functions.
The result follows from Pointwise Sum of Simple Functions is Simpl... | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $f: X \to \R$ be a [[Definition:Simple Function|simple function]].
Then $\size f: X \to \R$, the [[Definition:Absolute Value of Real-Valued Function|absolute value of $f$]], is also a [[Definition:Simple Function|simple function]]. | By [[Sum of Positive and Negative Parts]], we have:
:$\size f = f^+ + f^-$
We also have that [[Positive Part of Simple Function is Simple Function]] and [[Negative Part of Simple Function is Simple Function]].
Hence $\size f$ is a [[Definition:Pointwise Addition|pointwise sum]] of [[Definition:Simple Function|simple... | Absolute Value of Simple Function is Simple Function/Proof 1 | https://proofwiki.org/wiki/Absolute_Value_of_Simple_Function_is_Simple_Function | https://proofwiki.org/wiki/Absolute_Value_of_Simple_Function_is_Simple_Function/Proof_1 | [
"Simple Functions",
"Absolute Value of Simple Function is Simple Function"
] | [
"Definition:Measurable Space",
"Definition:Simple Function",
"Definition:Absolute Value of Mapping/Real-Valued Function",
"Definition:Simple Function"
] | [
"Sum of Positive and Negative Parts",
"Positive Part of Simple Function is Simple Function",
"Negative Part of Simple Function is Simple Function",
"Definition:Pointwise Addition",
"Definition:Simple Function",
"Pointwise Sum of Simple Functions is Simple Function"
] |
proofwiki-5317 | Absolute Value of Simple Function is Simple Function | Let $\struct {X, \Sigma}$ be a measurable space.
Let $f: X \to \R$ be a simple function.
Then $\size f: X \to \R$, the absolute value of $f$, is also a simple function. | By Simple Function has Standard Representation, $f$ has a standard representation, say:
:$(1): \quad f = \ds \sum_{k \mathop = 0}^n a_k \chi_{E_k}$
Then, for all $x \in X$:
:$\map {\size f} x = \ds \size {\sum_{k \mathop = 0}^n a_k \map {\chi_{E_k} } x}$
by definition of pointwise absolute value.
The fact that $(1)$ fo... | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $f: X \to \R$ be a [[Definition:Simple Function|simple function]].
Then $\size f: X \to \R$, the [[Definition:Absolute Value of Real-Valued Function|absolute value of $f$]], is also a [[Definition:Simple Function|simple function]]. | By [[Simple Function has Standard Representation]], $f$ has a [[Definition:Standard Representation of Simple Function|standard representation]], say:
:$(1): \quad f = \ds \sum_{k \mathop = 0}^n a_k \chi_{E_k}$
Then, for all $x \in X$:
:$\map {\size f} x = \ds \size {\sum_{k \mathop = 0}^n a_k \map {\chi_{E_k} } x}$
... | Absolute Value of Simple Function is Simple Function/Proof 2 | https://proofwiki.org/wiki/Absolute_Value_of_Simple_Function_is_Simple_Function | https://proofwiki.org/wiki/Absolute_Value_of_Simple_Function_is_Simple_Function/Proof_2 | [
"Simple Functions",
"Absolute Value of Simple Function is Simple Function"
] | [
"Definition:Measurable Space",
"Definition:Simple Function",
"Definition:Absolute Value of Mapping/Real-Valued Function",
"Definition:Simple Function"
] | [
"Measurable Function is Simple Function iff Finite Image Set/Corollary",
"Definition:Standard Representation of Simple Function",
"Definition:Pointwise Absolute Value",
"Definition:Standard Representation of Simple Function",
"Definition:Characteristic Function (Set Theory)/Set",
"Definition:Simple Functi... |
proofwiki-5318 | Measurable Function is Pointwise Limit of Simple Functions | Let $\struct {X, \Sigma}$ be a measurable space.
Let $f: X \to \overline \R$ be a $\Sigma$-measurable function.
Then there exists a sequence $\sequence {f_n}_{n \mathop \in \N} \in \map \EE \Sigma$ of simple functions, such that:
:$\forall x \in X: \map f x = \ds \lim_{n \mathop \to \infty} \map {f_n} x$
That is, such ... | First, let us prove the theorem when $f$ is a positive $\Sigma$-measurable function.
Now for any $n \in \N$, define for $0 \le k \le n 2^n$:
:$<nowiki>{A_k}^n := \begin{cases}
\set {k 2^{-n} \le f < \paren {k + 1} 2^{-n} } & : k \ne n 2^n \\
\set {f \ge n} & : k = n 2^n
\end{cases}</nowiki>$
where for example $\set {... | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $f: X \to \overline \R$ be a [[Definition:Measurable Function|$\Sigma$-measurable function]].
Then there exists a [[Definition:Sequence|sequence]] $\sequence {f_n}_{n \mathop \in \N} \in \map \EE \Sigma$ of [[Definition:Simple Funct... | First, let us prove the theorem when $f$ is a [[Definition:Positive Measurable Function|positive $\Sigma$-measurable function]].
Now for any $n \in \N$, define for $0 \le k \le n 2^n$:
:$<nowiki>{A_k}^n := \begin{cases}
\set {k 2^{-n} \le f < \paren {k + 1} 2^{-n} } & : k \ne n 2^n \\
\set {f \ge n} & : k = n 2^n
\... | Measurable Function is Pointwise Limit of Simple Functions | https://proofwiki.org/wiki/Measurable_Function_is_Pointwise_Limit_of_Simple_Functions | https://proofwiki.org/wiki/Measurable_Function_is_Pointwise_Limit_of_Simple_Functions | [
"Measurable Functions",
"Simple Functions"
] | [
"Definition:Measurable Space",
"Definition:Measurable Function",
"Definition:Sequence",
"Definition:Simple Function",
"Definition:Pointwise Limit",
"Definition:Sequence",
"Definition:Increasing Sequence of Real-Valued Functions"
] | [
"Definition:Measurable Function/Positive",
"Definition:Pairwise Disjoint",
"Definition:Characteristic Function (Set Theory)/Set",
"Definition:Pointwise Inequality of Extended Real-Valued Functions",
"Characterization of Measurable Functions",
"Sigma-Algebra Closed under Finite Intersection",
"Definition... |
proofwiki-5319 | Finite Cartesian Product of Non-Empty Sets is Non-Empty | Let $S_1, S_2, \ldots, S_n$ be non-empty sets.
Then their cartesian product $S_1 \times S_2 \times \cdots \times S_n$ is non-empty. | We use mathematical induction.
The base case $n = 2$ is proved in Kuratowski Formalization of Ordered Pair, and the induction step follows directly from the definition of an ordered tuple.
{{qed}}
{{finish}}
Category:Cartesian Product
dmytyq44dk054dtplnkbkhl67vh1by6 | Let $S_1, S_2, \ldots, S_n$ be [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|sets]].
Then their [[Definition:Finite Cartesian Product|cartesian product]] $S_1 \times S_2 \times \cdots \times S_n$ is [[Definition:Non-Empty Set|non-empty]]. | We use [[Principle of Mathematical Induction|mathematical induction]].
The [[Definition:Basis for the Induction|base case]] $n = 2$ is proved in [[Kuratowski Formalization of Ordered Pair]], and the [[Definition:Induction Step|induction step]] follows directly from the definition of an [[Definition:Ordered Tuple as Or... | Finite Cartesian Product of Non-Empty Sets is Non-Empty | https://proofwiki.org/wiki/Finite_Cartesian_Product_of_Non-Empty_Sets_is_Non-Empty | https://proofwiki.org/wiki/Finite_Cartesian_Product_of_Non-Empty_Sets_is_Non-Empty | [
"Cartesian Product"
] | [
"Definition:Non-Empty Set",
"Definition:Set",
"Definition:Cartesian Product/Finite",
"Definition:Non-Empty Set"
] | [
"Principle of Mathematical Induction",
"Definition:Basis for the Induction",
"Equivalence of Definitions of Ordered Pair",
"Definition:Induction Step",
"Definition:Ordered Tuple as Ordered Set",
"Category:Cartesian Product"
] |
proofwiki-5320 | Equivalent Conditions for Dedekind-Infinite Set | For a set $S$, the following conditions are equivalent:
:$(1): \quad$ $S$ is Dedekind-infinite.
:$(2): \quad$ $S$ has a countably infinite subset.
The above equivalence can be proven in Zermelo-Fraenkel set theory.
If the axiom of countable choice is accepted, then it can be proven that the following condition is also ... | {{Tidy}}
{{MissingLinks}} | For a [[Definition:Set|set]] $S$, the following conditions are [[Definition:Logical Equivalence|equivalent]]:
:$(1): \quad$ $S$ is [[Definition:Dedekind-Infinite|Dedekind-infinite]].
:$(2): \quad$ $S$ has a [[Definition:Countably Infinite Set|countably infinite]] [[Definition:Subset|subset]].
The above [[Definition:Log... | {{Tidy}}
{{MissingLinks}} | Equivalent Conditions for Dedekind-Infinite Set | https://proofwiki.org/wiki/Equivalent_Conditions_for_Dedekind-Infinite_Set | https://proofwiki.org/wiki/Equivalent_Conditions_for_Dedekind-Infinite_Set | [
"Infinite Sets"
] | [
"Definition:Set",
"Definition:Logical Equivalence",
"Definition:Dedekind-Infinite",
"Definition:Countably Infinite/Set",
"Definition:Subset",
"Definition:Logical Equivalence",
"Definition:Zermelo-Fraenkel Set Theory",
"Axiom:Axiom of Countable Choice",
"Definition:Logical Equivalence",
"Definition... | [] |
proofwiki-5321 | Relation Isomorphism Preserves Reflexivity | Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be relational structures.
Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be (relationally) isomorphic.
Then $\RR_1$ is a reflexive relation {{iff}} $\RR_2$ is also a reflexive relation. | {{WLOG}} it is necessary to prove only that if $\RR_1$ is reflexive then $\RR_2$ is reflexive.
Let $\phi: S \to T$ be a relation isomorphism.
Let $y \in T$.
Let $x = \map {\phi^{-1} } y$.
As $\phi$ is a bijection it follows from Inverse Element of Bijection that:
:$y = \map \phi x$
As $\RR_1$ is a reflexive relation it... | Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be [[Definition:Relational Structure|relational structures]].
Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be [[Definition:Relation Isomorphism|(relationally) isomorphic]].
Then $\RR_1$ is a [[Definition:Reflexive Relation|reflexive relation]] {{iff}} $\RR_2$ is... | {{WLOG}} it is necessary to prove only that if $\RR_1$ is [[Definition:Reflexive Relation|reflexive]] then $\RR_2$ is [[Definition:Reflexive Relation|reflexive]].
Let $\phi: S \to T$ be a [[Definition:Relation Isomorphism|relation isomorphism]].
Let $y \in T$.
Let $x = \map {\phi^{-1} } y$.
As $\phi$ is a [[Definit... | Relation Isomorphism Preserves Reflexivity | https://proofwiki.org/wiki/Relation_Isomorphism_Preserves_Reflexivity | https://proofwiki.org/wiki/Relation_Isomorphism_Preserves_Reflexivity | [
"Relation Isomorphisms",
"Reflexive Relations"
] | [
"Definition:Relational Structure",
"Definition:Relation Isomorphism",
"Definition:Reflexive Relation",
"Definition:Reflexive Relation"
] | [
"Definition:Reflexive Relation",
"Definition:Reflexive Relation",
"Definition:Relation Isomorphism",
"Definition:Bijection",
"Inverse Element of Bijection",
"Definition:Reflexive Relation",
"Definition:Relation Isomorphism"
] |
proofwiki-5322 | Preimage of Subset under Composite Mapping | Let $S_1, S_2, S_3$ be sets.
Let $f: S_1 \to S_2$ and $g: S_2 \to S_3$ be mappings.
Denote with $g \circ f: S_1 \to S_3$ the composition of $g$ and $f$.
Let $S_3' \subseteq S_3$ be a subset of $S_3$.
Then:
:$\paren {g \circ f}^{-1} \sqbrk {S_3'} = \paren {f^{-1} \circ g^{-1} } \sqbrk {S_3'}$
where $g^{-1} \sqbrk {S_3'}... | A mapping is a specific kind of relation.
Hence, Inverse of Composite Relation applies, and it follows that:
:$\paren {g \circ f}^{-1} \sqbrk {S_3'} = \paren {f^{-1} \circ g^{-1} } \sqbrk {S_3'}$
{{qed}} | Let $S_1, S_2, S_3$ be [[Definition:Set|sets]].
Let $f: S_1 \to S_2$ and $g: S_2 \to S_3$ be [[Definition:Mapping|mappings]].
Denote with $g \circ f: S_1 \to S_3$ the [[Definition:Composite Mapping|composition]] of $g$ and $f$.
Let $S_3' \subseteq S_3$ be a [[Definition:Subset|subset]] of $S_3$.
Then:
:$\paren {g... | A [[Definition:Mapping|mapping]] is a specific kind of [[Definition:Relation|relation]].
Hence, [[Inverse of Composite Relation]] applies, and it follows that:
:$\paren {g \circ f}^{-1} \sqbrk {S_3'} = \paren {f^{-1} \circ g^{-1} } \sqbrk {S_3'}$
{{qed}} | Preimage of Subset under Composite Mapping/Proof 1 | https://proofwiki.org/wiki/Preimage_of_Subset_under_Composite_Mapping | https://proofwiki.org/wiki/Preimage_of_Subset_under_Composite_Mapping/Proof_1 | [
"Composite Mappings",
"Preimages under Mappings",
"Preimage of Subset under Composite Mapping"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Composition of Mappings",
"Definition:Subset",
"Definition:Preimage/Mapping/Subset"
] | [
"Definition:Mapping",
"Definition:Relation",
"Inverse of Composite Relation"
] |
proofwiki-5323 | Preimage of Subset under Composite Mapping | Let $S_1, S_2, S_3$ be sets.
Let $f: S_1 \to S_2$ and $g: S_2 \to S_3$ be mappings.
Denote with $g \circ f: S_1 \to S_3$ the composition of $g$ and $f$.
Let $S_3' \subseteq S_3$ be a subset of $S_3$.
Then:
:$\paren {g \circ f}^{-1} \sqbrk {S_3'} = \paren {f^{-1} \circ g^{-1} } \sqbrk {S_3'}$
where $g^{-1} \sqbrk {S_3'}... | Let $x \in S_1$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \paren {g \circ f}^{-1} \sqbrk {S_3'}
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \map {\paren {g \circ f} } x
| o = \in
| r = S_3'
| c = {{Defof|Preimage of Subset under Mapping}}
}}
{{eqn | ll= \leadstoandfrom... | Let $S_1, S_2, S_3$ be [[Definition:Set|sets]].
Let $f: S_1 \to S_2$ and $g: S_2 \to S_3$ be [[Definition:Mapping|mappings]].
Denote with $g \circ f: S_1 \to S_3$ the [[Definition:Composite Mapping|composition]] of $g$ and $f$.
Let $S_3' \subseteq S_3$ be a [[Definition:Subset|subset]] of $S_3$.
Then:
:$\paren {g... | Let $x \in S_1$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = \in
| r = \paren {g \circ f}^{-1} \sqbrk {S_3'}
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = \map {\paren {g \circ f} } x
| o = \in
| r = S_3'
| c = {{Defof|Preimage of Subset under Mapping}}
}}
{{eqn | ll= \leadstoandfro... | Preimage of Subset under Composite Mapping/Proof 2 | https://proofwiki.org/wiki/Preimage_of_Subset_under_Composite_Mapping | https://proofwiki.org/wiki/Preimage_of_Subset_under_Composite_Mapping/Proof_2 | [
"Composite Mappings",
"Preimages under Mappings",
"Preimage of Subset under Composite Mapping"
] | [
"Definition:Set",
"Definition:Mapping",
"Definition:Composition of Mappings",
"Definition:Subset",
"Definition:Preimage/Mapping/Subset"
] | [] |
proofwiki-5324 | Infinite Set has Countably Infinite Subset/Proof 4 | If the axiom of countable choice is accepted, then it can be proven that every infinite set has a countably infinite subset. | Let $S$ be an infinite set.
For all $n \in \N$, let:
:$\FF_n = \set {T \subseteq S: \size T = n}$
where $\size T$ denotes the cardinality of $T$.
From Set is Infinite iff exist Subsets of all Finite Cardinalities:
:$\FF_n$ is non-empty.
Using the axiom of countable choice, we can obtain a sequence $\sequence {S_n}_{n \... | If the [[Axiom:Axiom of Countable Choice|axiom of countable choice]] is accepted, then it can be proven that every [[Definition:Infinite Set|infinite set]] has a [[Definition:Countably Infinite Set|countably infinite]] [[Definition:Subset|subset]]. | Let $S$ be an [[Definition:Infinite Set|infinite set]].
For all $n \in \N$, let:
:$\FF_n = \set {T \subseteq S: \size T = n}$
where $\size T$ denotes the [[Definition:Cardinality|cardinality]] of $T$.
From [[Set is Infinite iff exist Subsets of all Finite Cardinalities]]:
:$\FF_n$ is [[Definition:Non-Empty Set|non-e... | Infinite Set has Countably Infinite Subset/Proof 4 | https://proofwiki.org/wiki/Infinite_Set_has_Countably_Infinite_Subset/Proof_4 | https://proofwiki.org/wiki/Infinite_Set_has_Countably_Infinite_Subset/Proof_4 | [
"Infinite Set has Countably Infinite Subset"
] | [
"Axiom:Axiom of Countable Choice",
"Definition:Infinite Set",
"Definition:Countably Infinite/Set",
"Definition:Subset"
] | [
"Definition:Infinite Set",
"Definition:Cardinality",
"Set is Infinite iff exist Subsets of all Finite Cardinalities",
"Definition:Non-Empty Set",
"Axiom:Axiom of Countable Choice",
"Definition:Sequence",
"Definition:Subset",
"Definition:Cardinality",
"Set is Infinite iff exist Subsets of all Finite ... |
proofwiki-5325 | Set is Infinite iff exist Subsets of all Finite Cardinalities | A set $S$ is infinite {{iff}} for all $n \in \N$, there exists a subset of $S$ whose cardinality is $n$. | Let the cardinality of a set $S$ be denoted $\card S$. | A [[Definition:Set|set]] $S$ is [[Definition:Infinite Set|infinite]] {{iff}} for all $n \in \N$, there exists a [[Definition:Subset|subset]] of $S$ whose [[Definition:Cardinality|cardinality]] is $n$. | Let the [[Definition:Cardinality|cardinality]] of a [[Definition:Set|set]] $S$ be denoted $\card S$. | Set is Infinite iff exist Subsets of all Finite Cardinalities | https://proofwiki.org/wiki/Set_is_Infinite_iff_exist_Subsets_of_all_Finite_Cardinalities | https://proofwiki.org/wiki/Set_is_Infinite_iff_exist_Subsets_of_all_Finite_Cardinalities | [
"Infinite Sets"
] | [
"Definition:Set",
"Definition:Infinite Set",
"Definition:Subset",
"Definition:Cardinality"
] | [
"Definition:Cardinality",
"Definition:Set",
"Definition:Cardinality",
"Definition:Cardinality",
"Definition:Cardinality",
"Definition:Cardinality",
"Definition:Cardinality",
"Definition:Cardinality",
"Definition:Cardinality",
"Definition:Cardinality"
] |
proofwiki-5326 | Relation Isomorphism Preserves Symmetry | Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be relational structures.
Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be (relationally) isomorphic.
Then $\RR_1$ is a symmetric relation {{iff}} $\RR_2$ is also a symmetric relation. | Let $\phi: S \to T$ be a relation isomorphism.
By Inverse of Relation Isomorphism is Relation Isomorphism it follows that $\phi^{-1}: T \to S$ is also a relation isomorphism.
{{WLOG}}, it suffices to prove only that if $\RR_1$ is symmetric, then also $\RR_2$ is symmetric.
So, suppose $\RR_1$ is a symmetric relation.
Le... | Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be [[Definition:Relational Structure|relational structures]].
Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be [[Definition:Relation Isomorphism|(relationally) isomorphic]].
Then $\RR_1$ is a [[Definition:Symmetric Relation|symmetric relation]] {{iff}} $\RR_2$ is... | Let $\phi: S \to T$ be a [[Definition:Relation Isomorphism|relation isomorphism]].
By [[Inverse of Relation Isomorphism is Relation Isomorphism]] it follows that $\phi^{-1}: T \to S$ is also a [[Definition:Relation Isomorphism|relation isomorphism]].
{{WLOG}}, it suffices to prove only that if $\RR_1$ is [[Definition... | Relation Isomorphism Preserves Symmetry | https://proofwiki.org/wiki/Relation_Isomorphism_Preserves_Symmetry | https://proofwiki.org/wiki/Relation_Isomorphism_Preserves_Symmetry | [
"Relation Isomorphisms",
"Symmetric Relations"
] | [
"Definition:Relational Structure",
"Definition:Relation Isomorphism",
"Definition:Symmetric Relation",
"Definition:Symmetric Relation"
] | [
"Definition:Relation Isomorphism",
"Inverse of Relation Isomorphism is Relation Isomorphism",
"Definition:Relation Isomorphism",
"Definition:Symmetric Relation",
"Definition:Symmetric Relation",
"Definition:Symmetric Relation",
"Definition:Bijection",
"Inverse Element of Bijection",
"Definition:Rela... |
proofwiki-5327 | Set Difference with Proper Subset | Let $S$ be a set.
Let $T \subsetneq S$ be a proper subset of $S$.
Let $S \setminus T$ denote the set difference between $S$ and $T$.
Then:
:$S \setminus T \ne \O$
where $\O$ denotes the empty set. | {{AimForCont}} $S \setminus T = \O$.
Then:
:$\not \exists x \in S: x \notin T$
By De Morgan's laws:
:$\forall x \in S: x \in T$
By definition of subset:
:$S \subseteq T$
By definition of proper subset, we have that $T \subseteq S$ such that $T \ne S$.
But we have $T \subseteq S$ and $S \subseteq T$.
So by definition of... | Let $S$ be a [[Definition:Set|set]].
Let $T \subsetneq S$ be a [[Definition:Proper Subset|proper subset]] of $S$.
Let $S \setminus T$ denote the [[Definition:Set Difference|set difference]] between $S$ and $T$.
Then:
:$S \setminus T \ne \O$
where $\O$ denotes the [[Definition:Empty Set|empty set]]. | {{AimForCont}} $S \setminus T = \O$.
Then:
:$\not \exists x \in S: x \notin T$
By [[De Morgan's Laws (Predicate Logic)|De Morgan's laws]]:
:$\forall x \in S: x \in T$
By definition of [[Definition:Subset|subset]]:
:$S \subseteq T$
By definition of [[Definition:Proper Subset|proper subset]], we have that $T \subsete... | Set Difference with Proper Subset | https://proofwiki.org/wiki/Set_Difference_with_Proper_Subset | https://proofwiki.org/wiki/Set_Difference_with_Proper_Subset | [
"Set Difference",
"Proper Subsets"
] | [
"Definition:Set",
"Definition:Proper Subset",
"Definition:Set Difference",
"Definition:Empty Set"
] | [
"De Morgan's Laws (Predicate Logic)",
"Definition:Subset",
"Definition:Proper Subset",
"Definition:Set Equality/Definition 2",
"Proof by Contradiction",
"Category:Set Difference",
"Category:Proper Subsets"
] |
proofwiki-5328 | Order Isomorphism iff Strictly Increasing Surjection | Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be totally ordered sets.
A mapping $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ is an order isomorphism {{iff}}:
:$(1): \quad \phi$ is a surjection
:$(2): \quad \forall x, y \in S: x \mathop {\prec_1} y \implies \map \phi x \mathop {\prec_2} \map \p... | === Necessary Condition ===
Let $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ be an order isomorphism.
Then by definition $\phi$ is a bijection and so a surjection.
Suppose $x \mathop {\prec_1} y$.
That is:
:$x \mathop {\preceq_1} y$
:$x \ne y$
Then:
:$x \mathop {\prec_1} y \implies \map \phi x \mathop {\pr... | Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be [[Definition:Totally Ordered Set|totally ordered sets]].
A [[Definition:Mapping|mapping]] $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ is an [[Definition:Order Isomorphism|order isomorphism]] {{iff}}:
:$(1): \quad \phi$ is a [[Definition:Surjec... | === Necessary Condition ===
Let $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ be an [[Definition:Order Isomorphism|order isomorphism]].
Then by definition $\phi$ is a [[Definition:Bijection|bijection]] and so a [[Definition:Surjection|surjection]].
Suppose $x \mathop {\prec_1} y$.
That is:
:$x \mathop {... | Order Isomorphism iff Strictly Increasing Surjection | https://proofwiki.org/wiki/Order_Isomorphism_iff_Strictly_Increasing_Surjection | https://proofwiki.org/wiki/Order_Isomorphism_iff_Strictly_Increasing_Surjection | [
"Order Isomorphisms",
"Surjections"
] | [
"Definition:Totally Ordered Set",
"Definition:Mapping",
"Definition:Order Isomorphism",
"Definition:Surjection"
] | [
"Definition:Order Isomorphism",
"Definition:Bijection",
"Definition:Surjection",
"Definition:Order Isomorphism",
"Definition:Bijection",
"Definition:Injection",
"Definition:Surjection",
"Definition:Order Isomorphism"
] |
proofwiki-5329 | Relation Isomorphism Preserves Transitivity | Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be relational structures.
Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be (relationally) isomorphic.
Then $\RR_1$ is a transitive relation {{iff}} $\RR_2$ is a transitive relation. | Let $\phi: S \to T$ be a relation isomorphism.
By Inverse of Relation Isomorphism is Relation Isomorphism it follows that $\phi^{-1}: T \to S$ is also a relation isomorphism.
{{WLOG}}, it is therefore sufficient to prove only that if $\RR_1$ is transitive, then $\RR_2$ is also transitive.
So, suppose $\RR_1$ is a trans... | Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be [[Definition:Relational Structure|relational structures]].
Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be [[Definition:Relation Isomorphism|(relationally) isomorphic]].
Then $\RR_1$ is a [[Definition:Transitive Relation|transitive relation]] {{iff}} $\RR_2$ ... | Let $\phi: S \to T$ be a [[Definition:Relation Isomorphism|relation isomorphism]].
By [[Inverse of Relation Isomorphism is Relation Isomorphism]] it follows that $\phi^{-1}: T \to S$ is also a [[Definition:Relation Isomorphism|relation isomorphism]].
{{WLOG}}, it is therefore sufficient to prove only that if $\RR_1$ ... | Relation Isomorphism Preserves Transitivity | https://proofwiki.org/wiki/Relation_Isomorphism_Preserves_Transitivity | https://proofwiki.org/wiki/Relation_Isomorphism_Preserves_Transitivity | [
"Relation Isomorphisms",
"Transitive Relations"
] | [
"Definition:Relational Structure",
"Definition:Relation Isomorphism",
"Definition:Transitive Relation",
"Definition:Transitive Relation"
] | [
"Definition:Relation Isomorphism",
"Inverse of Relation Isomorphism is Relation Isomorphism",
"Definition:Relation Isomorphism",
"Definition:Transitive Relation",
"Definition:Transitive Relation",
"Definition:Transitive Relation",
"Definition:Bijection",
"Inverse Element of Bijection",
"Definition:R... |
proofwiki-5330 | Relation Isomorphism Preserves Antisymmetry | Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be relational structures.
Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be (relationally) isomorphic.
Then $\RR_1$ is an antisymmetric relation {{iff}} $\RR_2$ is also an antisymmetric relation. | Let $\phi: S \to T$ be a relation isomorphism.
By Inverse of Relation Isomorphism is Relation Isomorphism it follows that $\phi^{-1}: T \to S$ is also a relation isomorphism.
{{WLOG}}, it therefore suffices to prove only that if $\RR_1$ is antisymmetric, then also $\RR_2$ is antisymmetric.
So, suppose $\RR_1$ is an ant... | Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be [[Definition:Relational Structure|relational structures]].
Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be [[Definition:Relation Isomorphism|(relationally) isomorphic]].
Then $\RR_1$ is an [[Definition:Antisymmetric Relation|antisymmetric relation]] {{iff}} $... | Let $\phi: S \to T$ be a [[Definition:Relation Isomorphism|relation isomorphism]].
By [[Inverse of Relation Isomorphism is Relation Isomorphism]] it follows that $\phi^{-1}: T \to S$ is also a [[Definition:Relation Isomorphism|relation isomorphism]].
{{WLOG}}, it therefore suffices to prove only that if $\RR_1$ is [[... | Relation Isomorphism Preserves Antisymmetry | https://proofwiki.org/wiki/Relation_Isomorphism_Preserves_Antisymmetry | https://proofwiki.org/wiki/Relation_Isomorphism_Preserves_Antisymmetry | [
"Relation Isomorphisms",
"Antisymmetric Relations"
] | [
"Definition:Relational Structure",
"Definition:Relation Isomorphism",
"Definition:Antisymmetric Relation",
"Definition:Antisymmetric Relation"
] | [
"Definition:Relation Isomorphism",
"Inverse of Relation Isomorphism is Relation Isomorphism",
"Definition:Relation Isomorphism",
"Definition:Antisymmetric Relation",
"Definition:Antisymmetric Relation",
"Definition:Antisymmetric Relation",
"Definition:Bijection",
"Inverse Element of Bijection",
"Def... |
proofwiki-5331 | Relation Isomorphism Preserves Equivalence Relations | Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be relational structures.
Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be (relationally) isomorphic.
Then $\RR_1$ is an equivalence relation {{iff}} $\RR_2$ is also an equivalence relation. | Let $\phi: S \to T$ be a relation isomorphism.
By Inverse of Relation Isomorphism is Relation Isomorphism it follows that $\phi^{-1}: T \to S$ is also a relation isomorphism.
{{WLOG}}, it thus is necessary to prove only that if $\RR_1$ is an equivalence relation then $\RR_2$ is an equivalence relation.
So, suppose $\RR... | Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be [[Definition:Relational Structure|relational structures]].
Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be [[Definition:Relation Isomorphism|(relationally) isomorphic]].
Then $\RR_1$ is an [[Definition:Equivalence Relation|equivalence relation]] {{iff}} $\RR_... | Let $\phi: S \to T$ be a [[Definition:Relation Isomorphism|relation isomorphism]].
By [[Inverse of Relation Isomorphism is Relation Isomorphism]] it follows that $\phi^{-1}: T \to S$ is also a [[Definition:Relation Isomorphism|relation isomorphism]].
{{WLOG}}, it thus is necessary to prove only that if $\RR_1$ is an ... | Relation Isomorphism Preserves Equivalence Relations | https://proofwiki.org/wiki/Relation_Isomorphism_Preserves_Equivalence_Relations | https://proofwiki.org/wiki/Relation_Isomorphism_Preserves_Equivalence_Relations | [
"Relation Isomorphisms",
"Equivalence Relations"
] | [
"Definition:Relational Structure",
"Definition:Relation Isomorphism",
"Definition:Equivalence Relation",
"Definition:Equivalence Relation"
] | [
"Definition:Relation Isomorphism",
"Inverse of Relation Isomorphism is Relation Isomorphism",
"Definition:Relation Isomorphism",
"Definition:Equivalence Relation",
"Definition:Equivalence Relation",
"Definition:Equivalence Relation",
"Definition:Reflexive Relation",
"Definition:Symmetric Relation",
... |
proofwiki-5332 | Pointwise Supremum of Measurable Functions is Measurable | Let $\struct {X, \Sigma}$ be a measurable space, and let $I$ be a countable set.
Let $\family {f_i}_{i \mathop \in I}$, $f_i: X \to \overline \R$ be an $I$-indexed collection of $\Sigma$-measurable functions.
Then the pointwise supremum $\ds \sup_{i \mathop \in I} f_i: X \to \overline \R$ is also $\Sigma$-measurable. | Let $a \in \R$; for all $i \in I$, we have by Characterization of Measurable Functions that:
:$\set {f_i > a} \in \Sigma$
and as $\Sigma$ is a $\sigma$-algebra and $I$ is countable, also:
:$\ds \bigcup_{i \mathop \in I} \set {f_i > a} \in \Sigma$
We will now show that:
:$\ds \set {\sup_{i \mathop \in I} f_i > a} = \big... | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]], and let $I$ be a [[Definition:Countable Set|countable set]].
Let $\family {f_i}_{i \mathop \in I}$, $f_i: X \to \overline \R$ be an [[Definition:Indexed Set|$I$-indexed collection]] of [[Definition:Measurable Function|$\Sigma$-measurable ... | Let $a \in \R$; for all $i \in I$, we have by [[Characterization of Measurable Functions]] that:
:$\set {f_i > a} \in \Sigma$
and as $\Sigma$ is a [[Definition:Sigma-Algebra|$\sigma$-algebra]] and $I$ is [[Definition:Countable Set|countable]], also:
:$\ds \bigcup_{i \mathop \in I} \set {f_i > a} \in \Sigma$
We wil... | Pointwise Supremum of Measurable Functions is Measurable | https://proofwiki.org/wiki/Pointwise_Supremum_of_Measurable_Functions_is_Measurable | https://proofwiki.org/wiki/Pointwise_Supremum_of_Measurable_Functions_is_Measurable | [
"Measurable Functions"
] | [
"Definition:Measurable Space",
"Definition:Countable Set",
"Definition:Indexing Set/Indexed Set",
"Definition:Measurable Function",
"Definition:Pointwise Supremum",
"Definition:Measurable Function"
] | [
"Characterization of Measurable Functions",
"Definition:Sigma-Algebra",
"Definition:Countable Set",
"Union is Smallest Superset/Family of Sets",
"Definition:Set Union",
"Definition:Upper Bound of Mapping",
"Definition:Supremum of Set",
"Characterization of Measurable Functions",
"Definition:Measurab... |
proofwiki-5333 | Pointwise Infimum of Measurable Functions is Measurable | Let $\struct {X, \Sigma}$ be a measurable space.
Let $I$ be a countable set.
Let $\family {f_i}_{i \mathop \in I}$, $f_i: X \to \overline \R$ be an $I$-indexed family of $\Sigma$-measurable functions.
Then the pointwise infimum $\ds \inf_{i \mathop \in I} f_i: X \to \overline \R$ is also $\Sigma$-measurable. | From Infimum as Supremum, we have the Equality of Mappings:
:$\ds \inf_{i \mathop \in I} f_i = -\paren {\sup_{i \mathop \in I} \paren {-f_i} }$
Now, from Negative of Measurable Function is Measurable and Pointwise Supremum of Measurable Functions is Measurable, it follows that:
:$\ds - \paren {\sup_{i \mathop \in I} \p... | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $I$ be a [[Definition:Countable Set|countable set]].
Let $\family {f_i}_{i \mathop \in I}$, $f_i: X \to \overline \R$ be an [[Definition:Indexed Family|$I$-indexed family]] of [[Definition:Measurable Function|$\Sigma$-measurable func... | From [[Infimum as Supremum]], we have the [[Equality of Mappings]]:
:$\ds \inf_{i \mathop \in I} f_i = -\paren {\sup_{i \mathop \in I} \paren {-f_i} }$
Now, from [[Negative of Measurable Function is Measurable]] and [[Pointwise Supremum of Measurable Functions is Measurable]], it follows that:
:$\ds - \paren {\sup_... | Pointwise Infimum of Measurable Functions is Measurable | https://proofwiki.org/wiki/Pointwise_Infimum_of_Measurable_Functions_is_Measurable | https://proofwiki.org/wiki/Pointwise_Infimum_of_Measurable_Functions_is_Measurable | [
"Measurable Functions"
] | [
"Definition:Measurable Space",
"Definition:Countable Set",
"Definition:Indexing Set/Family",
"Definition:Measurable Function",
"Definition:Pointwise Infimum",
"Definition:Measurable Function"
] | [
"Infimum as Supremum",
"Equality of Mappings",
"Negative of Measurable Function is Measurable",
"Pointwise Supremum of Measurable Functions is Measurable",
"Definition:Measurable Function"
] |
proofwiki-5334 | Pointwise Upper Limit of Measurable Functions is Measurable | Let $\struct {X, \Sigma}$ be a measurable space.
Let $\sequence {f_n}_{n \mathop \in \N}$, $f_n: X \to \overline \R$ be a sequence of $\Sigma$-measurable functions.
Then the pointwise upper limit $\ds \limsup_{n \mathop \to \infty} f_n: X \to \overline \R$ is also $\Sigma$-measurable. | By definition of upper limit, we have:
:$\ds \limsup_{n \mathop \to \infty} f_n = \inf_{m \mathop \in \N} \sup_{n \mathop \ge m} f_n$
The result follows from combining:
:Pointwise Supremum of Measurable Functions is Measurable
:Pointwise Infimum of Measurable Functions is Measurable
{{qed}} | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $\sequence {f_n}_{n \mathop \in \N}$, $f_n: X \to \overline \R$ be a [[Definition:Sequence|sequence]] of [[Definition:Measurable Function|$\Sigma$-measurable functions]].
Then the [[Definition:Pointwise Upper Limit|pointwise upper l... | By definition of [[Definition:Upper Limit|upper limit]], we have:
:$\ds \limsup_{n \mathop \to \infty} f_n = \inf_{m \mathop \in \N} \sup_{n \mathop \ge m} f_n$
The result follows from combining:
:[[Pointwise Supremum of Measurable Functions is Measurable]]
:[[Pointwise Infimum of Measurable Functions is Measurable... | Pointwise Upper Limit of Measurable Functions is Measurable | https://proofwiki.org/wiki/Pointwise_Upper_Limit_of_Measurable_Functions_is_Measurable | https://proofwiki.org/wiki/Pointwise_Upper_Limit_of_Measurable_Functions_is_Measurable | [
"Measurable Functions"
] | [
"Definition:Measurable Space",
"Definition:Sequence",
"Definition:Measurable Function",
"Definition:Pointwise Upper Limit",
"Definition:Measurable Function"
] | [
"Definition:Limit Superior",
"Pointwise Supremum of Measurable Functions is Measurable",
"Pointwise Infimum of Measurable Functions is Measurable"
] |
proofwiki-5335 | Pointwise Lower Limit of Measurable Functions is Measurable | Let $\struct {X, \Sigma}$ be a measurable space.
Let $\sequence {f_n}_{n \mathop \in \N}$, $f_n: X \to \overline \R$ be a sequence of $\Sigma$-measurable functions.
Then the pointwise lower limit:
:$\ds \liminf_{n \mathop \to \infty} f_n: X \to \overline \R$
is also $\Sigma$-measurable. | By definition of limit inferior, we have:
:$\ds \liminf_{n \mathop \to \infty} f_n = \sup_{m \mathop \in \N} \ \inf_{n \mathop \ge m} f_n$
The result follows from combining:
:Pointwise Infimum of Measurable Functions is Measurable
:Pointwise Supremum of Measurable Functions is Measurable
{{qed}} | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $\sequence {f_n}_{n \mathop \in \N}$, $f_n: X \to \overline \R$ be a [[Definition:Sequence|sequence]] of [[Definition:Measurable Function|$\Sigma$-measurable functions]].
Then the [[Definition:Pointwise Lower Limit|pointwise lower l... | By definition of [[Definition:Limit Inferior|limit inferior]], we have:
:$\ds \liminf_{n \mathop \to \infty} f_n = \sup_{m \mathop \in \N} \ \inf_{n \mathop \ge m} f_n$
The result follows from combining:
:[[Pointwise Infimum of Measurable Functions is Measurable]]
:[[Pointwise Supremum of Measurable Functions is Me... | Pointwise Lower Limit of Measurable Functions is Measurable | https://proofwiki.org/wiki/Pointwise_Lower_Limit_of_Measurable_Functions_is_Measurable | https://proofwiki.org/wiki/Pointwise_Lower_Limit_of_Measurable_Functions_is_Measurable | [
"Measurable Functions"
] | [
"Definition:Measurable Space",
"Definition:Sequence",
"Definition:Measurable Function",
"Definition:Pointwise Lower Limit",
"Definition:Measurable Function"
] | [
"Definition:Limit Inferior",
"Pointwise Infimum of Measurable Functions is Measurable",
"Pointwise Supremum of Measurable Functions is Measurable"
] |
proofwiki-5336 | Pointwise Limit of Measurable Functions is Measurable | Let $\struct {X, \Sigma}$ be a measurable space.
Let $\sequence {f_n}_{n \mathop \in \N}$, $f_n: X \to \overline \R$ be a sequence of $\Sigma$-measurable functions.
Then the pointwise limit $\ds \lim_{n \mathop \to \infty} f_n: X \to \overline \R$ is also $\Sigma$-measurable. | From Convergence of Limsup and Liminf, it follows that:
:$\ds \lim_{n \mathop \to \infty} f_n = \limsup_{n \mathop \to \infty} f_n$
We have Pointwise Upper Limit of Measurable Functions is Measurable.
Hence the result.
{{qed}} | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $\sequence {f_n}_{n \mathop \in \N}$, $f_n: X \to \overline \R$ be a [[Definition:Sequence|sequence]] of [[Definition:Measurable Function|$\Sigma$-measurable functions]].
Then the [[Definition:Pointwise Limit|pointwise limit]] $\ds ... | From [[Convergence of Limsup and Liminf]], it follows that:
:$\ds \lim_{n \mathop \to \infty} f_n = \limsup_{n \mathop \to \infty} f_n$
We have [[Pointwise Upper Limit of Measurable Functions is Measurable]].
Hence the result.
{{qed}} | Pointwise Limit of Measurable Functions is Measurable | https://proofwiki.org/wiki/Pointwise_Limit_of_Measurable_Functions_is_Measurable | https://proofwiki.org/wiki/Pointwise_Limit_of_Measurable_Functions_is_Measurable | [
"Measurable Functions"
] | [
"Definition:Measurable Space",
"Definition:Sequence",
"Definition:Measurable Function",
"Definition:Pointwise Limit",
"Definition:Measurable Function"
] | [
"Convergence of Limsup and Liminf",
"Pointwise Upper Limit of Measurable Functions is Measurable"
] |
proofwiki-5337 | Pointwise Sum of Measurable Functions is Measurable | Let $\struct {X, \Sigma}$ be a measurable space.
Let $f, g: X \to \overline \R$ be $\Sigma$-measurable functions.
Assume that the pointwise sum $f + g: X \to \overline \R$ is well-defined.
Then $f + g$ is a $\Sigma$-measurable function. | By Measurable Function is Pointwise Limit of Simple Functions, we find sequences $\sequence {f_n}_{n \mathop \in \N}, \sequence {g_n}_{n \mathop \in \N}$ such that:
:$\ds f = \lim_{n \mathop \to \infty} f_n$
:$\ds g = \lim_{n \mathop \to \infty} g_n$
where the limits are pointwise.
It follows that for all $x \in X$:
:$... | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $f, g: X \to \overline \R$ be [[Definition:Measurable Function|$\Sigma$-measurable functions]].
Assume that the [[Definition:Pointwise Addition of Extended Real-Valued Functions|pointwise sum]] $f + g: X \to \overline \R$ is well-def... | By [[Measurable Function is Pointwise Limit of Simple Functions]], we find [[Definition:Sequence|sequences]] $\sequence {f_n}_{n \mathop \in \N}, \sequence {g_n}_{n \mathop \in \N}$ such that:
:$\ds f = \lim_{n \mathop \to \infty} f_n$
:$\ds g = \lim_{n \mathop \to \infty} g_n$
where the limits are [[Definition:Point... | Pointwise Sum of Measurable Functions is Measurable | https://proofwiki.org/wiki/Pointwise_Sum_of_Measurable_Functions_is_Measurable | https://proofwiki.org/wiki/Pointwise_Sum_of_Measurable_Functions_is_Measurable | [
"Measurable Functions",
"Pointwise Sum of Measurable Functions is Measurable"
] | [
"Definition:Measurable Space",
"Definition:Measurable Function",
"Definition:Pointwise Addition of Extended Real-Valued Functions",
"Definition:Measurable Function"
] | [
"Measurable Function is Pointwise Limit of Simple Functions",
"Definition:Sequence",
"Definition:Pointwise Limit",
"Definition:Pointwise Limit",
"Pointwise Sum of Simple Functions is Simple Function",
"Definition:Pointwise Limit",
"Definition:Simple Function",
"Simple Function is Measurable",
"Defin... |
proofwiki-5338 | Pointwise Difference of Measurable Functions is Measurable | Let $\struct {X, \Sigma}$ be a measurable space.
Let $f, g: X \to \overline \R$ be $\Sigma$-measurable functions.
Assume that the pointwise difference $f - g: X \to \overline \R$ is well-defined.
Then $f - g$ is a $\Sigma$-measurable function. | We have the apparent identity:
:$f - g = f + \paren {-g}$
By Negative of Measurable Function is Measurable, $-g$ is a measurable function.
Hence so is $f - g$, by Pointwise Sum of Measurable Functions is Measurable.
{{qed}} | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $f, g: X \to \overline \R$ be [[Definition:Measurable Function|$\Sigma$-measurable functions]].
Assume that the [[Definition:Pointwise Subtraction of Extended Real-Valued Functions|pointwise difference]] $f - g: X \to \overline \R$ i... | We have the apparent identity:
:$f - g = f + \paren {-g}$
By [[Negative of Measurable Function is Measurable]], $-g$ is a [[Definition:Measurable Function|measurable function]].
Hence so is $f - g$, by [[Pointwise Sum of Measurable Functions is Measurable]].
{{qed}} | Pointwise Difference of Measurable Functions is Measurable | https://proofwiki.org/wiki/Pointwise_Difference_of_Measurable_Functions_is_Measurable | https://proofwiki.org/wiki/Pointwise_Difference_of_Measurable_Functions_is_Measurable | [
"Measurable Functions"
] | [
"Definition:Measurable Space",
"Definition:Measurable Function",
"Definition:Pointwise Subtraction of Extended Real-Valued Functions",
"Definition:Measurable Function"
] | [
"Negative of Measurable Function is Measurable",
"Definition:Measurable Function",
"Pointwise Sum of Measurable Functions is Measurable"
] |
proofwiki-5339 | Pointwise Product of Measurable Functions is Measurable | Let $\struct {X, \Sigma}$ be a measurable space.
Let $f, g: X \to \overline \R$ be $\Sigma$-measurable functions.
Then the pointwise product $f \cdot g: X \to \overline \R$ is also $\Sigma$-measurable. | By Measurable Function is Pointwise Limit of Simple Functions, we find sequences $\sequence {f_n}_{n \mathop \in \N}, \sequence {g_n}_{n \mathop \in \N}$ such that:
:$\ds f = \lim_{n \mathop \to \infty} f_n$
:$\ds g = \lim_{n \mathop \to \infty} g_n$
where the limits are pointwise.
It follows that for all $x \in X$:
:$... | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $f, g: X \to \overline \R$ be [[Definition:Measurable Function|$\Sigma$-measurable functions]].
Then the [[Definition:Pointwise Product of Extended Real-Valued Functions|pointwise product]] $f \cdot g: X \to \overline \R$ is also [[... | By [[Measurable Function is Pointwise Limit of Simple Functions]], we find [[Definition:Sequence|sequences]] $\sequence {f_n}_{n \mathop \in \N}, \sequence {g_n}_{n \mathop \in \N}$ such that:
:$\ds f = \lim_{n \mathop \to \infty} f_n$
:$\ds g = \lim_{n \mathop \to \infty} g_n$
where the limits are [[Definition:Point... | Pointwise Product of Measurable Functions is Measurable | https://proofwiki.org/wiki/Pointwise_Product_of_Measurable_Functions_is_Measurable | https://proofwiki.org/wiki/Pointwise_Product_of_Measurable_Functions_is_Measurable | [
"Measurable Functions"
] | [
"Definition:Measurable Space",
"Definition:Measurable Function",
"Definition:Pointwise Product of Extended Real-Valued Functions",
"Definition:Measurable Function"
] | [
"Measurable Function is Pointwise Limit of Simple Functions",
"Definition:Sequence",
"Definition:Pointwise Limit",
"Definition:Pointwise Limit",
"Pointwise Product of Simple Functions is Simple Function",
"Definition:Pointwise Limit",
"Definition:Simple Function",
"Simple Function is Measurable",
"D... |
proofwiki-5340 | Pointwise Maximum of Measurable Functions is Measurable | Let $\struct {X, \Sigma}$ be a measurable space.
Let $f, g: X \to \overline \R$ be $\Sigma$-measurable functions.
Then the pointwise maximum $\max \set {f, g}: X \to \overline \R$ is also $\Sigma$-measurable. | For all $x \in X$ and $a \in \R$, we have by Max Operation Yields Supremum of Parameters that:
:$\max \set {\map f x, \map g x} \le a$
{{iff}} both $\map f x \le a$ and $\map g x \le a$.
That is, for all $a \in \R$:
:$\set {x \in X: \max \set {\map f x, \map g x} \le a} = \set {x \in X: \map f x \le a} \cap \set {x \in... | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $f, g: X \to \overline \R$ be [[Definition:Measurable Function|$\Sigma$-measurable functions]].
Then the [[Definition:Pointwise Maximum of Extended Real-Valued Functions|pointwise maximum]] $\max \set {f, g}: X \to \overline \R$ is ... | For all $x \in X$ and $a \in \R$, we have by [[Max Operation Yields Supremum of Parameters]] that:
:$\max \set {\map f x, \map g x} \le a$
{{iff}} both $\map f x \le a$ and $\map g x \le a$.
That is, for all $a \in \R$:
:$\set {x \in X: \max \set {\map f x, \map g x} \le a} = \set {x \in X: \map f x \le a} \cap \s... | Pointwise Maximum of Measurable Functions is Measurable | https://proofwiki.org/wiki/Pointwise_Maximum_of_Measurable_Functions_is_Measurable | https://proofwiki.org/wiki/Pointwise_Maximum_of_Measurable_Functions_is_Measurable | [
"Measurable Functions"
] | [
"Definition:Measurable Space",
"Definition:Measurable Function",
"Definition:Pointwise Maximum of Mappings/Extended Real-Valued Functions",
"Definition:Measurable Function"
] | [
"Max Operation Yields Supremum of Parameters",
"Characterization of Measurable Functions",
"Definition:Right Hand Side",
"Definition:Measurable Set",
"Sigma-Algebra Closed under Finite Intersection",
"Definition:Measurable Function",
"Characterization of Measurable Functions"
] |
proofwiki-5341 | Pointwise Minimum of Measurable Functions is Measurable | Let $\struct {X, \Sigma}$ be a measurable space.
Let $f, g: X \to \overline \R$ be $\Sigma$-measurable functions.
Then the pointwise minimum $\min \set {f, g}: X \to \overline \R$ is also $\Sigma$-measurable. | For all $x \in X$ and $a \in \R$, we have by Min Operation Yields Infimum of Parameters that:
:$a \le \min \set {\map f x, \map g x}$
{{iff}} both $a \le \map f x$ and $a \le \map g x$.
That is, for all $a \in \R$:
:$\set {x \in X: \min \set {\map f x, \map g x} \ge a} = \set {x \in X: \map f x \ge a} \cap \set {x \in... | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $f, g: X \to \overline \R$ be [[Definition:Measurable Function|$\Sigma$-measurable functions]].
Then the [[Definition:Pointwise Minimum of Extended Real-Valued Functions|pointwise minimum]] $\min \set {f, g}: X \to \overline \R$ is ... | For all $x \in X$ and $a \in \R$, we have by [[Min Operation Yields Infimum of Parameters]] that:
:$a \le \min \set {\map f x, \map g x}$
{{iff}} both $a \le \map f x$ and $a \le \map g x$.
That is, for all $a \in \R$:
:$\set {x \in X: \min \set {\map f x, \map g x} \ge a} = \set {x \in X: \map f x \ge a} \cap \s... | Pointwise Minimum of Measurable Functions is Measurable | https://proofwiki.org/wiki/Pointwise_Minimum_of_Measurable_Functions_is_Measurable | https://proofwiki.org/wiki/Pointwise_Minimum_of_Measurable_Functions_is_Measurable | [
"Measurable Functions"
] | [
"Definition:Measurable Space",
"Definition:Measurable Function",
"Definition:Pointwise Minimum of Mappings/Extended Real-Valued Functions",
"Definition:Measurable Function"
] | [
"Min Operation Yields Infimum of Parameters",
"Characterization of Measurable Functions",
"Definition:Right Hand Side",
"Definition:Measurable Set",
"Sigma-Algebra Closed under Finite Intersection",
"Definition:Measurable Function",
"Characterization of Measurable Functions"
] |
proofwiki-5342 | Function Measurable iff Positive and Negative Parts Measurable | Let $\struct {X, \Sigma}$ be a measurable space.
Let $f: X \to \overline \R$ be an extended real-valued function.
Let $f^+, f^-: X \to \overline \R$ be the positive and negative parts of $f$.
Then $f$ is $\Sigma$-measurable {{iff}} both $f^+$ and $f^-$ are $\Sigma$-measurable. | === Necessary Condition ===
Suppose $f$ is measurable.
By definition, its positive part $f^+$ equals the pointwise maximum:
:$f^+ = \max \set {f, 0}$
where $0$ denotes the zero function.
By Constant Function is Measurable, $0$ is a measurable function.
Thus, by Pointwise Maximum of Measurable Functions is Measurable, $... | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $f: X \to \overline \R$ be an [[Definition:Extended Real-Valued Function|extended real-valued function]].
Let $f^+, f^-: X \to \overline \R$ be the [[Definition:Positive Part|positive]] and [[Definition:Negative Part|negative parts]]... | === Necessary Condition ===
Suppose $f$ is [[Definition:Measurable Function|measurable]].
By definition, its [[Definition:Positive Part|positive part]] $f^+$ equals the [[Definition:Pointwise Maximum of Extended Real-Valued Functions|pointwise maximum]]:
:$f^+ = \max \set {f, 0}$
where $0$ denotes the [[Definition:... | Function Measurable iff Positive and Negative Parts Measurable | https://proofwiki.org/wiki/Function_Measurable_iff_Positive_and_Negative_Parts_Measurable | https://proofwiki.org/wiki/Function_Measurable_iff_Positive_and_Negative_Parts_Measurable | [
"Positive Parts",
"Negative Parts",
"Measurable Functions",
"Positive Parts",
"Negative Parts"
] | [
"Definition:Measurable Space",
"Definition:Extended Real-Valued Function",
"Definition:Positive Part",
"Definition:Negative Part",
"Definition:Measurable Function",
"Definition:Measurable Function"
] | [
"Definition:Measurable Function",
"Definition:Positive Part",
"Definition:Pointwise Maximum of Mappings/Extended Real-Valued Functions",
"Definition:Basic Primitive Recursive Function/Zero Function",
"Constant Function is Measurable",
"Definition:Measurable Function",
"Pointwise Maximum of Measurable Fu... |
proofwiki-5343 | Measurable Functions Determine Measurable Sets | Let $\struct {X, \Sigma}$ be a measurable space.
Let $f, g: X \to \overline \R$ be $\Sigma$-measurable functions.
Then the following sets are measurable:
:$\set {f < g}$
:$\set {f \le g}$
:$\set {f = g}$
:$\set {f \ne g}$
where, for example, $\set {f < g}$ is short for $\set {x \in X: \map f x < \map g x}$. | From Pointwise Difference of Measurable Functions is Measurable, $f - g: X \to \overline \R$ is $\Sigma$-measurable.
Now we have the following evident identities:
:$\set {f < g} = \set {f - g < 0}$
:$\set {f \ge g} = \set {f - g \le 0}$
:$\set {f = g} = \set {f - g = 0}$
:$\set {f \ne g} = \set {f - g \ne 0}$
Subsequen... | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $f, g: X \to \overline \R$ be [[Definition:Measurable Function|$\Sigma$-measurable functions]].
Then the following sets are [[Definition:Measurable Set|measurable]]:
:$\set {f < g}$
:$\set {f \le g}$
:$\set {f = g}$
:$\set {f \ne g... | From [[Pointwise Difference of Measurable Functions is Measurable]], $f - g: X \to \overline \R$ is [[Definition:Measurable Function|$\Sigma$-measurable]].
Now we have the following evident identities:
:$\set {f < g} = \set {f - g < 0}$
:$\set {f \ge g} = \set {f - g \le 0}$
:$\set {f = g} = \set {f - g = 0}$
:$\set ... | Measurable Functions Determine Measurable Sets | https://proofwiki.org/wiki/Measurable_Functions_Determine_Measurable_Sets | https://proofwiki.org/wiki/Measurable_Functions_Determine_Measurable_Sets | [
"Measurable Functions",
"Measurable Sets"
] | [
"Definition:Measurable Space",
"Definition:Measurable Function",
"Definition:Measurable Set"
] | [
"Pointwise Difference of Measurable Functions is Measurable",
"Definition:Measurable Function",
"Definition:Preimage/Mapping/Subset",
"Definition:Open Set/Topology",
"Definition:Closed Set/Topology",
"Definition:Euclidean Space/Euclidean Topology/Real Number Line",
"Definition:Borel Sigma-Algebra",
"C... |
proofwiki-5344 | Factorization Lemma/Extended Real-Valued Function | An extended real-valued function $g: X \to \overline \R$ is $\map \sigma f$-measurable {{iff}}:
:There exists a $\Sigma$-measurable mapping $\tilde g: Y \to \overline \R$ such that $g = \tilde g \circ f$
where:
:$\map \sigma f$ denotes the $\sigma$-algebra generated by $f$ | === Necessary Condition ===
Let $g$ be a $\map \sigma f \, / \, \overline \BB$-measurable function.
We need to construct a measurable $\tilde g$ such that $g = \tilde g \circ f$.
Let us proceed in the following fashion:
:Establish the result for $g$ a characteristic function;
:Establish the result for $g$ a simple func... | An [[Definition:Extended Real-Valued Function|extended real-valued function]] $g: X \to \overline \R$ is [[Definition:Measurable Function|$\map \sigma f$-measurable]] {{iff}}:
:There exists a [[Definition:Measurable Function|$\Sigma$-measurable mapping]] $\tilde g: Y \to \overline \R$ such that $g = \tilde g \circ f$
... | === Necessary Condition ===
Let $g$ be a [[Definition:Measurable Mapping|$\map \sigma f \, / \, \overline \BB$-measurable function]].
We need to construct a [[Definition:Measurable Mapping|measurable]] $\tilde g$ such that $g = \tilde g \circ f$.
Let us proceed in the following fashion:
:Establish the result for $... | Factorization Lemma/Extended Real-Valued Function | https://proofwiki.org/wiki/Factorization_Lemma/Extended_Real-Valued_Function | https://proofwiki.org/wiki/Factorization_Lemma/Extended_Real-Valued_Function | [
"Measure Theory"
] | [
"Definition:Extended Real-Valued Function",
"Definition:Measurable Function",
"Definition:Measurable Function",
"Definition:Sigma-Algebra Generated by Collection of Mappings"
] | [
"Definition:Measurable Mapping",
"Definition:Measurable Mapping",
"Definition:Characteristic Function (Set Theory)/Set",
"Definition:Simple Function",
"Definition:Characteristic Function (Set Theory)/Set",
"Characteristic Function Measurable iff Set Measurable",
"Definition:Measurable Set",
"Character... |
proofwiki-5345 | Piecewise Combination of Measurable Mappings is Measurable/Binary Case | Let $f, g: X \to X'$ be $\Sigma \, / \, \Sigma'$-measurable mappings.
Let $E \in \Sigma$ be a measurable set.
Define $h: X \to X'$ by:
:$<nowiki>\forall x \in X: \map h x := \begin{cases}
\map f x & : \text {if $x \in E$} \\
\map g x & : \text {if $x \notin E$}
\end{cases}</nowiki>$
Then $h$ is also a $\Sigma \, / \, \... | Let $E' \in \Sigma'$ be a $\Sigma'$-measurable set.
Then by definition of preimage:
:$\map {h^{-1} } {E'} = \set {x \in X: \map h x \in E'}$
Expanding the definition of $h$, this translates into:
:$\map {h^{-1} } {E'} = \set {x \in E: \map f x \in E'} \cup \set {x \in \relcomp X E: \map g x \in E'}$
where $\complement$... | Let $f, g: X \to X'$ be [[Definition:Measurable Mapping|$\Sigma \, / \, \Sigma'$-measurable mappings]].
Let $E \in \Sigma$ be a [[Definition:Measurable Set|measurable set]].
Define $h: X \to X'$ by:
:$<nowiki>\forall x \in X: \map h x := \begin{cases}
\map f x & : \text {if $x \in E$} \\
\map g x & : \text {if $x \... | Let $E' \in \Sigma'$ be a [[Definition:Measurable Set|$\Sigma'$-measurable set]].
Then by definition of [[Definition:Preimage of Subset under Mapping|preimage]]:
:$\map {h^{-1} } {E'} = \set {x \in X: \map h x \in E'}$
Expanding the definition of $h$, this translates into:
:$\map {h^{-1} } {E'} = \set {x \in E: \ma... | Piecewise Combination of Measurable Mappings is Measurable/Binary Case | https://proofwiki.org/wiki/Piecewise_Combination_of_Measurable_Mappings_is_Measurable/Binary_Case | https://proofwiki.org/wiki/Piecewise_Combination_of_Measurable_Mappings_is_Measurable/Binary_Case | [
"Piecewise Combination of Measurable Mappings is Measurable"
] | [
"Definition:Measurable Mapping",
"Definition:Measurable Set",
"Definition:Measurable Mapping"
] | [
"Definition:Measurable Set",
"Definition:Preimage/Mapping/Subset",
"Definition:Set Complement",
"Definition:Measurable Set",
"Sigma-Algebra Closed under Finite Intersection",
"Sigma-Algebra Closed under Union",
"Definition:Measurable Mapping"
] |
proofwiki-5346 | Piecewise Combination of Measurable Mappings is Measurable/General Case | Let $\sequence {E_n}_{n \mathop \in \N} \in \Sigma, \ds \bigcup_{n \mathop \in \N} E_n = X$ be a countable cover of $X$ by $\Sigma$-measurable sets.
For each $n \in \N$, let $f_n: E_n \to X'$ be a $\Sigma_{E_n} \, / \, \Sigma'$-measurable mapping.
Here, $\Sigma_{E_n}$ is the trace $\sigma$-algebra of $E_n$ in $\Sigma$.... | First, note that $f$ is well-defined, since if $x \in E_n$ and $x \in E_m$, we have that:
:$\map {f_n} x = \map f x = \map {f_m} x$
by $(1)$, since $x \in E_n \cap E_m$.
Let $E' \in \Sigma'$.
Then by definition of preimage, $f^{-1} \sqbrk {E'} \subseteq X$, and hence:
{{begin-eqn}}
{{eqn | l = f^{-1} \sqbrk {E'}
... | Let $\sequence {E_n}_{n \mathop \in \N} \in \Sigma, \ds \bigcup_{n \mathop \in \N} E_n = X$ be a [[Definition:Countable Cover|countable cover]] of $X$ by [[Definition:Measurable Set|$\Sigma$-measurable sets]].
For each $n \in \N$, let $f_n: E_n \to X'$ be a [[Definition:Measurable Mapping|$\Sigma_{E_n} \, / \, \Sigma'... | First, note that $f$ is [[Definition:Well-Defined Mapping|well-defined]], since if $x \in E_n$ and $x \in E_m$, we have that:
:$\map {f_n} x = \map f x = \map {f_m} x$
by $(1)$, since $x \in E_n \cap E_m$.
Let $E' \in \Sigma'$.
Then by definition of [[Definition:Preimage of Subset under Mapping|preimage]], $f^{-1}... | Piecewise Combination of Measurable Mappings is Measurable/General Case | https://proofwiki.org/wiki/Piecewise_Combination_of_Measurable_Mappings_is_Measurable/General_Case | https://proofwiki.org/wiki/Piecewise_Combination_of_Measurable_Mappings_is_Measurable/General_Case | [
"Piecewise Combination of Measurable Mappings is Measurable"
] | [
"Definition:Cover of Set/Countable",
"Definition:Measurable Set",
"Definition:Measurable Mapping",
"Definition:Trace Sigma-Algebra",
"Definition:Restriction/Mapping",
"Definition:Measurable Mapping"
] | [
"Definition:Well-Defined/Mapping",
"Definition:Preimage/Mapping/Subset",
"Intersection with Subset is Subset",
"Intersection Distributes over Union/General Result",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Measurable Mapping",
"Definition:Sigma-Algebra",
"Definition:Measurable Mapping"
] |
proofwiki-5347 | Function Simple iff Positive and Negative Parts Simple | Let $\left({X, \Sigma}\right)$ be a measurable space.
Let $g: X \to \overline{\R}$ be an extended real-valued function.
Then $g$ is a simple function {{iff}} its positive part $g^+$ and negative part $g^-$ are simple functions. | === Necessary Condition ===
Suppose $g$ is a simple function.
By Positive Part of Simple Function is Simple Function, so is $g^+$.
By Negative Part of Simple Function is Simple Function, so is $g^-$.
{{qed|lemma}} | Let $\left({X, \Sigma}\right)$ be a [[Definition:Measurable Space|measurable space]].
Let $g: X \to \overline{\R}$ be an [[Definition:Extended Real-Valued Function|extended real-valued function]].
Then $g$ is a [[Definition:Simple Function|simple function]] {{iff}} its [[Definition:Positive Part|positive part]] $g^+... | === Necessary Condition ===
Suppose $g$ is a [[Definition:Simple Function|simple function]].
By [[Positive Part of Simple Function is Simple Function]], so is $g^+$.
By [[Negative Part of Simple Function is Simple Function]], so is $g^-$.
{{qed|lemma}} | Function Simple iff Positive and Negative Parts Simple | https://proofwiki.org/wiki/Function_Simple_iff_Positive_and_Negative_Parts_Simple | https://proofwiki.org/wiki/Function_Simple_iff_Positive_and_Negative_Parts_Simple | [
"Positive Parts",
"Negative Parts",
"Simple Functions",
"Positive Parts",
"Negative Parts"
] | [
"Definition:Measurable Space",
"Definition:Extended Real-Valued Function",
"Definition:Simple Function",
"Definition:Positive Part",
"Definition:Negative Part",
"Definition:Simple Function"
] | [
"Definition:Simple Function",
"Positive Part of Simple Function is Simple Function",
"Negative Part of Simple Function is Simple Function",
"Definition:Simple Function",
"Definition:Simple Function"
] |
proofwiki-5348 | Bounded Measurable Function is Uniform Limit of Simple Functions | Let $\struct {X, \Sigma}$ be a measurable space.
Let $f: X \to \overline \R$ be a bounded $\Sigma$-measurable function.
Then there exists a sequence $\sequence {f_n}_{n \mathop \in \N} \in \map \EE \Sigma$ of simple functions, such that:
:$\forall \epsilon > 0: \exists n \in \N: \forall x \in X: \size {\map f x - \map ... | First, let us prove the theorem when $f$ is a positive $\Sigma$-measurable function.
Now for any $n \in \N$, define for $0 \le k \le n 2^n$:
:<nowiki>${A_k}^n := \begin {cases}
\set {k 2^{-n} \le f < \paren {k + 1} 2^{-n} } & : k \ne n 2^n \\
\set {f \ge n} & : k = n 2^n
\end {cases}$</nowiki>
where for example $\set... | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $f: X \to \overline \R$ be a [[Definition:Bounded Mapping|bounded]] [[Definition:Measurable Function|$\Sigma$-measurable function]].
Then there exists a [[Definition:Sequence|sequence]] $\sequence {f_n}_{n \mathop \in \N} \in \map \... | First, let us prove the theorem when $f$ is a [[Definition:Positive Measurable Function|positive $\Sigma$-measurable function]].
Now for any $n \in \N$, define for $0 \le k \le n 2^n$:
:<nowiki>${A_k}^n := \begin {cases}
\set {k 2^{-n} \le f < \paren {k + 1} 2^{-n} } & : k \ne n 2^n \\
\set {f \ge n} & : k = n 2^n
... | Bounded Measurable Function is Uniform Limit of Simple Functions | https://proofwiki.org/wiki/Bounded_Measurable_Function_is_Uniform_Limit_of_Simple_Functions | https://proofwiki.org/wiki/Bounded_Measurable_Function_is_Uniform_Limit_of_Simple_Functions | [
"Measurable Functions",
"Simple Functions"
] | [
"Definition:Measurable Space",
"Definition:Bounded Mapping",
"Definition:Measurable Function",
"Definition:Sequence",
"Definition:Simple Function",
"Definition:Uniform Limit",
"Definition:Sequence",
"Definition:Absolute Value of Mapping/Extended Real-Valued Function"
] | [
"Definition:Measurable Function/Positive",
"Definition:Pairwise Disjoint",
"Definition:Characteristic Function (Set Theory)/Set",
"Definition:Pointwise Inequality of Extended Real-Valued Functions",
"Characterization of Measurable Functions",
"Sigma-Algebra Closed under Finite Intersection",
"Definition... |
proofwiki-5349 | Integral of Positive Simple Function is Well-Defined | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \R, f \in \EE^+$ be a positive simple function.
Then the $\mu$-integral of $f$, $\map {I_\mu} f$, is well-defined.
That is, for any two standard representations for $f$, say:
:$\ds f = \sum_{i \mathop = 0}^n a_i \chi_{E_i} = \sum_{j \mathop = 0}^m b_j \ch... | The sets $F_0, \ldots, F_m$ are pairwise disjoint, and:
:$X = \ds \bigcup_{j \mathop = 0}^m F_j$
From Characteristic Function of Disjoint Union, we have:
:$\chi_X = \ds \sum_{j \mathop = 0}^m \chi_{F_j}$
Remark that $\map {\chi_X} x = 1$ for all $x \in X$, so that we have:
{{begin-eqn}}
{{eqn | l = f
| r = \sum_{... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $f: X \to \R, f \in \EE^+$ be a [[Definition:Positive Simple Function|positive simple function]].
Then the [[Definition:Integral of Positive Simple Function|$\mu$-integral of $f$]], $\map {I_\mu} f$, is well-defined.
That is, for an... | The sets $F_0, \ldots, F_m$ are [[Definition:Pairwise Disjoint|pairwise disjoint]], and:
:$X = \ds \bigcup_{j \mathop = 0}^m F_j$
From [[Characteristic Function of Disjoint Union]], we have:
:$\chi_X = \ds \sum_{j \mathop = 0}^m \chi_{F_j}$
Remark that $\map {\chi_X} x = 1$ for all $x \in X$, so that we have:
{{be... | Integral of Positive Simple Function is Well-Defined | https://proofwiki.org/wiki/Integral_of_Positive_Simple_Function_is_Well-Defined | https://proofwiki.org/wiki/Integral_of_Positive_Simple_Function_is_Well-Defined | [
"Integral of Positive Simple Function"
] | [
"Definition:Measure Space",
"Definition:Simple Function",
"Definition:Integral of Positive Simple Function",
"Definition:Standard Representation of Simple Function"
] | [
"Definition:Pairwise Disjoint",
"Characteristic Function of Disjoint Union",
"Characteristic Function of Intersection/Variant 1",
"Intersection Distributes over Union/General Result",
"Measure is Finitely Additive Function",
"Summation is Linear",
"Summation is Linear",
"Measure is Finitely Additive F... |
proofwiki-5350 | Integral of Characteristic Function | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $E \in \Sigma$ be a measurable set, and let $\chi_E: X \to \R$ be its characteristic function.
Then $\map {I_\mu} {\chi_E} = \map \mu E$, where $\map {I_\mu} {\chi_E}$ is the $\mu$-integral of $\chi_E$. | Let $a_1 = 1$ and $E_1 = E$.
As in the definition of standard representation, denote $a_0 = 0$ and $E_0 = X \setminus E_1$.
Then for $x \in X$, we have:
:$\map {\chi_E} x = 0 \cdot \map {\chi_{E_0} } x + 1 \cdot \map {\chi_{E_1} } x$
since $E_1 = E$.
Hence $\chi_E = a_0 \chi_{E_0} + a_1 \chi_{E_1}$ is a standard repres... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $E \in \Sigma$ be a [[Definition:Measurable Set|measurable set]], and let $\chi_E: X \to \R$ be its [[Definition:Characteristic Function of Set|characteristic function]].
Then $\map {I_\mu} {\chi_E} = \map \mu E$, where $\map {I_\mu}... | Let $a_1 = 1$ and $E_1 = E$.
As in the definition of [[Definition:Standard Representation of Simple Function|standard representation]], denote $a_0 = 0$ and $E_0 = X \setminus E_1$.
Then for $x \in X$, we have:
:$\map {\chi_E} x = 0 \cdot \map {\chi_{E_0} } x + 1 \cdot \map {\chi_{E_1} } x$
since $E_1 = E$.
Hence... | Integral of Characteristic Function | https://proofwiki.org/wiki/Integral_of_Characteristic_Function | https://proofwiki.org/wiki/Integral_of_Characteristic_Function | [
"Integral of Positive Simple Function",
"Characteristic Functions",
"Integral of Characteristic Function"
] | [
"Definition:Measure Space",
"Definition:Measurable Set",
"Definition:Characteristic Function (Set Theory)/Set",
"Definition:Integral of Positive Simple Function"
] | [
"Definition:Standard Representation of Simple Function",
"Definition:Standard Representation of Simple Function",
"Definition:Integral of Positive Simple Function"
] |
proofwiki-5351 | Integral of Positive Simple Function is Positive Homogeneous | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \R, f \in \EE^+$ be a positive simple function.
Let $\lambda \in \R_{\ge 0}$ be a positive real number.
Then:
:$\map {I_\mu} {\lambda \cdot f} = \map {\lambda \cdot I_\mu} f$
where:
:$\lambda \cdot f$ is the pointwise $\lambda$-multiple of $f$
:$I_\mu$ de... | Remark that $\lambda \cdot f$ is a positive simple function by Scalar Multiple of Simple Function is Simple Function.
Let:
:$f = \ds \sum_{i \mathop = 0}^n a_i \chi_{E_i}$
be a standard representation for $f$.
Then we also have, for all $x \in X$:
{{begin-eqn}}
{{eqn | l = \map {\lambda \cdot f} x
| r = \lambda \... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $f: X \to \R, f \in \EE^+$ be a [[Definition:Positive Simple Function|positive simple function]].
Let $\lambda \in \R_{\ge 0}$ be a [[Definition:Positive Real Number|positive real number]].
Then:
:$\map {I_\mu} {\lambda \cdot f} = \... | Remark that $\lambda \cdot f$ is a [[Definition:Positive Simple Function|positive simple function]] by [[Scalar Multiple of Simple Function is Simple Function]].
Let:
:$f = \ds \sum_{i \mathop = 0}^n a_i \chi_{E_i}$
be a [[Definition:Standard Representation of Simple Function|standard representation]] for $f$.
Then... | Integral of Positive Simple Function is Positive Homogeneous | https://proofwiki.org/wiki/Integral_of_Positive_Simple_Function_is_Positive_Homogeneous | https://proofwiki.org/wiki/Integral_of_Positive_Simple_Function_is_Positive_Homogeneous | [
"Integral of Positive Simple Function"
] | [
"Definition:Measure Space",
"Definition:Simple Function",
"Definition:Positive/Real Number",
"Definition:Pointwise Scalar Multiplication of Mappings/Real-Valued Functions",
"Definition:Integral of Positive Simple Function",
"Definition:Positive Homogeneous"
] | [
"Definition:Simple Function",
"Scalar Multiple of Simple Function is Simple Function",
"Definition:Standard Representation of Simple Function",
"Summation is Linear",
"Definition:Standard Representation of Simple Function",
"Summation is Linear"
] |
proofwiki-5352 | Strict Ordering Preserved under Product with Invertible Element | Let $\struct {S, \circ, \preceq}$ be an ordered semigroup.
Let $z \in S$ be invertible.
Suppose that either $x \circ z \prec y \circ z$ or $z \circ x \prec z \circ y$.
Then $x \prec y$. | Suppose $x \circ z \prec y \circ z$.
By Invertible Element of Monoid is Cancellable, $z^{-1}$ is cancellable.
Then from Strict Ordering Preserved under Product with Cancellable Element:
:$x = \paren {x \circ z} \circ z^{-1} \prec \paren {y \circ z} \circ z^{-1} = y$
Likewise, if $z \circ x \prec z \circ y$:
:$x = z^{-1... | Let $\struct {S, \circ, \preceq}$ be an [[Definition:Ordered Semigroup|ordered semigroup]].
Let $z \in S$ be [[Definition:Invertible Element|invertible]].
Suppose that either $x \circ z \prec y \circ z$ or $z \circ x \prec z \circ y$.
Then $x \prec y$. | Suppose $x \circ z \prec y \circ z$.
By [[Invertible Element of Monoid is Cancellable]], $z^{-1}$ is [[Definition:Cancellable Element|cancellable]].
Then from [[Strict Ordering Preserved under Product with Cancellable Element]]:
:$x = \paren {x \circ z} \circ z^{-1} \prec \paren {y \circ z} \circ z^{-1} = y$
Likew... | Strict Ordering Preserved under Product with Invertible Element | https://proofwiki.org/wiki/Strict_Ordering_Preserved_under_Product_with_Invertible_Element | https://proofwiki.org/wiki/Strict_Ordering_Preserved_under_Product_with_Invertible_Element | [
"Ordered Semigroups"
] | [
"Definition:Ordered Semigroup",
"Definition:Invertible Element"
] | [
"Invertible Element of Associative Structure is Cancellable/Corollary",
"Definition:Cancellable Element",
"Strict Ordering Preserved under Product with Cancellable Element"
] |
proofwiki-5353 | Strict Ordering Preserved under Cancellability in Totally Ordered Semigroup | Let $\struct {S, \circ, \preceq}$ be a totally ordered semigroup.
If either:
:$x \circ z \prec y \circ z$
or
:$z \circ x \prec z \circ y$
then $x \prec y$. | Let $x \circ z \prec y \circ z$.
{{AimForCont}} $x \succeq y$.
As $\struct {S, \circ, \preceq}$ is an ordered semigroup, $\preceq$ is compatible with $\circ$.
Hence we have:
:$x \succeq y \implies x \circ z \succeq y \circ z$
which contradicts $x \circ z \prec y \circ z$.
We have that $\preceq$ is a total ordering, and... | Let $\struct {S, \circ, \preceq}$ be a [[Definition:Totally Ordered Semigroup|totally ordered semigroup]].
If either:
:$x \circ z \prec y \circ z$
or
:$z \circ x \prec z \circ y$
then $x \prec y$. | Let $x \circ z \prec y \circ z$.
{{AimForCont}} $x \succeq y$.
As $\struct {S, \circ, \preceq}$ is an [[Definition:Ordered Semigroup|ordered semigroup]], $\preceq$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ$.
Hence we have:
:$x \succeq y \implies x \circ z \succeq y \circ z$
which [... | Strict Ordering Preserved under Cancellability in Totally Ordered Semigroup | https://proofwiki.org/wiki/Strict_Ordering_Preserved_under_Cancellability_in_Totally_Ordered_Semigroup | https://proofwiki.org/wiki/Strict_Ordering_Preserved_under_Cancellability_in_Totally_Ordered_Semigroup | [
"Ordered Semigroups"
] | [
"Definition:Totally Ordered Semigroup"
] | [
"Definition:Ordered Semigroup",
"Definition:Relation Compatible with Operation",
"Definition:Contradiction",
"Definition:Total Ordering",
"Trichotomy Law"
] |
proofwiki-5354 | Relation Compatibility in Totally Ordered Semigroup | Let $\left({S, \circ, \preceq}\right)$ be an ordered semigroup such that:
:$(1): \quad$ All the elements of $\left({S, \circ, \preceq}\right)$ are cancellable for $\circ$
:$(2): \quad \preceq$ is a total ordering.
Then:
:$\forall x, y, z \in S: x \circ z \preceq y \circ z \iff x \preceq y$ | From Strict Ordering Preserved under Cancellability in Totally Ordered Semigroup:
: $x \circ z \prec y \circ z \implies x \prec y$
From the definition of cancellable element:
: $x \circ z = y \circ z \implies x = y$
{{qed}}
Category:Ordered Semigroups
oafce8s0w7tw3us87tqoozua1fopu57 | Let $\left({S, \circ, \preceq}\right)$ be an [[Definition:Ordered Semigroup|ordered semigroup]] such that:
:$(1): \quad$ All the elements of $\left({S, \circ, \preceq}\right)$ are [[Definition:Cancellable Element|cancellable]] for $\circ$
:$(2): \quad \preceq$ is a [[Definition:Total Ordering|total ordering]].
Then:
... | From [[Strict Ordering Preserved under Cancellability in Totally Ordered Semigroup]]:
: $x \circ z \prec y \circ z \implies x \prec y$
From the definition of [[Definition:Cancellable Element|cancellable element]]:
: $x \circ z = y \circ z \implies x = y$
{{qed}}
[[Category:Ordered Semigroups]]
oafce8s0w7tw3us87tqoozu... | Relation Compatibility in Totally Ordered Semigroup | https://proofwiki.org/wiki/Relation_Compatibility_in_Totally_Ordered_Semigroup | https://proofwiki.org/wiki/Relation_Compatibility_in_Totally_Ordered_Semigroup | [
"Ordered Semigroups"
] | [
"Definition:Ordered Semigroup",
"Definition:Cancellable Element",
"Definition:Total Ordering"
] | [
"Strict Ordering Preserved under Cancellability in Totally Ordered Semigroup",
"Definition:Cancellable Element",
"Category:Ordered Semigroups"
] |
proofwiki-5355 | Integral of Positive Measurable Function Extends Integral of Positive Simple Function | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \R, f \in \EE^+$ be a positive simple function.
Then:
:$\ds \int f \rd \mu = \map {I_\mu} f$
where:
:$\ds \int \cdot \rd \mu$ denotes the $\mu$-integral of positive measurable functions
:$I_\mu$ denotes the $\mu$-integral of positive simple functions.
Tha... | From the definition of the integral of a positive measure function, we have:
:$\ds \int f \rd \mu = \sup \set {\map {I_\mu} g: g \le f, g \in \EE^+}$
Let $g \in \EE^+$ be such that $g \le f$.
Then, from Integral of Positive Simple Function is Increasing, we have:
:$\map {I_\mu} g \le \map {I_\mu} f$
So $\map {I_\mu... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $f: X \to \R, f \in \EE^+$ be a [[Definition:Positive Simple Function|positive simple function]].
Then:
:$\ds \int f \rd \mu = \map {I_\mu} f$
where:
:$\ds \int \cdot \rd \mu$ denotes the [[Definition:Integral of Positive Measurabl... | From the definition of the [[Definition:Integral of Positive Measurable Function|integral of a positive measure function]], we have:
:$\ds \int f \rd \mu = \sup \set {\map {I_\mu} g: g \le f, g \in \EE^+}$
Let $g \in \EE^+$ be such that $g \le f$.
Then, from [[Integral of Positive Simple Function is Increasing]], ... | Integral of Positive Measurable Function Extends Integral of Positive Simple Function | https://proofwiki.org/wiki/Integral_of_Positive_Measurable_Function_Extends_Integral_of_Positive_Simple_Function | https://proofwiki.org/wiki/Integral_of_Positive_Measurable_Function_Extends_Integral_of_Positive_Simple_Function | [
"Integral of Positive Measurable Function"
] | [
"Definition:Measure Space",
"Definition:Simple Function",
"Definition:Integral of Positive Measurable Function",
"Definition:Integral of Positive Simple Function",
"Definition:Restriction/Mapping"
] | [
"Definition:Integral of Positive Measurable Function",
"Integral of Positive Simple Function is Increasing",
"Definition:Upper Bound of Set/Real Numbers",
"Definition:Greatest Element",
"Greatest Element is Supremum"
] |
proofwiki-5356 | Beppo Levi's Theorem | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\sequence {f_n}_{n \mathop \in \N} \in \MM_{\overline \R}^+$ be an increasing sequence of positive $\Sigma$-measurable functions.
Let $\ds \sup_{n \mathop \in \N} f_n: X \to \overline \R$ be the pointwise supremum of $\sequence {f_n}_{n \mathop \in \N}$, where $\o... | {{tidy|Break down some of the long complex sentences into simple ones}}
{{MissingLinks}}
Since by definition $\ds \sup _{n \mathop \in \N} f_n \ge f_m$ for all $m$, we have:
:$\ds \int \sup_{n \mathop \in \N} f_n \rd \mu \ge \int f_m \rd \mu$
and hence the inequality holds for the supremum as well:
:$\ds \int \sup_{n \... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $\sequence {f_n}_{n \mathop \in \N} \in \MM_{\overline \R}^+$ be an [[Definition:Increasing Sequence of Extended Real-Valued Functions|increasing sequence]] of [[Definition:Positive Measurable Function|positive $\Sigma$-measurable func... | {{tidy|Break down some of the long complex sentences into simple ones}}
{{MissingLinks}}
Since by definition $\ds \sup _{n \mathop \in \N} f_n \ge f_m$ for all $m$, we have:
:$\ds \int \sup_{n \mathop \in \N} f_n \rd \mu \ge \int f_m \rd \mu$
and hence the inequality holds for the supremum as well:
:$\ds \int \sup_{... | Beppo Levi's Theorem | https://proofwiki.org/wiki/Beppo_Levi's_Theorem | https://proofwiki.org/wiki/Beppo_Levi's_Theorem | [
"Measure Theory"
] | [
"Definition:Measure Space",
"Definition:Increasing Sequence of Extended Real-Valued Functions",
"Definition:Measurable Function/Positive",
"Definition:Pointwise Supremum of Extended Real-Valued Functions",
"Definition:Extended Real Number Line",
"Definition:Supremum of Set",
"Definition:Ordering on Exte... | [
"Linear Combination of Measures",
"Definition:Increasing Sequence of Sets",
"Definition:Limit of Increasing Sequence of Sets",
"Measure of Limit of Increasing Sequence of Measurable Sets"
] |
proofwiki-5357 | Ordered Semigroup Isomorphism is Surjective Monomorphism | Let $\struct {S, \circ, \preceq}$ and $\struct {T, *, \preccurlyeq}$ be ordered semigroups.
Let $\phi: \struct {S, \circ, \preceq} \to \struct {T, *, \preccurlyeq}$ be a mapping.
Then $\phi$ is an ordered semigroup isomorphism {{iff}}:
:$(1): \quad \phi$ is an ordered semigroup monomorphism
:$(2): \quad \phi$ is a surj... | === Necessary Condition ===
Let $\phi: \struct {S, \circ, \preceq} \to \struct {T, *, \preccurlyeq}$ be an ordered semigroup isomorphism.
Then by definition:
:$\phi$ is a semigroup isomorphism from the semigroup $\struct {S, \circ}$ to the semigroup $\struct {T, *}$
:$\phi$ is an order isomorphism from the ordered set ... | Let $\struct {S, \circ, \preceq}$ and $\struct {T, *, \preccurlyeq}$ be [[Definition:Ordered Semigroup|ordered semigroups]].
Let $\phi: \struct {S, \circ, \preceq} \to \struct {T, *, \preccurlyeq}$ be a [[Definition:Mapping|mapping]].
Then $\phi$ is an [[Definition:Ordered Semigroup Isomorphism|ordered semigroup iso... | === Necessary Condition ===
Let $\phi: \struct {S, \circ, \preceq} \to \struct {T, *, \preccurlyeq}$ be an [[Definition:Ordered Semigroup Isomorphism|ordered semigroup isomorphism]].
Then by definition:
:$\phi$ is a [[Definition:Semigroup Isomorphism|semigroup isomorphism]] from the [[Definition:Semigroup|semigroup]... | Ordered Semigroup Isomorphism is Surjective Monomorphism | https://proofwiki.org/wiki/Ordered_Semigroup_Isomorphism_is_Surjective_Monomorphism | https://proofwiki.org/wiki/Ordered_Semigroup_Isomorphism_is_Surjective_Monomorphism | [
"Ordered Semigroups",
"Surjections",
"Monomorphisms (Abstract Algebra)",
"Isomorphisms (Abstract Algebra)"
] | [
"Definition:Ordered Semigroup",
"Definition:Mapping",
"Definition:Ordered Semigroup Isomorphism",
"Definition:Ordered Semigroup Monomorphism",
"Definition:Surjection"
] | [
"Definition:Ordered Semigroup Isomorphism",
"Definition:Isomorphism (Abstract Algebra)/Semigroup Isomorphism",
"Definition:Semigroup",
"Definition:Semigroup",
"Definition:Order Isomorphism",
"Definition:Ordered Set",
"Definition:Ordered Set",
"Definition:Isomorphism (Abstract Algebra)/Semigroup Isomor... |
proofwiki-5358 | Ordered Semigroup Monomorphism into Image is Isomorphism | Let $\struct {S, \circ, \preceq}$ and $\struct {T, *, \preccurlyeq}$ be ordered semigroups.
Let $\phi: \struct {S, \circ, \preceq} \to \struct {T, *, \preccurlyeq}$ be an ordered semigroup monomorphism.
Let $S'$ be the image of $\phi$.
Then $\phi$ is an ordered semigroup isomorphism from $\struct {S, \circ, \preceq}$ i... | Let $\phi: \struct {S, \circ, \preceq} \to \struct {T, *, \preccurlyeq}$ be an ordered semigroup monomorphism.
Then $\phi$ is an injection into $\struct {T, *, \preccurlyeq}$ by definition.
From Restriction of Mapping to Image is Surjection, a mapping from a set to the image of that mapping is a surjection.
Thus the su... | Let $\struct {S, \circ, \preceq}$ and $\struct {T, *, \preccurlyeq}$ be [[Definition:Ordered Semigroup|ordered semigroups]].
Let $\phi: \struct {S, \circ, \preceq} \to \struct {T, *, \preccurlyeq}$ be an [[Definition:Ordered Semigroup Monomorphism|ordered semigroup monomorphism]].
Let $S'$ be the [[Definition:Image o... | Let $\phi: \struct {S, \circ, \preceq} \to \struct {T, *, \preccurlyeq}$ be an [[Definition:Ordered Semigroup Monomorphism|ordered semigroup monomorphism]].
Then $\phi$ is an [[Definition:Injection|injection]] into $\struct {T, *, \preccurlyeq}$ by definition.
From [[Restriction of Mapping to Image is Surjection]], ... | Ordered Semigroup Monomorphism into Image is Isomorphism | https://proofwiki.org/wiki/Ordered_Semigroup_Monomorphism_into_Image_is_Isomorphism | https://proofwiki.org/wiki/Ordered_Semigroup_Monomorphism_into_Image_is_Isomorphism | [
"Semigroup Homomorphisms",
"Order Isomorphisms"
] | [
"Definition:Ordered Semigroup",
"Definition:Ordered Semigroup Monomorphism",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Ordered Semigroup Isomorphism",
"Definition:Restriction/Operation",
"Definition:Restriction/Relation"
] | [
"Definition:Ordered Semigroup Monomorphism",
"Definition:Injection",
"Restriction of Mapping to Image is Surjection",
"Definition:Mapping",
"Definition:Set",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Mapping",
"Definition:Surjection",
"Definition:Surjective Restriction",
"Defini... |
proofwiki-5359 | Order Completion is Unique up to Isomorphism | Let $\struct {S, \preceq_S}$ be an ordered set.
Suppose that both $\struct {T, \preceq_T}$ and $\struct {T', \preceq_{T'} }$ are order completions for $\struct {S, \preceq_S}$.
Then there exists a unique order isomorphism $\psi: T \to T'$.
In particular, $\struct {T, \preceq_T}$ and $\struct {T', \preceq_{T'} }$ are is... | Both $\struct {T, \preceq_T}$ and $\struct {T', \preceq_{T'} }$ are order completions for $\struct {S, \preceq_S}$.
Hence they both satisfy condition $(4)$ (and also $(1)$, $(2)$ and $(3)$).
Thus, applying condition $(4)$ to $\struct {T, \preceq_T}$ (with respect to $\struct {T', \preceq_{T'} }$), obtain a unique incre... | Let $\struct {S, \preceq_S}$ be an [[Definition:Ordered Set|ordered set]].
Suppose that both $\struct {T, \preceq_T}$ and $\struct {T', \preceq_{T'} }$ are [[Definition:Order Completion|order completions]] for $\struct {S, \preceq_S}$.
Then there exists a unique [[Definition:Order Isomorphism|order isomorphism]] $\p... | Both $\struct {T, \preceq_T}$ and $\struct {T', \preceq_{T'} }$ are [[Definition:Order Completion|order completions]] for $\struct {S, \preceq_S}$.
Hence they both satisfy condition $(4)$ (and also $(1)$, $(2)$ and $(3)$).
Thus, applying condition $(4)$ to $\struct {T, \preceq_T}$ (with respect to $\struct {T', \pre... | Order Completion is Unique up to Isomorphism | https://proofwiki.org/wiki/Order_Completion_is_Unique_up_to_Isomorphism | https://proofwiki.org/wiki/Order_Completion_is_Unique_up_to_Isomorphism | [
"Order Theory"
] | [
"Definition:Ordered Set",
"Definition:Order Completion",
"Definition:Order Isomorphism",
"Definition:Order Isomorphism"
] | [
"Definition:Order Completion",
"Definition:Unique",
"Definition:Increasing/Mapping",
"Definition:Unique",
"Definition:Increasing/Mapping",
"Composite of Increasing Mappings is Increasing",
"Definition:Composition of Mappings",
"Definition:Increasing/Mapping",
"Definition:Unique",
"Identity Mapping... |
proofwiki-5360 | Intersection of Strict Upper Closures in Toset | Let $\struct {S, \preceq}$ be a totally ordered set.
Let $a, b \in S$.
Then:
:$a^\succ \cap b^\succ = \paren {\map \max {a, b} }^\succ$
where:
:$a^\succ$ denotes strict upper closure of $a$
:$\max$ denotes the max operation. | As $\struct {S, \preceq}$ is a totally ordered set, have either $a \preceq b$ or $b \preceq a$.
Both sides are seen to be invariant upon interchanging $a$ and $b$.
{{WLOG}}, then, let $a \preceq b$.
Then it follows by definition of $\max$ that:
:$\map \max {a, b} = b$
Thus, from Intersection with Subset is Subset, it s... | Let $\struct {S, \preceq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Let $a, b \in S$.
Then:
:$a^\succ \cap b^\succ = \paren {\map \max {a, b} }^\succ$
where:
:$a^\succ$ denotes [[Definition:Strict Upper Closure of Element|strict upper closure]] of $a$
:$\max$ denotes the [[Definition:Max Operat... | As $\struct {S, \preceq}$ is a [[Definition:Totally Ordered Set|totally ordered set]], have either $a \preceq b$ or $b \preceq a$.
Both sides are seen to be invariant upon interchanging $a$ and $b$.
{{WLOG}}, then, let $a \preceq b$.
Then it follows by [[Definition:Max Operation|definition of $\max$]] that:
:$\map \... | Intersection of Strict Upper Closures in Toset | https://proofwiki.org/wiki/Intersection_of_Strict_Upper_Closures_in_Toset | https://proofwiki.org/wiki/Intersection_of_Strict_Upper_Closures_in_Toset | [
"Total Orderings"
] | [
"Definition:Totally Ordered Set",
"Definition:Strict Upper Closure/Element",
"Definition:Max Operation"
] | [
"Definition:Totally Ordered Set",
"Definition:Max Operation",
"Intersection with Subset is Subset",
"Definition:Strict Upper Closure/Element",
"Strictly Precedes is Strict Ordering"
] |
proofwiki-5361 | Intersection of Weak Upper Closures in Toset | Let $\struct {S, \preccurlyeq}$ be a totally ordered set.
Let $a, b \in S$.
Then:
:$a^\succcurlyeq \cap b^\succcurlyeq = \paren {\map \max {a, b} }^\succcurlyeq$
where:
:$a^\succcurlyeq$ denotes weak upper closure of $a$
:$\max$ denotes the max operation. | As $\struct {S, \preccurlyeq}$ is a totally ordered set, either $a \preccurlyeq b$ or $b \preccurlyeq a$.
Both sides are seen to be invariant upon interchanging $a$ and $b$.
{{WLOG}}, let $a \preccurlyeq b$.
Then it follows by definition of $\max$ that:
:$\map \max {a, b} = b$
Thus, from Intersection with Subset is Sub... | Let $\struct {S, \preccurlyeq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Let $a, b \in S$.
Then:
:$a^\succcurlyeq \cap b^\succcurlyeq = \paren {\map \max {a, b} }^\succcurlyeq$
where:
:$a^\succcurlyeq$ denotes [[Definition:Weak Upper Closure of Element|weak upper closure]] of $a$
:$\max$ denote... | As $\struct {S, \preccurlyeq}$ is a [[Definition:Totally Ordered Set|totally ordered set]], either $a \preccurlyeq b$ or $b \preccurlyeq a$.
Both sides are seen to be invariant upon interchanging $a$ and $b$.
{{WLOG}}, let $a \preccurlyeq b$.
Then it follows by [[Definition:Max Operation|definition of $\max$]] that:... | Intersection of Weak Upper Closures in Toset | https://proofwiki.org/wiki/Intersection_of_Weak_Upper_Closures_in_Toset | https://proofwiki.org/wiki/Intersection_of_Weak_Upper_Closures_in_Toset | [
"Total Orderings"
] | [
"Definition:Totally Ordered Set",
"Definition:Upper Closure/Element",
"Definition:Max Operation"
] | [
"Definition:Totally Ordered Set",
"Definition:Max Operation",
"Intersection with Subset is Subset",
"Definition:Upper Closure/Element",
"Definition:Total Ordering",
"Definition:Transitive Relation"
] |
proofwiki-5362 | Intersection of Weak Lower Closures in Toset | Let $\struct {S, \preccurlyeq}$ be a totally ordered set.
Let $a, b \in S$.
Then:
:$a^\preccurlyeq \cap b^\preccurlyeq = \paren {\min \set {a, b} }^\preccurlyeq$
where:
:$a^\preccurlyeq$ denotes the weak lower closure of $a$
:$\min$ denotes the min operation. | As $\struct {S, \preccurlyeq}$ is a totally ordered set, either $a \preccurlyeq b$ or $b \preccurlyeq a$.
Both sides are seen to be invariant upon interchanging $a$ and $b$.
{{WLOG}}, let $b \preccurlyeq a$.
Then it follows by definition of $\min$ that $\min \set {a, b} = b$.
Thus, from Intersection with Subset is Subs... | Let $\struct {S, \preccurlyeq}$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Let $a, b \in S$.
Then:
:$a^\preccurlyeq \cap b^\preccurlyeq = \paren {\min \set {a, b} }^\preccurlyeq$
where:
:$a^\preccurlyeq$ denotes the [[Definition:Weak Lower Closure of Element|weak lower closure of $a$]]
:$\min$ de... | As $\struct {S, \preccurlyeq}$ is a [[Definition:Totally Ordered Set|totally ordered set]], either $a \preccurlyeq b$ or $b \preccurlyeq a$.
Both sides are seen to be invariant upon interchanging $a$ and $b$.
{{WLOG}}, let $b \preccurlyeq a$.
Then it follows by [[Definition:Min Operation|definition of $\min$]] that ... | Intersection of Weak Lower Closures in Toset | https://proofwiki.org/wiki/Intersection_of_Weak_Lower_Closures_in_Toset | https://proofwiki.org/wiki/Intersection_of_Weak_Lower_Closures_in_Toset | [
"Total Orderings"
] | [
"Definition:Totally Ordered Set",
"Definition:Lower Closure/Element",
"Definition:Min Operation"
] | [
"Definition:Totally Ordered Set",
"Definition:Min Operation",
"Intersection with Subset is Subset",
"Definition:Lower Closure/Element",
"Definition:Total Ordering",
"Definition:Transitive Relation"
] |
proofwiki-5363 | Intersection of Strict Lower Closures in Toset | Let $\left({S, \preceq}\right)$ be a totally ordered set.
Let $a,b \in S$.
Then:
:$a^\prec \cap b^\prec = \left({\min \left({a, b}\right)}\right)^\prec$
where:
: $a^\prec$ denotes strict lower closure of $a$
: $\min$ denotes the min operation. | As $\left({S, \preceq}\right)$ is a totally ordered set, have either $a \preceq b$ or $b \preceq a$.
Since both sides are seen to be invariant upon interchanging $a$ and $b$, let WLOG $b \preceq a$.
Then it follows by definition of $\min$ that $\min \left({a, b}\right) = b$.
Thus, from Intersection with Subset is Subse... | Let $\left({S, \preceq}\right)$ be a [[Definition:Totally Ordered Set|totally ordered set]].
Let $a,b \in S$.
Then:
:$a^\prec \cap b^\prec = \left({\min \left({a, b}\right)}\right)^\prec$
where:
: $a^\prec$ denotes [[Definition:Strict Lower Closure of Element|strict lower closure]] of $a$
: $\min$ denotes the [[De... | As $\left({S, \preceq}\right)$ is a [[Definition:Totally Ordered Set|totally ordered set]], have either $a \preceq b$ or $b \preceq a$.
Since both sides are seen to be invariant upon interchanging $a$ and $b$, let [[Definition:WLOG|WLOG]] $b \preceq a$.
Then it follows by [[Definition:Max Operation|definition of $\mi... | Intersection of Strict Lower Closures in Toset | https://proofwiki.org/wiki/Intersection_of_Strict_Lower_Closures_in_Toset | https://proofwiki.org/wiki/Intersection_of_Strict_Lower_Closures_in_Toset | [
"Total Orderings"
] | [
"Definition:Totally Ordered Set",
"Definition:Strict Lower Closure/Element",
"Definition:Min Operation"
] | [
"Definition:Totally Ordered Set",
"Definition:WLOG",
"Definition:Max Operation",
"Intersection with Subset is Subset",
"Definition:Strict Lower Closure/Element",
"Strictly Precedes is Strict Ordering"
] |
proofwiki-5364 | Naturally Ordered Semigroup Exists | Let the Zermelo-Fraenkel axioms be accepted as axiomatic.
Then there exists a '''naturally ordered semigroup'''. | We take as axiomatic the Zermelo-Fraenkel axioms.
From these, Minimally Inductive Set Exists is demonstrated.
This proves the existence of a minimally inductive set.
Then we have that the Minimally Inductive Set forms Peano Structure.
It follows that the existence of a Peano structure depends upon the existence of such... | Let the [[Axiom:Zermelo-Fraenkel Axioms|Zermelo-Fraenkel axioms]] be accepted as [[Definition:Axiom|axiomatic]].
Then there exists a '''[[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]]'''. | We take as [[Definition:Axiom|axiomatic]] the [[Axiom:Zermelo-Fraenkel Axioms|Zermelo-Fraenkel axioms]].
From these, [[Minimally Inductive Set Exists]] is demonstrated.
This proves the existence of a [[Definition:Minimally Inductive Set|minimally inductive set]].
Then we have that the [[Minimally Inductive Set forms... | Naturally Ordered Semigroup Exists | https://proofwiki.org/wiki/Naturally_Ordered_Semigroup_Exists | https://proofwiki.org/wiki/Naturally_Ordered_Semigroup_Exists | [
"Naturally Ordered Semigroup"
] | [
"Axiom:Zermelo-Fraenkel Axioms",
"Definition:Axiom",
"Definition:Naturally Ordered Semigroup"
] | [
"Definition:Axiom",
"Axiom:Zermelo-Fraenkel Axioms",
"Minimally Inductive Set Exists",
"Definition:Minimally Inductive Set",
"Minimally Inductive Set forms Peano Structure",
"Definition:Peano Structure",
"Definition:Minimally Inductive Set",
"Naturally Ordered Semigroup forms Peano Structure"
] |
proofwiki-5365 | Strict Upper Closure in Restricted Ordering | Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$ be a subset of $S$, and let $\preceq \restriction_T$ be the restricted ordering on $T$.
Then for all $t \in T$:
:$t^{\succ T} = T \cap t^{\succ S}$
where:
:$t^{\succ T}$ is the strict upper closure of $t$ in $\struct {T, \preceq \restriction_T}$
:$t^{\su... | Let $t \in T$, and suppose that $t' \in t^{\succ T}$.
By definition of strict upper closure, this is equivalent to:
:$t \preceq \restriction_T t' \land t \ne t'$
By definition of $\preceq \restriction_T$, the first condition comes down to:
:$t \preceq t' \land t' \in T$
as it is assumed that $t \in T$.
In conclusion, $... | Let $\struct {S, \preceq}$ be an [[Definition:Ordered Set|ordered set]].
Let $T \subseteq S$ be a [[Definition:Subset|subset]] of $S$, and let $\preceq \restriction_T$ be the [[Definition:Restricted Ordering|restricted ordering]] on $T$.
Then for all $t \in T$:
:$t^{\succ T} = T \cap t^{\succ S}$
where:
:$t^{\succ... | Let $t \in T$, and suppose that $t' \in t^{\succ T}$.
By definition of [[Definition:Strict Upper Closure of Element|strict upper closure]], this is equivalent to:
:$t \preceq \restriction_T t' \land t \ne t'$
By [[Definition:Restricted Ordering|definition of $\preceq \restriction_T$]], the first condition comes down... | Strict Upper Closure in Restricted Ordering | https://proofwiki.org/wiki/Strict_Upper_Closure_in_Restricted_Ordering | https://proofwiki.org/wiki/Strict_Upper_Closure_in_Restricted_Ordering | [
"Upper Closures"
] | [
"Definition:Ordered Set",
"Definition:Subset",
"Definition:Restriction of Ordering",
"Definition:Strict Upper Closure/Element",
"Definition:Strict Upper Closure/Element"
] | [
"Definition:Strict Upper Closure/Element",
"Definition:Restriction of Ordering",
"Definition:Set Intersection",
"Definition:Set Equality",
"Category:Upper Closures"
] |
proofwiki-5366 | Weak Upper Closure in Restricted Ordering | Let $\struct {S, \preccurlyeq}$ be an ordered set.
Let $T \subseteq S$ be a subset of $S$.
Let $\preccurlyeq \restriction_T$ be the restricted ordering on $T$.
Then for all $t \in T$:
:$t^{\succcurlyeq T} = T \cap t^{\succcurlyeq S}$
where:
:$t^{\succcurlyeq T}$ is the weak upper closure of $t$ in $\struct {T, \preccur... | Let $t \in T$.
Suppose that:
:$t' \in t^{\succcurlyeq T}$
By definition of weak upper closure $t^{\succcurlyeq T}$, this is equivalent to:
:$t \preccurlyeq \restriction_T t'$
By definition of $\preccurlyeq \restriction_T$, this comes down to:
:$t \preccurlyeq t' \land t' \in T$
as it is assumed that $t \in T$.
The firs... | Let $\struct {S, \preccurlyeq}$ be an [[Definition:Ordered Set|ordered set]].
Let $T \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
Let $\preccurlyeq \restriction_T$ be the [[Definition:Restricted Ordering|restricted ordering]] on $T$.
Then for all $t \in T$:
:$t^{\succcurlyeq T} = T \cap t^{\succcurlyeq S... | Let $t \in T$.
Suppose that:
:$t' \in t^{\succcurlyeq T}$
By definition of [[Definition:Weak Upper Closure of Element|weak upper closure $t^{\succcurlyeq T}$]], this is equivalent to:
:$t \preccurlyeq \restriction_T t'$
By [[Definition:Restricted Ordering|definition of $\preccurlyeq \restriction_T$]], this comes do... | Weak Upper Closure in Restricted Ordering | https://proofwiki.org/wiki/Weak_Upper_Closure_in_Restricted_Ordering | https://proofwiki.org/wiki/Weak_Upper_Closure_in_Restricted_Ordering | [
"Upper Closures"
] | [
"Definition:Ordered Set",
"Definition:Subset",
"Definition:Restriction of Ordering",
"Definition:Upper Closure/Element",
"Definition:Upper Closure/Element"
] | [
"Definition:Upper Closure/Element",
"Definition:Restriction of Ordering",
"Definition:Set Intersection",
"Definition:Set Equality",
"Category:Upper Closures"
] |
proofwiki-5367 | Strict Lower Closure in Restricted Ordering | Let $\left({S, \preceq}\right)$ be an ordered set.
Let $T \subseteq S$ be a subset of $S$, and let $\preceq \restriction_T$ be the restricted ordering on $T$.
Then for all $t \in T$:
:$t^{\prec T} = T \cap t^{\prec S}$
where:
: $t^{\prec T}$ is the strict lower closure of $t$ in $\left({T, \preceq \restriction_T}\right... | Let $t \in T$, and suppose that $t' \in t^{\prec T}$.
By definition of strict lower closure, this is equivalent to:
:$t' \preceq \restriction_T t \land t \ne t'$
By definition of $\preceq \restriction_T$, the first condition comes down to:
:$t' \preceq t \land t' \in T$
as it is assumed that $t \in T$.
In conclusion, $... | Let $\left({S, \preceq}\right)$ be an [[Definition:Ordered Set|ordered set]].
Let $T \subseteq S$ be a [[Definition:Subset|subset]] of $S$, and let $\preceq \restriction_T$ be the [[Definition:Restricted Ordering|restricted ordering]] on $T$.
Then for all $t \in T$:
:$t^{\prec T} = T \cap t^{\prec S}$
where:
: $t^... | Let $t \in T$, and suppose that $t' \in t^{\prec T}$.
By definition of [[Definition:Strict Lower Closure of Element|strict lower closure]], this is equivalent to:
:$t' \preceq \restriction_T t \land t \ne t'$
By [[Definition:Restricted Ordering|definition of $\preceq \restriction_T$]], the first condition comes down... | Strict Lower Closure in Restricted Ordering | https://proofwiki.org/wiki/Strict_Lower_Closure_in_Restricted_Ordering | https://proofwiki.org/wiki/Strict_Lower_Closure_in_Restricted_Ordering | [
"Lower Closures"
] | [
"Definition:Ordered Set",
"Definition:Subset",
"Definition:Restriction of Ordering",
"Definition:Strict Lower Closure/Element",
"Definition:Strict Lower Closure/Element"
] | [
"Definition:Strict Lower Closure/Element",
"Definition:Restriction of Ordering",
"Definition:Set Intersection",
"Definition:Set Equality",
"Category:Lower Closures"
] |
proofwiki-5368 | Weak Lower Closure in Restricted Ordering | Let $\left({S, \preccurlyeq}\right)$ be an ordered set.
Let $T \subseteq S$ be a subset of $S$.
Let $\preccurlyeq \restriction_T$ be the restricted ordering on $T$.
Then for all $t \in T$:
:$t^{\preccurlyeq T} = T \cap t^{\preccurlyeq S}$
where:
:$t^{\preccurlyeq T}$ is the weak lower closure of $t$ in $\left({T, \prec... | Let $t \in T$, and suppose that $t' \in t^{\preccurlyeq T}$.
By definition of weak lower closure $t^{\preccurlyeq T}$, this is equivalent to:
:$t' \preccurlyeq \restriction_T t$
By definition of $\preccurlyeq \restriction_T$, this comes down to:
:$t' \preccurlyeq t \land t' \in T$
as it is assumed that $t \in T$.
The f... | Let $\left({S, \preccurlyeq}\right)$ be an [[Definition:Ordered Set|ordered set]].
Let $T \subseteq S$ be a [[Definition:Subset|subset]] of $S$.
Let $\preccurlyeq \restriction_T$ be the [[Definition:Restricted Ordering|restricted ordering]] on $T$.
Then for all $t \in T$:
:$t^{\preccurlyeq T} = T \cap t^{\preccurl... | Let $t \in T$, and suppose that $t' \in t^{\preccurlyeq T}$.
By definition of [[Definition:Weak Lower Closure of Element|weak lower closure $t^{\preccurlyeq T}$]], this is equivalent to:
:$t' \preccurlyeq \restriction_T t$
By [[Definition:Restricted Ordering|definition of $\preccurlyeq \restriction_T$]], this comes ... | Weak Lower Closure in Restricted Ordering | https://proofwiki.org/wiki/Weak_Lower_Closure_in_Restricted_Ordering | https://proofwiki.org/wiki/Weak_Lower_Closure_in_Restricted_Ordering | [
"Lower Closures"
] | [
"Definition:Ordered Set",
"Definition:Subset",
"Definition:Restriction of Ordering",
"Definition:Lower Closure/Element",
"Definition:Lower Closure/Element"
] | [
"Definition:Lower Closure/Element",
"Definition:Restriction of Ordering",
"Definition:Set Intersection",
"Definition:Set Equality",
"Category:Lower Closures"
] |
proofwiki-5369 | Order Topology on Natural Numbers is Discrete Topology | Let $\le$ be the standard ordering on the natural numbers $\N$.
Then the order topology $\tau$ on $\N$ is the discrete topology. | By Topology is Discrete iff All Singletons are Open, it suffices to show that for all $n \in \N$, the singleton $\set n$ is an open of $\tau$.
Now observe that $\map \downarrow 1 = \set 0$, since for all $n \in \N$, $n < 1 \implies n = 0$.
It follows that $\set 0$ is an open set of $\tau$.
Suppose now that $n \in \N$ a... | Let $\le$ be the standard ordering on the [[Definition:Natural Number|natural numbers]] $\N$.
Then the [[Definition:Order Topology|order topology]] $\tau$ on $\N$ is the [[Definition:Discrete Topology|discrete topology]]. | By [[Topology is Discrete iff All Singletons are Open]], it suffices to show that for all $n \in \N$, the [[Definition:Singleton|singleton]] $\set n$ is an [[Definition:Open Set (Topology)|open]] of $\tau$.
Now observe that $\map \downarrow 1 = \set 0$, since for all $n \in \N$, $n < 1 \implies n = 0$.
It follows th... | Order Topology on Natural Numbers is Discrete Topology | https://proofwiki.org/wiki/Order_Topology_on_Natural_Numbers_is_Discrete_Topology | https://proofwiki.org/wiki/Order_Topology_on_Natural_Numbers_is_Discrete_Topology | [
"Order Topologies",
"Discrete Topologies",
"Natural Numbers"
] | [
"Definition:Natural Numbers",
"Definition:Order Topology",
"Definition:Discrete Topology"
] | [
"Topology is Discrete iff All Singletons are Open",
"Definition:Singleton",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Definition:Open Set/Topology",
"Proof by Cases",
"Category:Order Topologies",
"Category:Discrete Topologies",
"Category:Natural Numbers"
] |
proofwiki-5370 | Natural Numbers under Multiplication form Ordered Commutative Semigroup | Let $\N$ be the natural numbers.
Let $\times$ be multiplication.
Let $\le$ be the ordering on $\N$.
Then $\struct {\N, \times, \le}$ is an ordered commutative semigroup. | By Natural Numbers under Multiplication form Semigroup, $\struct {\N, \times, \le}$ is a semigroup.
By Natural Number Multiplication is Commutative, $\times$ is commutative.
By Ordering on Natural Numbers is Compatible with Multiplication, $\le$ is compatible with $\times$.
The result follows.
{{qed}} | Let $\N$ be the [[Definition:Natural Numbers|natural numbers]].
Let $\times$ be [[Definition:Natural Number Multiplication|multiplication]].
Let $\le$ be the [[Definition:Ordering on Natural Numbers|ordering on $\N$]].
Then $\struct {\N, \times, \le}$ is an [[Definition:Ordered Commutative Semigroup|ordered commuta... | By [[Natural Numbers under Multiplication form Semigroup]], $\struct {\N, \times, \le}$ is a [[Definition:Semigroup|semigroup]].
By [[Natural Number Multiplication is Commutative]], $\times$ is [[Definition:Commutative Operation|commutative]].
By [[Ordering on Natural Numbers is Compatible with Multiplication]], $\le... | Natural Numbers under Multiplication form Ordered Commutative Semigroup | https://proofwiki.org/wiki/Natural_Numbers_under_Multiplication_form_Ordered_Commutative_Semigroup | https://proofwiki.org/wiki/Natural_Numbers_under_Multiplication_form_Ordered_Commutative_Semigroup | [
"Natural Number Multiplication",
"Examples of Commutative Semigroups",
"Examples of Ordered Semigroups"
] | [
"Definition:Natural Numbers",
"Definition:Multiplication/Natural Numbers",
"Definition:Ordering on Natural Numbers",
"Definition:Ordered Commutative Semigroup"
] | [
"Natural Numbers under Multiplication form Semigroup",
"Definition:Semigroup",
"Natural Number Multiplication is Commutative",
"Definition:Commutative/Operation",
"Ordering on Natural Numbers is Compatible with Multiplication",
"Definition:Relation Compatible with Operation"
] |
proofwiki-5371 | Invertible Elements under Natural Number Multiplication | Let $\N$ be the natural numbers.
Let $\times$ denote multiplication.
Then the only invertible element of $\N$ for $\times$ is $1$. | $m \in \N$ is invertible for $\times$.
Let $n \in \N: m \times n = 1$.
Then from Natural Numbers have No Proper Zero Divisors:
:$m \ne 0$ and $n \ne 0$
Thus, $1 \le m$ and $1 \le n$.
If $1 \le m$ then from Ordering on Natural Numbers is Compatible with Multiplication:
:$1 \le n < m \times n$
This contradicts $m \times ... | Let $\N$ be the [[Definition:Natural Numbers|natural numbers]].
Let $\times$ denote [[Definition:Natural Number Multiplication|multiplication]].
Then the only [[Definition:Invertible Element|invertible element]] of $\N$ for $\times$ is $1$. | $m \in \N$ is [[Definition:Invertible Element|invertible]] for $\times$.
Let $n \in \N: m \times n = 1$.
Then from [[Natural Numbers have No Proper Zero Divisors]]:
:$m \ne 0$ and $n \ne 0$
Thus, $1 \le m$ and $1 \le n$.
If $1 \le m$ then from [[Ordering on Natural Numbers is Compatible with Multiplication]]:
:$1 \... | Invertible Elements under Natural Number Multiplication | https://proofwiki.org/wiki/Invertible_Elements_under_Natural_Number_Multiplication | https://proofwiki.org/wiki/Invertible_Elements_under_Natural_Number_Multiplication | [
"Natural Number Multiplication"
] | [
"Definition:Natural Numbers",
"Definition:Multiplication/Natural Numbers",
"Definition:Invertible Element"
] | [
"Definition:Invertible Element",
"Natural Numbers have No Proper Zero Divisors",
"Ordering on Natural Numbers is Compatible with Multiplication"
] |
proofwiki-5372 | Graph containing Closed Walk of Odd Length also contains Odd Cycle | Let $G$ be a graph.
{{explain|This proof works for a simple graph, but the theorem may hold for loop graphs and/or multigraphs. Clarification needed as to what applies.}}
Let $G$ have a closed walk of odd length.
Then $G$ has an odd cycle. | Let $G = \struct {V, E}$ be a graph with closed walk whose length is odd.
From Closed Walk of Odd Length contains Odd Circuit, such a walk contains a circuit whose length is odd.
Let $C_1 = \tuple {v_1, \ldots, v_{2 n + 1} = v_1}$ be such a circuit.
{{AimForCont}} $G$ has no odd cycles.
Then $C_1$ is not a cycle.
Hence... | Let $G$ be a [[Definition:Graph (Graph Theory)|graph]].
{{explain|This proof works for a [[Definition:Simple Graph|simple graph]], but the theorem may hold for loop graphs and/or multigraphs. Clarification needed as to what applies.}}
Let $G$ have a [[Definition:Closed Walk|closed walk]] of [[Definition:Odd Integer|od... | Let $G = \struct {V, E}$ be a [[Definition:Graph (Graph Theory)|graph]] with [[Definition:Closed Walk|closed walk]] whose [[Definition:Length of Walk|length]] is [[Definition:Odd Integer|odd]].
From [[Closed Walk of Odd Length contains Odd Circuit]], such a walk contains a [[Definition:Circuit (Graph Theory)|circuit]]... | Graph containing Closed Walk of Odd Length also contains Odd Cycle | https://proofwiki.org/wiki/Graph_containing_Closed_Walk_of_Odd_Length_also_contains_Odd_Cycle | https://proofwiki.org/wiki/Graph_containing_Closed_Walk_of_Odd_Length_also_contains_Odd_Cycle | [
"Graph Theory"
] | [
"Definition:Graph (Graph Theory)",
"Definition:Simple Graph",
"Definition:Walk (Graph Theory)/Closed",
"Definition:Odd Integer",
"Definition:Walk (Graph Theory)/Length",
"Definition:Cycle (Graph Theory)/Odd"
] | [
"Definition:Graph (Graph Theory)",
"Definition:Walk (Graph Theory)/Closed",
"Definition:Walk (Graph Theory)/Length",
"Definition:Odd Integer",
"Closed Walk of Odd Length contains Odd Circuit",
"Definition:Circuit (Graph Theory)",
"Definition:Walk (Graph Theory)/Length",
"Definition:Odd Integer",
"De... |
proofwiki-5373 | Homomorphism of Powers/Naturally Ordered Semigroup | Let $\struct {S, \circ, \preceq}$ be a naturally ordered semigroup.
For a given $a \in T_1$, let $\map {\odot^n} a$ be the $n$th power of $a$ in $T_1$.
For a given $a \in T_2$, let $\map {\oplus^n} a$ be the $n$th power of $a$ in $T_2$.
Then:
:$\forall a \in T_1: \forall n \in \struct {S^*, \circ, \preceq}: \map \phi {... | The proof proceeds by the Principle of Mathematical Induction for a Naturally Ordered Semigroup.
Let $A := \set {n \in S^*: \forall a \in T_1: \map \phi {\map {\odot^n} a} = \map {\oplus^n} {\map \phi a} }$
That is, $A$ is defined as the set of all $n$ such that:
:$\forall a \in T_1 \map \phi {\map {\odot^n} a} = \map ... | Let $\struct {S, \circ, \preceq}$ be a [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]].
For a given $a \in T_1$, let $\map {\odot^n} a$ be the [[Definition:Power of Element of Magma|$n$th power of $a$]] in $T_1$.
For a given $a \in T_2$, let $\map {\oplus^n} a$ be the [[Definition:Power of Elem... | The proof proceeds by the [[Principle of Mathematical Induction for Naturally Ordered Semigroup|Principle of Mathematical Induction for a Naturally Ordered Semigroup]].
Let $A := \set {n \in S^*: \forall a \in T_1: \map \phi {\map {\odot^n} a} = \map {\oplus^n} {\map \phi a} }$
That is, $A$ is defined as the set of ... | Homomorphism of Powers/Naturally Ordered Semigroup | https://proofwiki.org/wiki/Homomorphism_of_Powers/Naturally_Ordered_Semigroup | https://proofwiki.org/wiki/Homomorphism_of_Powers/Naturally_Ordered_Semigroup | [
"Homomorphism of Powers",
"Naturally Ordered Semigroup"
] | [
"Definition:Naturally Ordered Semigroup",
"Definition:Power of Element/Magma",
"Definition:Power of Element/Magma"
] | [
"Principle of Mathematical Induction/Naturally Ordered Semigroup",
"Principle of Mathematical Induction/Naturally Ordered Semigroup"
] |
proofwiki-5374 | Homomorphism of Powers/Natural Numbers | Let $n \in \N$.
Let $\odot^n$ and $\oplus^n$ be the $n$th powers of $\odot$ and $\oplus$, respectively.
Then:
:$\forall a \in T_1: \forall n \in \N: \map \phi {\map {\odot^n} a} = \map {\oplus^n} {\map \phi a}$ | Consider the natural numbers $\N$ defined as a naturally ordered semigroup.
Then the result follows from Homomorphism of Powers: Naturally Ordered Semigroup.
{{qed}}
Category:Homomorphism of Powers
Category:Natural Numbers
lxnxcsdzoy5a8ayu5kvqqyojyw99bn8 | Let $n \in \N$.
Let $\odot^n$ and $\oplus^n$ be the [[Definition:Power of Element of Semigroup|$n$th powers]] of $\odot$ and $\oplus$, respectively.
Then:
:$\forall a \in T_1: \forall n \in \N: \map \phi {\map {\odot^n} a} = \map {\oplus^n} {\map \phi a}$ | Consider the [[Definition:Natural Numbers|natural numbers]] $\N$ defined as a [[Definition:Naturally Ordered Semigroup|naturally ordered semigroup]].
Then the result follows from [[Homomorphism of Powers/Naturally Ordered Semigroup|Homomorphism of Powers: Naturally Ordered Semigroup]].
{{qed}}
[[Category:Homomorphism... | Homomorphism of Powers/Natural Numbers | https://proofwiki.org/wiki/Homomorphism_of_Powers/Natural_Numbers | https://proofwiki.org/wiki/Homomorphism_of_Powers/Natural_Numbers | [
"Homomorphism of Powers",
"Natural Numbers"
] | [
"Definition:Power of Element/Semigroup"
] | [
"Definition:Natural Numbers",
"Definition:Naturally Ordered Semigroup",
"Homomorphism of Powers/Naturally Ordered Semigroup",
"Category:Homomorphism of Powers",
"Category:Natural Numbers"
] |
proofwiki-5375 | Homomorphism of Powers/Integers | Let $\struct {T_1, \odot}$ and $\struct {T_2, \oplus}$ be monoids.
Let $\phi: \struct {T_1, \odot} \to \struct {T_2, \oplus}$ be a (semigroup) homomorphism.
Let $a$ be an invertible element of $T_1$.
Let $n \in \Z$.
Let $\odot^n$ and $\oplus^n$ be as defined as in Index Laws for Monoids.
Then:
:$\forall n \in \Z: \map ... | By Homomorphism of Powers: Natural Numbers, we need show this only for negative $n$, that is:
:$\forall n \in \N^*: \map \phi {\map {\odot^{-n} } a} = \map {\oplus^{-n} } {\map \phi a}$
But by Homomorphism with Identity Preserves Inverses:
:$\map \phi {a^{-1} } = \paren {\map \phi a}^{-1}$
Hence by Homomorphism of Powe... | Let $\struct {T_1, \odot}$ and $\struct {T_2, \oplus}$ be [[Definition:Monoid|monoids]].
Let $\phi: \struct {T_1, \odot} \to \struct {T_2, \oplus}$ be a [[Definition:Semigroup Homomorphism|(semigroup) homomorphism]].
Let $a$ be an [[Definition:Invertible Element|invertible element]] of $T_1$.
Let $n \in \Z$.
Let $\... | By [[Homomorphism of Powers/Natural Numbers|Homomorphism of Powers: Natural Numbers]], we need show this only for negative $n$, that is:
:$\forall n \in \N^*: \map \phi {\map {\odot^{-n} } a} = \map {\oplus^{-n} } {\map \phi a}$
But by [[Homomorphism with Identity Preserves Inverses]]:
:$\map \phi {a^{-1} } = \paren ... | Homomorphism of Powers/Integers | https://proofwiki.org/wiki/Homomorphism_of_Powers/Integers | https://proofwiki.org/wiki/Homomorphism_of_Powers/Integers | [
"Homomorphism of Powers",
"Integers"
] | [
"Definition:Monoid",
"Definition:Semigroup Homomorphism",
"Definition:Invertible Element",
"Index Laws for Monoids"
] | [
"Homomorphism of Powers/Natural Numbers",
"Homomorphism with Identity Preserves Inverses",
"Homomorphism of Powers/Natural Numbers"
] |
proofwiki-5376 | Right Operation is Left Distributive over All Operations | Let $\struct {S, \circ, \rightarrow}$ be an algebraic structure where:
:$\rightarrow$ is the right operation
:$\circ$ is any arbitrary binary operation.
Then $\rightarrow$ is left distributive over $\circ$. | By definition of the right operation:
{{begin-eqn}}
{{eqn | l = a \rightarrow \paren {b \circ c}
| r = b \circ c
| c =
}}
{{eqn | r = \paren {a \rightarrow b} \circ \paren {a \rightarrow c}
| c =
}}
{{end-eqn}}
The result follows by definition of left distributivity.
{{qed}} | Let $\struct {S, \circ, \rightarrow}$ be an [[Definition:Algebraic Structure|algebraic structure]] where:
:$\rightarrow$ is the [[Definition:Right Operation|right operation]]
:$\circ$ is any arbitrary [[Definition:Binary Operation|binary operation]].
Then $\rightarrow$ is [[Definition:Left Distributive Operation|left ... | By definition of the [[Definition:Right Operation|right operation]]:
{{begin-eqn}}
{{eqn | l = a \rightarrow \paren {b \circ c}
| r = b \circ c
| c =
}}
{{eqn | r = \paren {a \rightarrow b} \circ \paren {a \rightarrow c}
| c =
}}
{{end-eqn}}
The result follows by definition of [[Definition:Left Di... | Right Operation is Left Distributive over All Operations | https://proofwiki.org/wiki/Right_Operation_is_Left_Distributive_over_All_Operations | https://proofwiki.org/wiki/Right_Operation_is_Left_Distributive_over_All_Operations | [
"Right Operation"
] | [
"Definition:Algebraic Structure",
"Definition:Right Operation",
"Definition:Operation/Binary Operation",
"Definition:Distributive Operation/Left"
] | [
"Definition:Right Operation",
"Definition:Distributive Operation/Left"
] |
proofwiki-5377 | Left Operation is Right Distributive over All Operations | Let $\struct {S, \circ, \leftarrow}$ be an algebraic structure where:
:$\leftarrow$ is the left operation
:$\circ$ is any arbitrary binary operation.
Then $\leftarrow$ is right distributive over $\circ$. | {{begin-eqn}}
{{eqn | l = \forall a, b, c \in S: \paren {a \circ b} \leftarrow c
| r = a \circ b
| c = {{Defof|Left Operation}}
}}
{{eqn | r = \paren {a \leftarrow c} \circ \paren {b \leftarrow c}
| c = {{Defof|Left Operation}}
}}
{{end-eqn}}
The result follows by definition of right distributivity.
{... | Let $\struct {S, \circ, \leftarrow}$ be an [[Definition:Algebraic Structure|algebraic structure]] where:
:$\leftarrow$ is the [[Definition:Left Operation|left operation]]
:$\circ$ is any arbitrary [[Definition:Binary Operation|binary operation]].
Then $\leftarrow$ is [[Definition:Right Distributive Operation|right dis... | {{begin-eqn}}
{{eqn | l = \forall a, b, c \in S: \paren {a \circ b} \leftarrow c
| r = a \circ b
| c = {{Defof|Left Operation}}
}}
{{eqn | r = \paren {a \leftarrow c} \circ \paren {b \leftarrow c}
| c = {{Defof|Left Operation}}
}}
{{end-eqn}}
The result follows by definition of [[Definition:Right Dis... | Left Operation is Right Distributive over All Operations | https://proofwiki.org/wiki/Left_Operation_is_Right_Distributive_over_All_Operations | https://proofwiki.org/wiki/Left_Operation_is_Right_Distributive_over_All_Operations | [
"Left Operation"
] | [
"Definition:Algebraic Structure",
"Definition:Left Operation",
"Definition:Operation/Binary Operation",
"Definition:Distributive Operation/Right"
] | [
"Definition:Distributive Operation/Right"
] |
proofwiki-5378 | Right Operation is Distributive over Idempotent Operation | Let $\struct {S, \circ, \rightarrow}$ be an algebraic structure where:
:$\rightarrow$ is the right operation
:$\circ$ is any arbitrary binary operation.
Then:
:$\rightarrow$ is distributive over $\circ$
{{iff}}
:$\circ$ is idempotent. | From Right Operation is Left Distributive over All Operations:
:$\forall a, b, c \in S: a \rightarrow \paren {b \circ c} = \paren {a \rightarrow b} \circ \paren {a \rightarrow c}$
for all binary operations $\circ$.
It remains to show that $\rightarrow$ is right distributive over $\circ$ {{iff}} $\circ$ is idempotent. | Let $\struct {S, \circ, \rightarrow}$ be an [[Definition:Algebraic Structure|algebraic structure]] where:
:$\rightarrow$ is the [[Definition:Right Operation|right operation]]
:$\circ$ is any arbitrary [[Definition:Binary Operation|binary operation]].
Then:
:$\rightarrow$ is [[Definition:Distributive Operation|distrib... | From [[Right Operation is Left Distributive over All Operations]]:
:$\forall a, b, c \in S: a \rightarrow \paren {b \circ c} = \paren {a \rightarrow b} \circ \paren {a \rightarrow c}$
for all [[Definition:Binary Operation|binary operations]] $\circ$.
It remains to show that $\rightarrow$ is [[Definition:Right Distrib... | Right Operation is Distributive over Idempotent Operation | https://proofwiki.org/wiki/Right_Operation_is_Distributive_over_Idempotent_Operation | https://proofwiki.org/wiki/Right_Operation_is_Distributive_over_Idempotent_Operation | [
"Right Operation",
"Distributive Operations",
"Idempotence"
] | [
"Definition:Algebraic Structure",
"Definition:Right Operation",
"Definition:Operation/Binary Operation",
"Definition:Distributive Operation",
"Definition:Idempotence/Operation"
] | [
"Right Operation is Left Distributive over All Operations",
"Definition:Operation/Binary Operation",
"Definition:Distributive Operation/Right",
"Definition:Idempotence/Operation",
"Definition:Idempotence/Operation",
"Definition:Distributive Operation/Right",
"Definition:Distributive Operation/Right",
... |
proofwiki-5379 | Left Operation is Distributive over Idempotent Operation | Let $\struct {S, \circ, \leftarrow}$ be an algebraic structure where:
:$\leftarrow$ is the left operation
:$\circ$ is any arbitrary binary operation.
Then:
:$\leftarrow$ is distributive over $\circ$
{{iff}}:
:$\circ$ is idempotent. | From Left Operation is Right Distributive over All Operations:
:$\forall a, b, c \in S: \paren {a \circ b} \leftarrow c = \paren {a \leftarrow c} \circ \paren {b \leftarrow c}$
for all binary operations $\circ$.
It remains to show that $\leftarrow$ is left distributive over $\circ$ {{iff}} $\circ$ is idempotent. | Let $\struct {S, \circ, \leftarrow}$ be an [[Definition:Algebraic Structure|algebraic structure]] where:
:$\leftarrow$ is the [[Definition:Left Operation|left operation]]
:$\circ$ is any arbitrary [[Definition:Binary Operation|binary operation]].
Then:
:$\leftarrow$ is [[Definition:Distributive Operation|distributive... | From [[Left Operation is Right Distributive over All Operations]]:
:$\forall a, b, c \in S: \paren {a \circ b} \leftarrow c = \paren {a \leftarrow c} \circ \paren {b \leftarrow c}$
for all [[Definition:Binary Operation|binary operations]] $\circ$.
It remains to show that $\leftarrow$ is [[Definition:Left Distributive... | Left Operation is Distributive over Idempotent Operation | https://proofwiki.org/wiki/Left_Operation_is_Distributive_over_Idempotent_Operation | https://proofwiki.org/wiki/Left_Operation_is_Distributive_over_Idempotent_Operation | [
"Left Operation",
"Distributive Operations",
"Idempotence"
] | [
"Definition:Algebraic Structure",
"Definition:Left Operation",
"Definition:Operation/Binary Operation",
"Definition:Distributive Operation",
"Definition:Idempotence/Operation"
] | [
"Left Operation is Right Distributive over All Operations",
"Definition:Operation/Binary Operation",
"Definition:Distributive Operation/Left",
"Definition:Idempotence/Operation",
"Definition:Idempotence/Operation",
"Definition:Distributive Operation/Left",
"Definition:Distributive Operation/Left",
"De... |
proofwiki-5380 | Integral of Positive Measurable Function as Limit of Integrals of Positive Simple Functions | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \overline \R$ be a positive $\Sigma$-measurable function.
For each $n \in \N$, let $f_n : X \to \R$ be a positive simple function, such that:
:$\ds \lim_{n \mathop \to \infty} f_n = f$
and:
:for each $x \in X$, the sequence $\sequence {\map {f_n} x}_{n \m... | Let $\EE^+$ be the space of positive simple functions.
Note that since:
:for each $x \in X$, the sequence $\sequence {\map {f_n} x}$ is increasing
we have that:
:$f_i \le f_j$
whenever $i \le j$.
Since $f_n \to f$, from Monotone Convergence Theorem (Real Analysis): Increasing Sequence, we further obtain:
:$f_i \le... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $f: X \to \overline \R$ be a [[Definition:Positive Measurable Function|positive $\Sigma$-measurable function]].
For each $n \in \N$, let $f_n : X \to \R$ be a [[Definition:Positive Simple Function|positive simple function]], such that... | Let $\EE^+$ be the [[Definition:Space of Positive Simple Functions|space of positive simple functions]].
Note that since:
:for each $x \in X$, the [[Definition:Sequence|sequence]] $\sequence {\map {f_n} x}$ is [[Definition:Increasing Sequence|increasing]]
we have that:
:$f_i \le f_j$
whenever $i \le j$.
Since... | Integral of Positive Measurable Function as Limit of Integrals of Positive Simple Functions | https://proofwiki.org/wiki/Integral_of_Positive_Measurable_Function_as_Limit_of_Integrals_of_Positive_Simple_Functions | https://proofwiki.org/wiki/Integral_of_Positive_Measurable_Function_as_Limit_of_Integrals_of_Positive_Simple_Functions | [
"Integral of Positive Measurable Function"
] | [
"Definition:Measure Space",
"Definition:Measurable Function/Positive",
"Definition:Simple Function",
"Definition:Sequence",
"Definition:Increasing/Sequence",
"Definition:Pointwise Limit",
"Definition:Integral Sign",
"Definition:Integral of Positive Measurable Function"
] | [
"Definition:Space of Simple Functions",
"Definition:Sequence",
"Definition:Increasing/Sequence",
"Monotone Convergence Theorem (Real Analysis)/Increasing Sequence",
"Integral of Positive Simple Function is Increasing",
"Definition:Sequence",
"Definition:Increasing/Sequence",
"Definition:Bounded Sequen... |
proofwiki-5381 | Integral of Characteristic Function/Corollary | :$\ds \int \chi_E \rd \mu = \map \mu E$
where the integral sign denotes the $\mu$-integral of $\chi_E$. | By Integral of Characteristic Function, have:
:$\map {I_\mu} {\chi_E} = \map \mu E$
where $\map {I_\mu} {\chi_E}$ is the $\mu$-integral of $\chi_E$.
From Integral of Positive Measurable Function Extends Integral of Positive Simple Function, it also holds that:
:$\ds \int \chi_E \rd \mu = \map {I_\mu} {\chi_E}$
Combinin... | :$\ds \int \chi_E \rd \mu = \map \mu E$
where the [[Definition:Integral Sign|integral sign]] denotes the [[Definition:Integral of Positive Measurable Function|$\mu$-integral of $\chi_E$]]. | By [[Integral of Characteristic Function]], have:
:$\map {I_\mu} {\chi_E} = \map \mu E$
where $\map {I_\mu} {\chi_E}$ is the [[Definition:Integral of Positive Simple Function|$\mu$-integral of $\chi_E$]].
From [[Integral of Positive Measurable Function Extends Integral of Positive Simple Function]], it also holds th... | Integral of Characteristic Function/Corollary | https://proofwiki.org/wiki/Integral_of_Characteristic_Function/Corollary | https://proofwiki.org/wiki/Integral_of_Characteristic_Function/Corollary | [
"Integral of Characteristic Function",
"Integral of Positive Measurable Function"
] | [
"Definition:Integral Sign",
"Definition:Integral of Positive Measurable Function"
] | [
"Integral of Characteristic Function",
"Definition:Integral of Positive Simple Function",
"Integral of Positive Measurable Function Extends Integral of Positive Simple Function"
] |
proofwiki-5382 | Integral of Positive Measurable Function is Positive Homogeneous | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f : X \to \overline \R$ be a positive $\Sigma$-measurable function.
Let $\lambda \in \overline \R$ be an extended real number with $\lambda \ge 0$.
Then:
:$\ds \int \lambda f \rd \mu = \lambda \int f \rd \mu$
where:
:$\lambda f$ is the pointwise $\lambda$-multipl... | Suppose that $\lambda < \infty$.
From Measurable Function is Pointwise Limit of Simple Functions, there exists an increasing sequence $\sequence {f_n}_{n \mathop \in \N}$ of positive simple functions such that:
:$\ds \map f x = \lim_{n \mathop \to \infty} \map {f_n} x$
From the Multiple Rule for Real Sequences, we ha... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $f : X \to \overline \R$ be a [[Definition:Positive Measurable Function|positive $\Sigma$-measurable function]].
Let $\lambda \in \overline \R$ be an [[Definition:Extended Real Number Line|extended real number]] with $\lambda \ge 0$. ... | Suppose that $\lambda < \infty$.
From [[Measurable Function is Pointwise Limit of Simple Functions]], there exists an [[Definition:Increasing Sequence of Real-Valued Functions|increasing sequence]] $\sequence {f_n}_{n \mathop \in \N}$ of [[Definition:Positive Simple Function|positive simple functions]] such that:
:... | Integral of Positive Measurable Function is Positive Homogeneous | https://proofwiki.org/wiki/Integral_of_Positive_Measurable_Function_is_Positive_Homogeneous | https://proofwiki.org/wiki/Integral_of_Positive_Measurable_Function_is_Positive_Homogeneous | [
"Integral of Positive Measurable Function",
"Integral of Positive Measurable Function is Positive Homogeneous"
] | [
"Definition:Measure Space",
"Definition:Measurable Function/Positive",
"Definition:Extended Real Number Line",
"Definition:Pointwise Scalar Multiplication of Extended Real-Valued Functions",
"Definition:Integral Sign",
"Definition:Integral of Positive Measurable Function",
"Definition:Positive Homogeneo... | [
"Measurable Function is Pointwise Limit of Simple Functions",
"Definition:Increasing Sequence of Real-Valued Functions",
"Definition:Simple Function",
"Combination Theorem for Sequences/Real/Multiple Rule",
"Integral of Positive Measurable Function as Limit of Integrals of Positive Simple Functions",
"Int... |
proofwiki-5383 | Linear Transformation as Matrix Product | Let $T: \R^n \to \R^m, \mathbf x \mapsto \map T {\mathbf x}$ be a linear transformation.
Then:
:$\map T {\mathbf x} = \mathbf A_T \mathbf x$
where $\mathbf A_T$ is the $m \times n$ matrix defined as:
:$\mathbf A_T = \begin {bmatrix} \map T {\mathbf e_1} & \map T {\mathbf e_2} & \cdots & \map T {\mathbf e_n} \end {bmatr... | Let $\mathbf x = \begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix}$.
Let $\mathbf I_n$ be the unit matrix of order $n$.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf x_{n \times 1}
| r = \mathbf I_n \mathbf x_{n \times 1}
| c = {{Defof|Left Identity}}
}}
{{eqn | r = \begin {bmatrix} 1 & 0 & \cdots & 0 \... | Let $T: \R^n \to \R^m, \mathbf x \mapsto \map T {\mathbf x}$ be a [[Definition:Linear Transformation on Vector Space|linear transformation]].
Then:
:$\map T {\mathbf x} = \mathbf A_T \mathbf x$
where $\mathbf A_T$ is the [[Definition:Matrix|$m \times n$ matrix]] defined as:
:$\mathbf A_T = \begin {bmatrix} \map T {\... | Let $\mathbf x = \begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix}$.
Let $\mathbf I_n$ be the [[Definition:Unit Matrix|unit matrix of order $n$]].
Then:
{{begin-eqn}}
{{eqn | l = \mathbf x_{n \times 1}
| r = \mathbf I_n \mathbf x_{n \times 1}
| c = {{Defof|Left Identity}}
}}
{{eqn | r = \begin {... | Linear Transformation as Matrix Product | https://proofwiki.org/wiki/Linear_Transformation_as_Matrix_Product | https://proofwiki.org/wiki/Linear_Transformation_as_Matrix_Product | [
"Linear Transformations"
] | [
"Definition:Linear Transformation/Vector Space",
"Definition:Matrix",
"Definition:Standard Ordered Basis/Vector Space"
] | [
"Definition:Unit Matrix",
"Unit Matrix is Unity of Ring of Square Matrices",
"Definition:Element",
"Definition:Matrix/Row"
] |
proofwiki-5384 | Integral of Positive Measurable Function is Additive | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f : X \to \overline \R$ and $g : X \to \overline \R$ be positive $\Sigma$-measurable functions.
Then:
:$\ds \int \paren {f + g} \rd \mu = \int f \rd \mu + \int g \rd \mu$
where:
:$f + g$ is the pointwise sum of $f$ and $g$
:the integral sign denotes $\mu$-integrat... | We are given that $f : X \to \overline \R$ and $g : X \to \overline \R$ is a positive $\Sigma$-measurable functions, which is {{afortiori}} a measurable function, so we can apply Measurable Function is Pointwise Limit of Simple Functions.
From Measurable Function is Pointwise Limit of Simple Functions, there exists an ... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $f : X \to \overline \R$ and $g : X \to \overline \R$ be [[Definition:Positive Measurable Function|positive $\Sigma$-measurable functions]].
Then:
:$\ds \int \paren {f + g} \rd \mu = \int f \rd \mu + \int g \rd \mu$
where:
:$f + g$... | We are [[Definition:Given|given]] that $f : X \to \overline \R$ and $g : X \to \overline \R$ is a [[Definition:Positive Measurable Function|positive $\Sigma$-measurable functions]], which is {{afortiori}} a [[Definition:Measurable Function|measurable function]], so we can apply [[Measurable Function is Pointwise Limit ... | Integral of Positive Measurable Function is Additive | https://proofwiki.org/wiki/Integral_of_Positive_Measurable_Function_is_Additive | https://proofwiki.org/wiki/Integral_of_Positive_Measurable_Function_is_Additive | [
"Integral of Positive Measurable Function",
"Integral of Positive Measurable Function is Additive"
] | [
"Definition:Measure Space",
"Definition:Measurable Function/Positive",
"Definition:Pointwise Addition",
"Definition:Integral Sign",
"Definition:Integral of Positive Measurable Function",
"Definition:Additive Function (Algebra)"
] | [
"Definition:Given",
"Definition:Measurable Function/Positive",
"Definition:Measurable Function",
"Measurable Function is Pointwise Limit of Simple Functions",
"Measurable Function is Pointwise Limit of Simple Functions",
"Definition:Increasing Sequence of Real-Valued Functions",
"Definition:Simple Funct... |
proofwiki-5385 | Integral of Positive Simple Function is Increasing | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g: X \to \R$, $f, g \in \EE^+$ be positive simple functions.
Suppose that:
: $f \le g$
where $\le$ denotes pointwise inequality.
Then:
:$\map {I_\mu} f \le \map {I_\mu} g$
where $I_\mu$ denotes $\mu$-integration
This can be summarized by saying that $I_\mu$ is ... | Note that:
:$g - f \ge 0$
From Scalar Multiple of Simple Function is Simple Function and Pointwise Sum of Simple Functions is Simple Function, we then have that:
:$g - f \in \EE^+$
Write:
:$g = f + \paren {g - f}$
From Integral of Positive Simple Function is Additive, we then have:
:$\map {I_\mu} g = \map {I_\mu... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $f, g: X \to \R$, $f, g \in \EE^+$ be [[Definition:Positive Simple Function|positive simple functions]].
Suppose that:
: $f \le g$
where $\le$ denotes [[Definition:Pointwise Inequality of Real-Valued Functions|pointwise inequality]].... | Note that:
:$g - f \ge 0$
From [[Scalar Multiple of Simple Function is Simple Function]] and [[Pointwise Sum of Simple Functions is Simple Function]], we then have that:
:$g - f \in \EE^+$
Write:
:$g = f + \paren {g - f}$
From [[Integral of Positive Simple Function is Additive]], we then have:
:$\map {I_\... | Integral of Positive Simple Function is Increasing | https://proofwiki.org/wiki/Integral_of_Positive_Simple_Function_is_Increasing | https://proofwiki.org/wiki/Integral_of_Positive_Simple_Function_is_Increasing | [
"Integral of Positive Simple Function"
] | [
"Definition:Measure Space",
"Definition:Simple Function",
"Definition:Pointwise Inequality of Real-Valued Functions",
"Definition:Integral of Positive Simple Function",
"Definition:Increasing/Mapping"
] | [
"Scalar Multiple of Simple Function is Simple Function",
"Pointwise Sum of Simple Functions is Simple Function",
"Integral of Positive Simple Function is Additive"
] |
proofwiki-5386 | Integral of Positive Measurable Function is Monotone | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g: X \to \overline \R$ be positive $\Sigma$-measurable functions.
Suppose that $f \le g$, where $\le$ denotes pointwise inequality.
Then:
:$\ds \int f \rd \mu \le \int g \rd \mu$
where the integral sign denotes $\mu$-integration.
This can be summarized by sayin... | By the definition of $\mu$-integration, we have:
:$\ds \int f \rd \mu = \sup \set {\map {I_\mu} h: h \le f, h \in \EE^+}$
and:
:$\ds \int g \rd \mu = \sup \set {\map {I_\mu} h : h \le g, h \in \EE^+}$
where:
:$\EE^+$ denotes the space of positive simple functions
:$\map {I_\mu} g$ denotes the $\mu$-integral of the pos... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $f, g: X \to \overline \R$ be [[Definition:Positive Measurable Function|positive $\Sigma$-measurable functions]].
Suppose that $f \le g$, where $\le$ denotes [[Definition:Pointwise Inequality of Extended Real-Valued Functions|pointwi... | By the definition of [[Definition:Integral of Positive Measurable Function|$\mu$-integration]], we have:
:$\ds \int f \rd \mu = \sup \set {\map {I_\mu} h: h \le f, h \in \EE^+}$
and:
:$\ds \int g \rd \mu = \sup \set {\map {I_\mu} h : h \le g, h \in \EE^+}$
where:
:$\EE^+$ denotes the [[Definition:Space of Positive... | Integral of Positive Measurable Function is Monotone | https://proofwiki.org/wiki/Integral_of_Positive_Measurable_Function_is_Monotone | https://proofwiki.org/wiki/Integral_of_Positive_Measurable_Function_is_Monotone | [
"Integral of Positive Measurable Function",
"Integral of Positive Measurable Function is Monotone"
] | [
"Definition:Measure Space",
"Definition:Measurable Function/Positive",
"Definition:Pointwise Inequality of Extended Real-Valued Functions",
"Definition:Integral Sign",
"Definition:Integral of Positive Measurable Function",
"Definition:Monotone (Order Theory)/Mapping"
] | [
"Definition:Integral of Positive Measurable Function",
"Definition:Space of Simple Functions",
"Definition:Integral of Positive Simple Function",
"Supremum of Subset"
] |
proofwiki-5387 | Matrix Multiplication is Homogeneous of Degree 1 | Let $\mathbf A$ be an $m \times n$ matrix and $\mathbf B$ be an $n \times p$ matrix.
Let $\lambda$ be a scalar in a field $\mathbb F$.
Then, the multiplication of these matrices is homogenous of degree one:
:$\mathbf A \paren {\lambda \mathbf B} = \lambda \paren {\mathbf A \mathbf B}$ | Let $\mathbf A = \sqbrk a_{m n}, \mathbf B = \sqbrk b_{n p}$.
Let $i \in \closedint 1 m$.
Let $j \in \closedint 1 p$.
{{begin-eqn}}
{{eqn | l = \left[ \lambda \mathbf A \mathbf B \right]_{i j}
| r = \lambda \sum_{k \mathop = 1}^n a_{i k} b_{k j}
| c = {{Defof|Matrix Product (Conventional)}}
}}
{{eqn | r = \... | Let $\mathbf A$ be an [[Definition:Matrix|$m \times n$ matrix]] and $\mathbf B$ be an [[Definition:Matrix|$n \times p$ matrix]].
Let $\lambda$ be a [[Definition:Scalar (Matrix Theory)|scalar]] in a [[Definition:Field (Abstract Algebra)|field]] $\mathbb F$.
Then, the [[Definition:Matrix Multiplication|multiplication]]... | Let $\mathbf A = \sqbrk a_{m n}, \mathbf B = \sqbrk b_{n p}$.
Let $i \in \closedint 1 m$.
Let $j \in \closedint 1 p$.
{{begin-eqn}}
{{eqn | l = \left[ \lambda \mathbf A \mathbf B \right]_{i j}
| r = \lambda \sum_{k \mathop = 1}^n a_{i k} b_{k j}
| c = {{Defof|Matrix Product (Conventional)}}
}}
{{eqn | r ... | Matrix Multiplication is Homogeneous of Degree 1 | https://proofwiki.org/wiki/Matrix_Multiplication_is_Homogeneous_of_Degree_1 | https://proofwiki.org/wiki/Matrix_Multiplication_is_Homogeneous_of_Degree_1 | [
"Conventional Matrix Multiplication"
] | [
"Definition:Matrix",
"Definition:Matrix",
"Definition:Scalar (Matrix Theory)",
"Definition:Field (Abstract Algebra)",
"Definition:Matrix Product",
"Definition:Matrix",
"Definition:Homogeneous Function",
"Definition:Homogeneous Function/Degree"
] | [
"Category:Conventional Matrix Multiplication"
] |
proofwiki-5388 | Integral of Series of Positive Measurable Functions | Let $\struct {X, \Sigma}$ be a measurable space.
Let $\sequence {f_n}_{n \mathop \in \N} \in \MM_{\overline \R}^+$, $f_n: X \to \overline \R$ be a sequence of positive measurable functions.
Let $\ds \sum_{n \mathop \in \N} f_n: X \to \overline \R$ be the pointwise series of the $f_n$.
Then:
:$\ds \int \sum_{n \mathop \... | Define the sequence $\sequence {g_N}_{n \mathop \in \N}$ of functions $g_N : X \to \overline \R$ by:
:$\ds \map {g_N} x = \sum_{n \mathop = 1}^N \map {f_n} x$
Since $f_n \ge 0$ for each $n$, we have:
{{begin-eqn}}
{{eqn | l = \map {g_{N + 1} } x
| r = \sum_{n \mathop = 1}^{N + 1} \map {f_n} x
}}
{{eqn | r = \sum_{... | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $\sequence {f_n}_{n \mathop \in \N} \in \MM_{\overline \R}^+$, $f_n: X \to \overline \R$ be a [[Definition:Sequence|sequence]] of [[Definition:Positive Measurable Function|positive measurable functions]].
Let $\ds \sum_{n \mathop \in... | Define the [[Definition:Sequence|sequence]] $\sequence {g_N}_{n \mathop \in \N}$ of [[Definition:Extended Real-Valued Function|functions]] $g_N : X \to \overline \R$ by:
:$\ds \map {g_N} x = \sum_{n \mathop = 1}^N \map {f_n} x$
Since $f_n \ge 0$ for each $n$, we have:
{{begin-eqn}}
{{eqn | l = \map {g_{N + 1} } x... | Integral of Series of Positive Measurable Functions | https://proofwiki.org/wiki/Integral_of_Series_of_Positive_Measurable_Functions | https://proofwiki.org/wiki/Integral_of_Series_of_Positive_Measurable_Functions | [
"Integral of Positive Measurable Function"
] | [
"Definition:Measurable Space",
"Definition:Sequence",
"Definition:Measurable Function/Positive",
"Definition:Pointwise Series",
"Definition:Integral Sign",
"Definition:Integral of Positive Measurable Function"
] | [
"Definition:Sequence",
"Definition:Extended Real-Valued Function",
"Definition:Increasing Sequence of Real-Valued Functions",
"Pointwise Sum of Measurable Functions is Measurable/General Result",
"Definition:Measurable Function",
"Monotone Convergence Theorem (Measure Theory)",
"Definition:Series/Real",... |
proofwiki-5389 | Integral with respect to Dirac Measure | Let $\struct {X, \Sigma}$ be a measurable space.
Let $x \in X$, and let $\delta_x$ be the Dirac measure at $x$.
Let $f \in \MM _{\overline \R}, f: X \to \overline \R$ be a measurable function.
Then:
:$\ds \int f \rd \delta_x = \map f x$
where the integral sign denotes the $\delta_x$-integral. | Define the constant function $g : X \to \overline \R$ by:
:$\map g {x'} = \map f x$
for each $x' \in X$.
From Constant Function is Measurable, we have:
:$g$ is $\Sigma$-measurable.
From Measurable Functions Determine Measurable Sets:
:$\set {x' \in X : \map g {x'} \ne \map f {x'} } \in \Sigma$
Further:
:$x \not \i... | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $x \in X$, and let $\delta_x$ be the [[Definition:Dirac Measure|Dirac measure]] at $x$.
Let $f \in \MM _{\overline \R}, f: X \to \overline \R$ be a [[Definition:Measurable Function|measurable function]].
Then:
:$\ds \int f \rd \d... | Define the [[Definition:Constant Function|constant function]] $g : X \to \overline \R$ by:
:$\map g {x'} = \map f x$
for each $x' \in X$.
From [[Constant Function is Measurable]], we have:
:$g$ is [[Definition:Measurable Function|$\Sigma$-measurable]].
From [[Measurable Functions Determine Measurable Sets]]:
... | Integral with respect to Dirac Measure/Proof 1 | https://proofwiki.org/wiki/Integral_with_respect_to_Dirac_Measure | https://proofwiki.org/wiki/Integral_with_respect_to_Dirac_Measure/Proof_1 | [
"Integral with respect to Dirac Measure",
"Dirac Measure",
"Integrals of Integrable Functions"
] | [
"Definition:Measurable Space",
"Definition:Dirac Measure",
"Definition:Measurable Function",
"Definition:Integral Sign",
"Definition:Integral of Measurable Function"
] | [
"Definition:Constant Mapping",
"Constant Function is Measurable",
"Definition:Measurable Function",
"Measurable Functions Determine Measurable Sets",
"Definition:Dirac Measure",
"Definition:Almost Everywhere",
"A.E. Equal Positive Measurable Functions have Equal Integrals/Corollary 1",
"Integral of In... |
proofwiki-5390 | Integral with respect to Dirac Measure | Let $\struct {X, \Sigma}$ be a measurable space.
Let $x \in X$, and let $\delta_x$ be the Dirac measure at $x$.
Let $f \in \MM _{\overline \R}, f: X \to \overline \R$ be a measurable function.
Then:
:$\ds \int f \rd \delta_x = \map f x$
where the integral sign denotes the $\delta_x$-integral. | We first prove the result for positive simple functions.
Let $g : X \to \R$ be a positive simple function.
From Simple Function has Standard Representation, there exists:
:a finite sequence $a_1, \ldots, a_n$ of real numbers
:a partition $E_0, E_1, \ldots, E_n$ of $X$ into $\Sigma$-measurable sets
such that:
:$\ds g =... | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $x \in X$, and let $\delta_x$ be the [[Definition:Dirac Measure|Dirac measure]] at $x$.
Let $f \in \MM _{\overline \R}, f: X \to \overline \R$ be a [[Definition:Measurable Function|measurable function]].
Then:
:$\ds \int f \rd \d... | We first prove the result for [[Definition:Positive Simple Function|positive simple functions]].
Let $g : X \to \R$ be a [[Definition:Positive Simple Function|positive simple function]].
From [[Simple Function has Standard Representation]], there exists:
:a [[Definition:Finite Sequence|finite sequence]] $a_1, \ldots... | Integral with respect to Dirac Measure/Proof 2 | https://proofwiki.org/wiki/Integral_with_respect_to_Dirac_Measure | https://proofwiki.org/wiki/Integral_with_respect_to_Dirac_Measure/Proof_2 | [
"Integral with respect to Dirac Measure",
"Dirac Measure",
"Integrals of Integrable Functions"
] | [
"Definition:Measurable Space",
"Definition:Dirac Measure",
"Definition:Measurable Function",
"Definition:Integral Sign",
"Definition:Integral of Measurable Function"
] | [
"Definition:Simple Function",
"Definition:Simple Function",
"Measurable Function is Simple Function iff Finite Image Set/Corollary",
"Definition:Finite Sequence",
"Definition:Real Number",
"Definition:Set Partition",
"Definition:Measurable Set",
"Definition:Integral of Positive Simple Function",
"De... |
proofwiki-5391 | Integral with respect to Discrete Measure | Let $\struct {X, \Sigma}$ be a measurable space.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $X$.
Let $\ds \mu = \sum_{n \mathop \in \N} \lambda_n \delta_{x_n}$ be a discrete measure on $\struct {X, \Sigma}$.
Let $f \in \MM_{\overline \R}^+, f: X \to \overline \R$ be a positive measurable function.
Then:... | We have:
{{begin-eqn}}
{{eqn | l = \int f \rd \mu
| r = \sum_{n \mathop \in \N} \lambda_n \int f \rd \delta_{x_n}
| c = Integral with respect to Series of Measures
}}
{{eqn | r = \sum_{n \mathop \in \N} \lambda_n \map f {x_n}
| c = Integral with respect to Dirac Measure
}}
{{end-eqn}}
{{qed}} | Let $\struct {X, \Sigma}$ be a [[Definition:Measurable Space|measurable space]].
Let $\sequence {x_n}_{n \mathop \in \N}$ be a [[Definition:Sequence|sequence]] in $X$.
Let $\ds \mu = \sum_{n \mathop \in \N} \lambda_n \delta_{x_n}$ be a [[Definition:Discrete Measure|discrete measure]] on $\struct {X, \Sigma}$.
Let $... | We have:
{{begin-eqn}}
{{eqn | l = \int f \rd \mu
| r = \sum_{n \mathop \in \N} \lambda_n \int f \rd \delta_{x_n}
| c = [[Integral with respect to Series of Measures]]
}}
{{eqn | r = \sum_{n \mathop \in \N} \lambda_n \map f {x_n}
| c = [[Integral with respect to Dirac Measure]]
}}
{{end-eqn}}
{{qed}} | Integral with respect to Discrete Measure | https://proofwiki.org/wiki/Integral_with_respect_to_Discrete_Measure | https://proofwiki.org/wiki/Integral_with_respect_to_Discrete_Measure | [
"Discrete Measure",
"Discrete Measures",
"Integral of Positive Measurable Function",
"Discrete Measures"
] | [
"Definition:Measurable Space",
"Definition:Sequence",
"Definition:Discrete Measure",
"Definition:Measurable Function/Positive",
"Definition:Integral Sign",
"Definition:Integral of Positive Measurable Function"
] | [
"Integral with respect to Series of Measures",
"Integral with respect to Dirac Measure"
] |
proofwiki-5392 | Matrix Product as Linear Transformation | Let:
{{begin-eqn}}
{{eqn | l = \mathbf A_{m \times n}
| r = \begin {bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \\ \end {bmatrix}
}}
{{eqn | l = \mathbf x_{n \times 1}
| r = \begin {bmatrix} x_1 \\ ... | From Matrix Multiplication is Homogeneous of Degree $1$:
:$\forall \lambda \in \mathbb F \in \set {\R, \C}: \mathbf A \paren {\lambda \mathbf x} = \lambda \paren {\mathbf A \mathbf x}$
From Matrix Multiplication Distributes over Matrix Addition:
:$\forall \mathbf x, \mathbf y \in \R^m: \mathbf A \paren {\mathbf x + \ma... | Let:
{{begin-eqn}}
{{eqn | l = \mathbf A_{m \times n}
| r = \begin {bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \\ \end {bmatrix}
}}
{{eqn | l = \mathbf x_{n \times 1}
| r = \begin {bmatrix} x_1 \\... | From [[Matrix Multiplication is Homogeneous of Degree 1|Matrix Multiplication is Homogeneous of Degree $1$]]:
:$\forall \lambda \in \mathbb F \in \set {\R, \C}: \mathbf A \paren {\lambda \mathbf x} = \lambda \paren {\mathbf A \mathbf x}$
From [[Matrix Multiplication Distributes over Matrix Addition]]:
:$\forall \mathb... | Matrix Product as Linear Transformation | https://proofwiki.org/wiki/Matrix_Product_as_Linear_Transformation | https://proofwiki.org/wiki/Matrix_Product_as_Linear_Transformation | [
"Linear Transformations"
] | [
"Definition:Matrix",
"Definition:Matrix/Column",
"Definition:Element",
"Definition:Real Vector Space",
"Definition:Mapping",
"Definition:Linear Transformation/Vector Space"
] | [
"Matrix Multiplication is Homogeneous of Degree 1",
"Matrix Multiplication Distributes over Matrix Addition",
"Definition:Linear Transformation/Vector Space"
] |
proofwiki-5393 | Linear Transformation Maps Zero Vector to Zero Vector | Let $\mathbf V$ be a vector space, with zero $\mathbf 0$.
Likewise let $\mathbf V\,'$ be another vector space, with zero $\mathbf 0'$.
Let $T: \mathbf V \to \mathbf V\,'$ be a linear transformation.
Then:
:$T: \mathbf 0 \mapsto \mathbf 0'$ | From the vector space axioms we have that $\exists \mathbf 0 \in \mathbf V$.
It remains to be proved that $\map T {\mathbf 0} = \mathbf 0'$:
{{begin-eqn}}
{{eqn | l = \map T {\mathbf 0}
| r = \map T {\mathbf 0 + \mathbf 0}
}}
{{eqn | r = \map T {\mathbf 0} + \map T {\mathbf 0}
| c = {{Defof|Linear Transform... | Let $\mathbf V$ be a [[Definition:Vector Space|vector space]], with [[Definition:Zero Vector|zero]] $\mathbf 0$.
Likewise let $\mathbf V\,'$ be another [[Definition:Vector Space|vector space]], with [[Definition:Zero Vector|zero]] $\mathbf 0'$.
Let $T: \mathbf V \to \mathbf V\,'$ be a [[Definition:Linear Transformati... | From the [[Axiom:Vector Space Axioms|vector space axioms]] we have that $\exists \mathbf 0 \in \mathbf V$.
It remains to be proved that $\map T {\mathbf 0} = \mathbf 0'$:
{{begin-eqn}}
{{eqn | l = \map T {\mathbf 0}
| r = \map T {\mathbf 0 + \mathbf 0}
}}
{{eqn | r = \map T {\mathbf 0} + \map T {\mathbf 0}
... | Linear Transformation Maps Zero Vector to Zero Vector/Proof 1 | https://proofwiki.org/wiki/Linear_Transformation_Maps_Zero_Vector_to_Zero_Vector | https://proofwiki.org/wiki/Linear_Transformation_Maps_Zero_Vector_to_Zero_Vector/Proof_1 | [
"Linear Transformations",
"Linear Transformation Maps Zero Vector to Zero Vector"
] | [
"Definition:Vector Space",
"Definition:Zero Vector",
"Definition:Vector Space",
"Definition:Zero Vector",
"Definition:Linear Transformation/Vector Space"
] | [
"Axiom:Vector Space Axioms"
] |
proofwiki-5394 | Linear Transformation Maps Zero Vector to Zero Vector | Let $\mathbf V$ be a vector space, with zero $\mathbf 0$.
Likewise let $\mathbf V\,'$ be another vector space, with zero $\mathbf 0'$.
Let $T: \mathbf V \to \mathbf V\,'$ be a linear transformation.
Then:
:$T: \mathbf 0 \mapsto \mathbf 0'$ | From the vector space axioms we have that $\exists \mathbf 0 \in \mathbf V$.
What remains is to prove that $\map T {\mathbf 0} = \mathbf 0'$:
{{begin-eqn}}
{{eqn | l = \map T {\mathbf 0}
| r = \map T {0 \, \mathbf 0}
| c = Zero Vector Scaled is Zero Vector
}}
{{eqn | r = 0 \, \map T {\mathbf 0}
| c = ... | Let $\mathbf V$ be a [[Definition:Vector Space|vector space]], with [[Definition:Zero Vector|zero]] $\mathbf 0$.
Likewise let $\mathbf V\,'$ be another [[Definition:Vector Space|vector space]], with [[Definition:Zero Vector|zero]] $\mathbf 0'$.
Let $T: \mathbf V \to \mathbf V\,'$ be a [[Definition:Linear Transformati... | From the [[Axiom:Vector Space Axioms|vector space axioms]] we have that $\exists \mathbf 0 \in \mathbf V$.
What remains is to prove that $\map T {\mathbf 0} = \mathbf 0'$:
{{begin-eqn}}
{{eqn | l = \map T {\mathbf 0}
| r = \map T {0 \, \mathbf 0}
| c = [[Zero Vector Scaled is Zero Vector]]
}}
{{eqn | r = ... | Linear Transformation Maps Zero Vector to Zero Vector/Proof 2 | https://proofwiki.org/wiki/Linear_Transformation_Maps_Zero_Vector_to_Zero_Vector | https://proofwiki.org/wiki/Linear_Transformation_Maps_Zero_Vector_to_Zero_Vector/Proof_2 | [
"Linear Transformations",
"Linear Transformation Maps Zero Vector to Zero Vector"
] | [
"Definition:Vector Space",
"Definition:Zero Vector",
"Definition:Vector Space",
"Definition:Zero Vector",
"Definition:Linear Transformation/Vector Space"
] | [
"Axiom:Vector Space Axioms",
"Zero Vector Scaled is Zero Vector",
"Vector Scaled by Zero is Zero Vector"
] |
proofwiki-5395 | Infimum of Subset Product in Ordered Group | Let $\struct {G, \circ, \preceq}$ be an ordered group.
Let subsets $A$ and $B$ of $G$ admit infima in $G$.
Then:
:$\map \inf {A \circ_\PP B} = \inf A \circ \inf B$
where $\circ_\PP$ denotes subset product. | This follows from Supremum of Subset Product in Ordered Group and the Duality Principle.
{{qed}} | Let $\struct {G, \circ, \preceq}$ be an [[Definition:Ordered Group|ordered group]].
Let [[Definition:Subset|subsets]] $A$ and $B$ of $G$ admit [[Definition:Infimum of Set|infima]] in $G$.
Then:
:$\map \inf {A \circ_\PP B} = \inf A \circ \inf B$
where $\circ_\PP$ denotes [[Definition:Subset Product|subset product]]. | This follows from [[Supremum of Subset Product in Ordered Group]] and the [[Duality Principle (Order Theory)/Global Duality|Duality Principle]].
{{qed}} | Infimum of Subset Product in Ordered Group | https://proofwiki.org/wiki/Infimum_of_Subset_Product_in_Ordered_Group | https://proofwiki.org/wiki/Infimum_of_Subset_Product_in_Ordered_Group | [
"Infima",
"Subset Products",
"Ordered Groups"
] | [
"Definition:Ordered Group",
"Definition:Subset",
"Definition:Infimum of Set",
"Definition:Subset Product"
] | [
"Supremum of Subset Product in Ordered Group",
"Duality Principle (Order Theory)/Global Duality"
] |
proofwiki-5396 | Supremum of Subset Product in Ordered Group | Let $\struct {G, \circ, \preceq}$ be an ordered group.
Suppose that subsets $A$ and $B$ of $G$ admit suprema in $G$.
Then:
:$\sup \paren {A \circ_\PP B} = \sup A \circ \sup B$
where $\circ_\PP$ denotes subset product. | Let $a \in A$, $b \in B$.
Then:
{{begin-eqn}}
{{eqn | l = a \circ b
| o = \preceq
| r = \sup A \circ b
| c = {{Defof|Supremum of Set}}
}}
{{eqn | o = \preceq
| r = \sup A \circ \sup B
| c = {{Defof|Supremum of Set}}
}}
{{end-eqn}}
Hence $\sup A \circ \sup B$ is an upper bound for $A \circ_... | Let $\struct {G, \circ, \preceq}$ be an [[Definition:Ordered Group|ordered group]].
Suppose that [[Definition:Subset|subsets]] $A$ and $B$ of $G$ admit [[Definition:Supremum of Set|suprema]] in $G$.
Then:
:$\sup \paren {A \circ_\PP B} = \sup A \circ \sup B$
where $\circ_\PP$ denotes [[Definition:Subset Product|subs... | Let $a \in A$, $b \in B$.
Then:
{{begin-eqn}}
{{eqn | l = a \circ b
| o = \preceq
| r = \sup A \circ b
| c = {{Defof|Supremum of Set}}
}}
{{eqn | o = \preceq
| r = \sup A \circ \sup B
| c = {{Defof|Supremum of Set}}
}}
{{end-eqn}}
Hence $\sup A \circ \sup B$ is an [[Definition:Upper Boun... | Supremum of Subset Product in Ordered Group | https://proofwiki.org/wiki/Supremum_of_Subset_Product_in_Ordered_Group | https://proofwiki.org/wiki/Supremum_of_Subset_Product_in_Ordered_Group | [
"Suprema",
"Subset Products",
"Ordered Groups"
] | [
"Definition:Ordered Group",
"Definition:Subset",
"Definition:Supremum of Set",
"Definition:Subset Product"
] | [
"Definition:Upper Bound of Set",
"Definition:Upper Bound of Set"
] |
proofwiki-5397 | Fatou's Lemma for Measures | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\sequence {E_n}_{n \mathop \in \N} \in \Sigma$ be a sequence of $\Sigma$-measurable sets.
Then:
:$\ds \map \mu {\liminf_{n \mathop \to \infty} E_n} \le \liminf_{n \mathop \to \infty} \map \mu {E_n}$
where:
:$\ds \liminf_{n \mathop \to \infty} E_n$ is the limit inf... | Let:
:$\ds E := \liminf_{n \mathop \to \infty} E_n$
Then:
{{begin-eqn}}
{{eqn | l = \map \mu E
| r = \int \chi_E \rd \mu
| c = {{Defof|Lebesgue Integral|subdef = Simple Function}}
}}
{{eqn | r = \int \liminf_{n \mathop \to \infty} \chi_{E_n} \rd \mu
| c = Characteristic Function of Limit Inferior of S... | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $\sequence {E_n}_{n \mathop \in \N} \in \Sigma$ be a [[Definition:Sequence|sequence]] of [[Definition:Measurable Set|$\Sigma$-measurable sets]].
Then:
:$\ds \map \mu {\liminf_{n \mathop \to \infty} E_n} \le \liminf_{n \mathop \to \i... | Let:
:$\ds E := \liminf_{n \mathop \to \infty} E_n$
Then:
{{begin-eqn}}
{{eqn | l = \map \mu E
| r = \int \chi_E \rd \mu
| c = {{Defof|Lebesgue Integral|subdef = Simple Function}}
}}
{{eqn | r = \int \liminf_{n \mathop \to \infty} \chi_{E_n} \rd \mu
| c = [[Characteristic Function of Limit Inferior o... | Fatou's Lemma for Measures | https://proofwiki.org/wiki/Fatou's_Lemma_for_Measures | https://proofwiki.org/wiki/Fatou's_Lemma_for_Measures | [
"Fatou's Lemma for Measures",
"Measure Theory",
"Fatou's Lemma"
] | [
"Definition:Measure Space",
"Definition:Sequence",
"Definition:Measurable Set",
"Definition:Limit Inferior of Sequence of Sets",
"Definition:Limit Inferior",
"Definition:Extended Real Number Line"
] | [
"Characteristic Function of Limit Inferior of Sequence of Sets",
"Fatou's Lemma for Integrals/Positive Measurable Functions"
] |
proofwiki-5398 | Kernel Transformation of Measure is Measure | Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $N: X \times \Sigma \to \overline \R_{\ge0}$ be a kernel.
Then $\mu N: X \to \overline \R$, the kernel transformation of $\mu$, is a measure. | From the definition of the kernel transformation of $\mu$, we have:
:$\ds \map {\paren {\mu N} } E = \int \map N {x, E} \rd \map \mu x$
for each $E \in \Sigma$.
We verify each of the conditions for a measure in turn. | Let $\struct {X, \Sigma, \mu}$ be a [[Definition:Measure Space|measure space]].
Let $N: X \times \Sigma \to \overline \R_{\ge0}$ be a [[Definition:Kernel (Measure Theory)|kernel]].
Then $\mu N: X \to \overline \R$, the [[Definition:Kernel Transformation of Measure|kernel transformation of $\mu$]], is a [[Definition:... | From the definition of the [[Definition:Kernel Transformation of Measure|kernel transformation of $\mu$]], we have:
:$\ds \map {\paren {\mu N} } E = \int \map N {x, E} \rd \map \mu x$
for each $E \in \Sigma$.
We verify each of the conditions for a [[Definition:Measure (Measure Theory)|measure]] in turn. | Kernel Transformation of Measure is Measure | https://proofwiki.org/wiki/Kernel_Transformation_of_Measure_is_Measure | https://proofwiki.org/wiki/Kernel_Transformation_of_Measure_is_Measure | [
"Measures",
"Kernel Transformation of Measure",
"Measures"
] | [
"Definition:Measure Space",
"Definition:Kernel (Measure Theory)",
"Definition:Kernel Transformation of Measure",
"Definition:Measure (Measure Theory)"
] | [
"Definition:Kernel Transformation of Measure",
"Definition:Measure (Measure Theory)",
"Definition:Measure (Measure Theory)"
] |
proofwiki-5399 | Canonical Injection is Injection | Let $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$ be algebraic structures with identities $e_1, e_2$ respectively.
The canonical injections:
:$\inj_1: \struct {S_1, \circ_1} \to \struct {S_1, \circ_1} \times \struct {S_2, \circ_2}: \forall x \in S_1: \map {\inj_1} x = \tuple {x, e_2}$
:$\inj_2: \struct {S_2, \c... | Let $x, x' \in S_1$.
Suppose that:
:$\map {\inj_1} x = \map {\inj_1} {x'}$
Then by definition of canonical injection:
:$\tuple {x, e_2} = \tuple {x', e_2}$
By Equality of Ordered Pairs:
:$x = x'$
That is, $\inj_1$ is an injection.
{{qed|lemma}}
Similarly, let $x, x' \in S_2$.
Suppose that:
:$\map {\inj_2} x = \map {\in... | Let $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]] with [[Definition:Identity Element|identities]] $e_1, e_2$ respectively.
The [[Definition:Canonical Injection (Abstract Algebra)|canonical injections]]:
:$\inj_1: \struct {S_1, \cir... | Let $x, x' \in S_1$.
Suppose that:
:$\map {\inj_1} x = \map {\inj_1} {x'}$
Then by definition of [[Definition:Canonical Injection (Abstract Algebra)|canonical injection]]:
:$\tuple {x, e_2} = \tuple {x', e_2}$
By [[Equality of Ordered Pairs]]:
:$x = x'$
That is, $\inj_1$ is an [[Definition:Injection|injection]].
{{... | Canonical Injection is Injection | https://proofwiki.org/wiki/Canonical_Injection_is_Injection | https://proofwiki.org/wiki/Canonical_Injection_is_Injection | [
"Canonical Injections"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Canonical Injection (Abstract Algebra)",
"Definition:Injection"
] | [
"Definition:Canonical Injection (Abstract Algebra)",
"Equality of Ordered Pairs",
"Definition:Injection",
"Definition:Canonical Injection (Abstract Algebra)",
"Equality of Ordered Pairs",
"Definition:Injection",
"Definition:Injection"
] |
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