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If $M$ is a finite measure space, then the extended measure of a set $A$ is equal to the real measure of $A$.
If $M$ is a finite measure space, then the measure of any set $A$ is a nonnegative real number.
If $M$ is a finite measure space, then the measure of any set is bounded above by the measure of the whole space.
If $A$ and $B$ are measurable sets with $B \subseteq A$, then the measure of the difference $A - B$ is the measure of $A$ minus the measure of $B$.
If $A$ and $B$ are measurable sets with disjoint union, then the measure of their union is the sum of their measures.
If $A_1, \ldots, A_n$ are disjoint measurable sets, then $\mu(\bigcup_{i=1}^n A_i) = \sum_{i=1}^n \mu(A_i)$.
If $A$ is a disjoint family of measurable sets, then the measure of the union of the sets in $A$ is the sum of the measures of the sets in $A$.
If $A \subseteq B$ and $B$ is measurable, then the measure of $A$ is less than or equal to the measure of $B$.
If $A$ and $B$ are measurable sets, then the measure of their union is less than or equal to the sum of their measures.
If $I$ is a finite set and $A_i \in \mathcal{M}$ for all $i \in I$, then $\mu(\bigcup_{i \in I} A_i) \leq \sum_{i \in I} \mu(A_i)$.
If $A$ is a countable collection of measurable sets, then the measure of the union of the sets in $A$ is less than or equal to the sum of the measures of the sets in $A$.
If $S$ is a finite set and each singleton $\{x\}$ is measurable, then the measure of $S$ is the sum of the measures of the singletons.
If $A_1 \subseteq A_2 \subseteq \cdots$ is an increasing sequence of measurable sets, then the sequence of measures of the $A_i$ converges to the measure of the union of the $A_i$.
If $A_1, A_2, \ldots$ is a decreasing sequence of measurable sets, then the sequence of measures of the sets converges to the measure of the intersection of the sets.
If $S$ is a measurable set, then the measure of the complement of $S$ is the total measure of the space minus the measure of $S$.
If $A$ is a subset of $B$ almost everywhere, then the measure of $A$ is less than or equal to the measure of $B$.
If two sets $A$ and $B$ are equal almost everywhere, then they have the same measure.
The measure of a set is an increasing function of the set.
If $t$ is a set of measure zero, then the measure of the union of $s$ and $t$ is the same as the measure of $s$.
If two sets have the same measure of their complements, then they have the same measure.
If two families of disjoint sets have the same measure, then the union of the two families also have the same measure.
If a collection of sets has measure zero, then the union of the sets also has measure zero.
If $t$ is a measurable set with measure equal to the measure of the whole space, then the intersection of $t$ with any measurable set $s$ has measure equal to the measure of $s$.
If $s$ is a finite set of measurable sets, each of which has the same measure, then the measure of the union of the sets is equal to the number of sets times the measure of any one of the sets.
If $f$ is a finite set of disjoint measurable sets, then the measure of the union of the $f$ is the sum of the measures of the $f$.
If $A$ and $B$ are measurable sets such that $A$ has measure $1$ and $A \cap B = \emptyset$, then $B$ has measure $0$.
If $f$ is a measurable function from a measure space $(X, \mathcal{A}, \mu)$ to a measure space $(Y, \mathcal{B}, \nu)$, then the pushforward measure $\mu_*$ is a finite measure.
If $F$ is a monotone function on the set of measurable subsets of a measurable space $N$, and $f$ is a monotone function on the set of functions from a set $S$ to the set of nonnegative real numbers, then the greatest fixed point of $F$ is equal to the greatest fixed point of $f$.
If $f$ is a function such that $f(n) < f(n+1)$ for all $n$, then $f$ is strictly monotone.
The measure of a finite subset of a countable set is the cardinality of the subset.
If $f$ is a bijection from $A$ to $B$, then the distribution of $B$ is the same as the distribution of $A$ under $f$.
If $X$ is a finite subset of $A$, then the measure of $X$ is the number of elements in $X$.
If $X$ is an infinite subset of $A$, then the measure of $X$ is infinite.
The measure of a subset of a countable set is the number of elements in the subset.
The measure of a subset of a countable set is zero if and only if the subset is empty.
The only null set in a countable space is the empty set.
For any countable set $A$, the set of points $x \in A$ for which a property $P$ holds is either all of $A$ or none of $A$.
If $A$ is a countable set, then the counting measure on $A$ is $\sigma$-finite.
The counting measure on a countable set is sigma-finite.
If $A$ is a finite set, then the counting measure on $A$ is a finite measure.
If $A$ is a finite set, then the counting measure on $A$ is $\sigma$-finite.
If $\Omega$ is a measurable set, then the restriction of a measure to $\Omega$ is the same as the original measure.
If $\Omega$ is a measurable set, then the restriction of a measure to $\Omega$ is the same as the original measure.
If $\Omega$ is a measurable subset of the space $M$, then the following are equivalent: $\forall \epsilon > 0, \exists A \in \mathcal{M}$ such that $A \subseteq \Omega$ and $\mu(A) < \epsilon$ $\forall \epsilon > 0, \exists A \in \mathcal{M}$ such that $A \subseteq \Omega$ and $\mu(A) < \epsilon$
If $A$ and $B$ are subsets of a measurable space $M$, then the restriction of $M$ to $A$ and then to $B$ is the same as the restriction of $M$ to $A \cap B$.
The restriction of a countable space to a subset is a countable space.
If $M$ is a $\sigma$-finite measure and $A$ is a measurable set, then the restriction of $M$ to $A$ is also $\sigma$-finite.
If $M$ is a finite measure and $A$ is a measurable set, then the restriction of $M$ to $A$ is a finite measure.
If $f$ is a measurable function from a measure space $(M, \mathcal{M})$ to a measure space $(N, \mathcal{N})$, and $\Omega \in \mathcal{N}$, then the distribution of $f$ restricted to $\Omega$ is the same as the distribution of $f$ restricted to $(M, \mathcal{M})$ and $(N, \mathcal{N})$ restricted to $\Omega$.
If $M$ and $N$ are two measures on the same $\sigma$-algebra, and if $M$ and $N$ agree on all sets in a certain collection $E$, then $M$ and $N$ are equal.
The space of a null measure is the same as the space of the original measure.
The sets of a null measure are the same as the sets of the underlying measure space.
The measure of any set is zero under the null measure.
The measure of any set with respect to the null measure is zero.
The null measure of a null measure is the null measure.
The space of a scaled measure is the same as the space of the original measure.
The sets of a scaled measure are the same as the sets of the original measure.
The measure of a set $A$ under the scaled measure $rM$ is $r$ times the measure of $A$ under $M$.
The measure obtained by scaling a measure by 1 is the same as the original measure.
If you scale a measure by 0, you get the null measure.
If $r \geq 0$, then the measure of a set $A$ under the measure $rM$ is $r$ times the measure of $A$ under the measure $M$.
If $M$ is a measure and $r, r'$ are positive real numbers, then $r \cdot (r' \cdot M) = (r \cdot r') \cdot M$.
If $r$ is a real number and $M$ is a measure space, then $r \cdot \mu$ is the null measure if and only if $\mu$ is the null measure.
If $A$ and $B$ are measurable sets, then the measure of the difference $A - B$ is equal to the measure of $A$ minus the measure of the intersection $A \<inter> B$.
If $A$ and $B$ are measurable sets, then the measure of $A \cup B$ is the sum of the measure of $A$ and the measure of $B \setminus A$.
If $M$ and $N$ are finite measures on the same measurable space, then there exists a measurable set $Y$ such that $N(X) \leq M(X)$ for all measurable sets $X$ contained in $Y$, and $M(X) \leq N(X)$ for all measurable sets $X$ disjoint from $Y$.
If $A$ is a measurable set with finite measure in both $M$ and $N$, then there exists a measurable set $Y \subseteq A$ such that $Y$ is $N$-smaller than $A$ and $A \setminus Y$ is $M$-smaller than $A$.
The measure $M$ is less than or equal to the measure $N$ if and only if either the spaces of $M$ and $N$ are equal and the sets of $M$ and $N$ are equal and the emeasure of $M$ is less than or equal to the emeasure of $N$, or the space of $M$ is a subset of the space of $N$.
The empty measure space is the measure space with empty space and empty sets.
If two measures have the same sets, then one is less than or equal to the other if and only if the measure of every set is less than or equal to the measure of the same set under the other measure.
If $B$ is a sub-$\sigma$-algebra of $A$, then the space and sets of the measure space $(A, B, \mu)$ are the same as the space and sets of the measure space $(A, A, \mu)$.
If $A$ and $B$ are two $\sigma$-algebras on the same set, then the measure of a set $X$ in the $\sigma$-algebra generated by $A$ and $B$ is the supremum of the measures of $X \cap Y$ in $A$ and $X \cap (Y^c)$ in $B$ over all sets $Y$ in $A$.
If $B$ is a $\sigma$-algebra that contains $A$, then the outer measure of $X$ with respect to $A$ is less than or equal to the outer measure of $X$ with respect to $B$.
If $B$ is a sub-$\sigma$-algebra of $A$, then the outer measure of $X$ with respect to $B$ is less than or equal to the outer measure of $X$ with respect to the smallest $\sigma$-algebra containing $A$ and $B$.
If $A$ and $B$ are two measures on the same space $C$ such that $A(Y) \leq C(Y)$ and $B(Y) \leq C(Y)$ for all $Y \subseteq X$, then $A \vee B(X) \leq C(X)$.
If $k(A) < k(B)$, then $P(B)$; if $k(B) < k(A)$, then $P(A)$; if $k(A) = k(B)$, then $P(c)$; if $k(B) \not\leq k(A)$ and $k(A) \not\leq k(B)$, then $P(s)$. Then $P(\sup_\text{lexord}(A, B, k, s, c))$.
If $a$ is a real number, then $a \leq \sup(S)$ if and only if $a \leq s$ for all $s \in S$.
If $k(A) = k(B)$, then $\sup\{A, B\} = c$.
The lexicographic order on pairs of sets is commutative.
A set of sets is a $\sigma$-algebra if and only if it is closed under countable unions.
If $\mathcal{A}$ is a collection of subsets of $\Omega$, then $\sigma(\Omega, \mathcal{A}) \leq x$ if and only if $\Omega \subseteq \text{space}(x)$ and if $\text{space}(x) = \Omega$, then $\mathcal{A} \subseteq \text{sets}(x)$.
A measure space is empty if and only if it is the trivial measure space.
If $A \subseteq B \subseteq C$ and $\sigma(A) = \sigma(C)$, then $\sigma(B) = \sigma(A)$.
If $A$ and $B$ are $\sigma$-algebras on a set $X$, then $A \cup B$ is also a $\sigma$-algebra on $X$.
If $A \le B$, then $A$ is a subset of $B$.
If $A \leq B$, then $A$ and $B$ have the same underlying set and $A$ is a sub-algebra of $B$.
If $A$ is a sub-$\sigma$-algebra of $B$ and $X$ is a measurable set with respect to both $A$ and $B$, then the measure of $X$ with respect to $A$ is less than or equal to the measure of $X$ with respect to $B$.
The union of the sets in a family of topological spaces is a subset of the union of the spaces in the family.
If $A$ is a set and $k$ is a function from $A$ to some ordered set, then the supremum of $k$ in $A$ is either a constant value $k(a)$ for some $a \in A$, or it is the supremum of $k$ in some subset of $A$.
If $A$ is a nonempty set, $k$ is a function from $A$ to the set of subsets of $A$, $c$ is a function from $A$ to the set of subsets of $A$, and $P$ is a predicate on the set of subsets of $A$, then if $P$ holds for $c(A)$, it holds for $\sup_{\mathrm{lex}}(k, c, s, A)$.
If $I$ and $J$ are finite sets, then $\sup_{i \in I} f(i) \leq \sup_{i \in I \cup J} f(i)$.
If $I$ is a finite set and $J$ is a subset of $I$, then the measure of $J$ is less than or equal to the measure of $I$.
If $I$ is a nonempty finite set and $i \in I$ implies that $i$ is a measure space with the same sets as $M$, then the supremum of the $i$'s is a measure space with the same sets as $M$.
The space of the supremum measure is the union of the spaces of the measures in the family.
The sets of the supremum measure of a family of measures are the sigma-algebra generated by the union of the sets of the measures in the family.
If all measures in a non-empty set of measures have the same measurable sets, then the supremum of these measures also has the same measurable sets.
If $M$ is a nonempty set of measures on a measurable space $(X, \mathcal{A})$, then the space of the supremum measure of $M$ is $X$.
If $M$ is a nonempty set of measures on a common $\sigma$-algebra, then the supremum measure of $M$ is the measure that assigns to each measurable set $X$ the supremum of the measures of $X$ over all finite subsets of $M$.
If the measurable spaces $M_i$ are all equal to $N$, then the supremum of the $M_i$ is also equal to $N$.
If $M_i$ is a family of measures on a common measurable space $N$, then the supremum of the $M_i$ is the supremum of the suprema of the finite subfamilies of $M_i$.