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If $P(x)$ and $Q(x)$ are almost everywhere equal, then $Q(x)$ and $P(x)$ are almost everywhere equal.
If $P$ implies that $Q$ holds almost everywhere, then $P$ and $Q$ hold almost everywhere.
If $P$ holds almost everywhere, then the measure of the set of points where $P$ holds is equal to the measure of the whole space.
Almost every point in a measurable space is in the space.
If $P$ holds for all $x$ in the space $M$, then $P$ holds almost everywhere in $M$.
If $P$ holds for all $x$ in the space $M$, and $P$ implies $Q$ almost everywhere, then $Q$ holds almost everywhere.
If $P$ and $Q$ are two properties of elements of a measurable space $M$ such that $P(x)$ and $Q(x)$ are equivalent for all $x \in M$, then the set of elements of $M$ satisfying $P$ is equal to the set of elements of $M$ satisfying $Q$.
If two measures are equal, then two sets are almost equal with respect to one measure if and only if they are almost equal with respect to the other measure.
For all countable $i$, $P_i$ holds almost everywhere if and only if $P_i$ holds almost everywhere for all countable $i$.
If $X$ is countable, then the almost everywhere statement $\forall y \in X. P(x, y)$ is equivalent to the statement $\forall y \in X. \text{almost everywhere } P(x, y)$.
If $P(N,x)$ holds for almost every $x$ for each $N$, and $I$ is countable, then $P(N,x)$ holds for almost every $x$ for all $N \in I$.
If $F$ is a countable family of sets, then $F$ is pairwise disjoint if and only if for almost every $x$, the family $F$ is pairwise disjoint.
If $X$ is a countable set and $M$ is a measure space such that each point in $X$ has measure zero, then almost every point in $M$ is not in $X$.
If $S$ is a finite set, then the statement "$\forall i \in S, P_i(x)$ holds for almost every $x$" is equivalent to the statement "$P_i(x)$ holds for almost every $x$ for every $i \in S$".
If $S$ is a finite set and for each $s \in S$, $Q(s,x)$ holds almost everywhere, then $\forall s \in S, Q(s,x)$ holds almost everywhere.
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 almost equal, then they have the same measure.
If two predicates $P$ and $Q$ are almost everywhere equal, then the measure of the set of points where $P$ holds is equal to the measure of the set of points where $Q$ holds.
If $P$ is a property that holds for almost every element of a measure space, then the set of elements for which $P$ holds has measure zero.
If the measure of the entire space is zero, then almost every point satisfies $P$.
If $A$ and $B$ are measurable sets such that $C = A \cup B$ and $A \cap B = \emptyset$, then $\mu(C) = \mu(A) + \mu(B)$.
Any $\sigma$-finite measure space is a countable union of sets of finite measure.
If $\mu$ is a $\sigma$-finite measure, then there exists a countable family of disjoint measurable sets whose union is the whole space and whose measures are all finite.
If $\mu$ is a $\sigma$-finite measure, then there exists a sequence of sets $A_1, A_2, \ldots$ such that $\mu(A_i) < \infty$ for all $i$, $\bigcup_{i=1}^\infty A_i = X$, and $A_1 \subseteq A_2 \subseteq \cdots$.
If $W$ is a measurable set with infinite measure, then there exists a measurable subset $Z$ of $W$ with finite measure greater than $C$.
The distribution of a measurable function is the same as the distribution of the codomain.
The measurable space of a distribution is the same as the measurable space of the function.
If two measures are defined on the same space and have the same sets, and if two functions are equal on the space, then the distributions of the functions with respect to the measures are equal.
If $f$ is a measurable function from a measure space $(X, \mathcal{A}, \mu)$ to a measure space $(Y, \mathcal{B},
If $X$ is a measurable function from a measure space $(M, \mathcal{M})$ to a measure space $(N, \mathcal{N})$, then the pushforward measure of $X$ is the same as the pushforward measure of the indicator function of the preimage of $P$ under $X$.
If $f$ is a measurable function from a measurable space $M'$ to a measurable space $M$, and $F$ is a sup-continuous function from the set of measurable subsets of $M$ to itself, then the Lebesgue measure of the set $\{x \in M' : \text{lfp}(F)(f(x))\}$ is equal to the supremum of the Lebesgue measures of the sets $\{x \...
The distribution of a random variable that is equal to itself is equal to the distribution of the random variable.
If two measures have the same sets, then the identity map is a measure isomorphism.
If $f$ is a measurable function from a measure space $(X, \mathcal{A}, \mu)$ to a measure space $(Y, \mathcal{B}, \nu)$, then for any $B \in \mathcal{B}$, the measure of $f^{-1}(B)$ in $(X, \mathcal{A}, \mu)$ is equal to the measure of $B$ in $(Y, \mathcal{B}, \nu)$.
If two measures $M$ and $K$ are equal and two functions $f$ and $g$ are equal almost everywhere, then the distributions of $f$ and $g$ are equal.
If $f$ is a measurable function from $M$ to $M'$ and $P$ holds almost everywhere on the image of $f$, then $P$ holds almost everywhere on $M$.
If $f$ is a measurable function from a measure space $(M, \mathcal{M})$ to a measure space $(N, \mathcal{N})$, then for any set $A \in \mathcal{N}$, the set of points $x \in M$ such that $f(x) \in A$ is in $\mathcal{M}$.
If $f$ is a measurable function from a measure space $(M, \mathcal{M}, \mu)$ to a measure space $(N, \mathcal{N},
If $f$ is a measurable function from $M$ to $N$ and $g$ is a measurable function from $N$ to $L$, then the composition $g \circ f$ is a measurable function from $M$ to $L$.
The collection of measurable sets of finite measure is a ring.
The measure of a set is nonnegative.
The measure of a set is never negative.
The measure of a set $A$ is positive if and only if the measure of $A$ is not zero.
The measure of a set is non-negative, and is zero if and only if the set has measure zero.
The measure of the empty set is zero.
If the measure of a set $A$ is not infinite, then the extended measure of $A$ is equal to the measure of $A$.
If the outer measure of a set $A$ is $+\infty$, then the measure of $A$ is $0$.
If $x$ is a non-negative real number and $\mu(A) = x$, then $\mu^*(A) = x$.
If $a$ and $b$ are positive real numbers, then $a + b$ is also a positive real number.
If $f(i) < \top$ for all $i \in I$, then $\sum_{i \in I} f(i) = \sum_{i \in I} \text{enn2real}(f(i))$.
If two sets $A$ and $B$ are equal almost everywhere, then they have the same measure.
If $A$ and $B$ are disjoint measurable sets with finite measure, then the measure of their union is the sum of their measures.
If $i$ is not in $I$, then the family of sets $\{A_i\}_{i \in I \cup \{i\}}$ is disjoint if and only if $A_i$ is disjoint from $\bigcup_{i \in I} A_i$ and the family of sets $\{A_i\}_{i \in I}$ is disjoint.
If $S$ is a finite set, $A$ is a family of measurable sets indexed by $S$, and the sets in $A$ are pairwise disjoint, then the measure of the union of the sets in $A$ is the sum of the measures of the sets in $A$.
If $A$ and $B$ are measurable sets with $B \subseteq A$ and $A$ has finite measure, then $A - B$ has measure $m(A) - m(B)$.
If $A_1, A_2, \ldots$ is a disjoint family of measurable sets, then $\sum_{i=1}^\infty \mu(A_i) = \mu(\bigcup_{i=1}^\infty A_i)$.
If $A$ and $B$ are measurable sets with finite measure, then the measure of their union is less than or equal to the sum of their measures.
If $A_1, \ldots, A_n$ are finite sets, then $\sum_{i=1}^n |A_i| \geq |\bigcup_{i=1}^n A_i|$.
If $A_1, A_2, \ldots$ are measurable sets, then the measure of their union is less than or equal to the sum of their measures.
If $A$ is a measurable set and $B$ is a null set, then $A \cup B$ is measurable and has the same measure as $A$.
If $A$ is a measurable set and $B$ is a null set, then the measure of the difference $A - B$ is the same as the measure of $A$.
If $S$ is a finite set and $M$ is a measure space, then the measure of $S$ is the sum of the measures of the singletons in $S$.
If $A_1 \subseteq A_2 \subseteq \cdots$ is an increasing sequence of measurable sets, then the sequence of measures of the sets converges to the measure of the union of the sets.
If $A_1, A_2, \ldots$ is a decreasing sequence of measurable sets, then the sequence of measures of the $A_i$ converges to the measure of their intersection.
If $A$ is a $f$-measurable set, then $A$ is a measurable set.
If $A$ is $f$-measurable, then $A$ has finite measure.
If $A$ is a measurable set with finite measure, then $A$ is a finite-measurable set.
If $A$ is a null set, then $A$ is $f$-measurable.
If $A$ is a $f$-measurable set and $B \subseteq A$, then $B$ is $f$-measurable.
If $A \subseteq B$ and $A$ is measurable, then $B$ is measurable and $\mu(A) \leq \mu(B)$.
If $A$ is a measurable set, then the outer measure of $A$ is equal to the measure of $A$.
If $A$ is a $\sigma$-finite measurable set and $B$ is a measurable set, then $A - B$ is a $\sigma$-finite measurable set.
If $S$ is a measurable set and $T$ is any set, then $S \cap T$ is measurable.
If $A$ is a $\sigma$-finite measure space, $I$ is a countable index set, and $F_i$ is a measurable subset of $A$ for each $i \in I$, then $\bigcup_{i \in I} F_i$ is measurable.
If $I$ is a countable set, $F_i \in \mathcal{F}$ for all $i \in I$, and $F_i \in \mathcal{F}^*$ for some $i \in I$, then $\bigcap_{i \in I} F_i \in \mathcal{F}^*$.
If $A$ is a measurable set and $B$ is a subset of $A$, then the measure of the difference $A - B$ is equal to the measure of $A$ minus the measure of $B$.
If $B$ is a null set, then $A \cup B$ is measurable if and only if $A$ is measurable.
If $B$ is a null set, then $A - B$ is measurable if and only if $A$ is measurable.
If $S$ and $T - S$ are measurable, and $S \subseteq T$, then $T$ is measurable.
If $A$ and $B$ are measurable sets, then the measure of their union is the sum of the measures of $A$ and $B$ minus the measure of their intersection.
If $A$ and $B$ are measurable sets, then the measure of their union is the sum of their measures minus the measure of their intersection.
If $A$ and $B$ are measurable sets and almost every point of $M$ is in at most one of $A$ and $B$, then the measure of $A \cup B$ is the sum of the measures of $A$ and $B$.
If $F_1, \ldots, F_n$ are pairwise disjoint measurable sets, then $\mu(\bigcup_{i=1}^n F_i) = \sum_{i=1}^n \mu(F_i)$.
If $I$ is a finite set, $F_i$ is a measurable set for each $i \in I$, and the sets $F_i$ are pairwise disjoint, then the measure of the union of the $F_i$ is the sum of the measures of the $F_i$.
If $F$ is a finite collection of measurable sets such that any two sets in $F$ are almost disjoint, then the measure of the union of the sets in $F$ is equal to the sum of the measures of the sets in $F$.
If $F$ is a finite collection of pairwise disjoint sets that are all measurable, then the measure of the union of the sets in $F$ is the sum of the measures of the sets in $F$.
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 $F_i$ is a measurable set for each $i \in I$, then the measure of the union of the $F_i$ is less than or equal to the sum of the measures of the $F_i$.
If $F$ is a finite collection of measurable sets, then the measure of the union of the sets in $F$ is less than or equal to the sum of the measures of the sets in $F$.
If $I$ is a countable index set, $A_i$ is a family of measurable sets, and $\sum_{i \in I} \mu(A_i) \leq B$ for all finite subsets $I' \subseteq I$, then $\bigcup_{i \in I} A_i$ is measurable and $\mu(\bigcup_{i \in I} A_i) \leq B$.
If $\mathcal{D}$ is a countable collection of measurable sets such that the union of any finite subcollection has measure at most $B$, then $\bigcup \mathcal{D}$ is measurable and has measure at most $B$.
If $S_i$ is a sequence of sets in $M$ such that $\sum_{i=1}^\infty \mu(S_i) \leq B$, then $\bigcup_{i=1}^\infty S_i \in M$ and $\mu(\bigcup_{i=1}^\infty S_i) \leq B$.
If $S$ and $T$ are measurable sets, then the measure of the symmetric difference of $S$ and $T$ is less than or equal to the measure of the difference of $S$ and $T$.
If the series $\sum_{n=0}^\infty f(n)$ converges, then the series $\sum_{n=0}^\infty \sum_{k=0}^\infty f(k+n)$ converges.
If $A_1, A_2, \ldots$ are measurable sets with finite measure, and $\sum_{n=1}^\infty \mu(A_n) < \infty$, then $\mu(\bigcup_{n=1}^\infty A_n) < \infty$ and $\mu(\bigcup_{n=1}^\infty A_n) \leq \sum_{n=1}^\infty \mu(A_n)$.
If the series $\sum_{n=1}^{\infty} \mu(A_n)$ converges, then the set $\limsup A_n$ is a null set.
If the series $\sum_{n=1}^\infty \mu(A_n)$ converges, then almost every point $x$ is in all but finitely many of the sets $A_n$.
If the measure of the whole space is finite, then the measure is finite.
If $M$ is a finite measure space, then the measure of any set is finite.
In a finite measure space, the $\sigma$-algebra of measurable sets is the same as the $\sigma$-algebra of sets.