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If $a$ is an element of $M$ and $b$ is an element of the smallest closed, convex, and directed set containing $M$, then $a \cap b$ is an element of the smallest closed, convex, and directed set containing $M$.

If $b$ is a smallest closed-and-discrete set in $\Omega$ with respect to $M$, then for any smallest closed-and-discrete set $a$ in $\Omega$ with respect to $M$, $a \cap b$ is a smallest closed-and-discrete set in $\Omega$ with respect to $M$.

If $M$ is a subset of $C$ and $C$ is a closed $\sigma$-algebra, then the $\sigma$-algebra generated by $M$ is a subset of $C$.

If $M$ is a $\sigma$-algebra, $C$ is a class of subsets of $\Omega$ that contains $M$, and $C$ is closed under complements, countable unions of increasing sequences, and countable disjoint unions, then $C$ is a $\sigma$-algebra.

The empty set is measurable.

If $D$ and $E$ are measurable sets and $D \subseteq E$, then $E - D$ is measurable.

A Dynkin system is a collection of subsets of a set $\Omega$ that satisfies the following properties: $\Omega$ is in the collection. If $A$ is in the collection, then $\Omega - A$ is in the collection. If $A$ is a disjoint family of sets in the collection, then $\bigcup_{i=1}^{\infty} A_i$ is in the collection.

A collection $M$ of subsets of a set $\Omega$ is a Dynkin system if it satisfies the following conditions: $\Omega \in M$ $\emptyset \in M$ $\Omega - A \in M$ for all $A \in M$ If $A$ is a disjoint family of sets in $M$, then $\bigcup_{i=1}^{\infty} A_i \in M$.

The power set of a set $A$ is a Dynkin system.

If $\mathcal{M}$ is a $\sigma$-algebra, then it is a Dynkin system.

The intersection of two submodules of a module is a submodule.

If for any two elements $i$ and $j$ in $I$, there exists an element $k$ in $I$ such that $A_i \cap A_j = A_k$, then the image of $I$ under $A$ is an intersection-stable set.

If the intersection of any two sets in $A$ is also in $A$, then $A$ is an intersection-stable set.

If $M$ is an intersection-stable set of sets, and $a, b \in M$, then $a \cap b \in M$.

A Dynkin system is a sigma-algebra if and only if it is closed under intersections.

If $M$ is a subset of the power set of $\Omega$, then the Dynkin system of $\Omega$ generated by $M$ is a Dynkin system.

If $A$ is a member of a $\sigma$-algebra $M$, then $A$ is a member of the Dynkin $\sigma$-algebra of $M$.

If $D$ is a Dynkin system, then the collection of all subsets of $\Omega$ that intersect $D$ in a set in $D$ is also a Dynkin system.

If $N$ is a subset of $M$, then the Dynkin system of $N$ is a subset of $M$.

If $M$ is a collection of subsets of $\Omega$ that is closed under finite intersections, then the $\sigma$-algebra generated by $M$ is equal to the Dynkin $\sigma$-algebra generated by $M$.

In a Dynkin system, the Dynkin $\sigma$-algebra is equal to the $\sigma$-algebra of the system.

If $E$ is a Dynkin system, then $\sigma(E) = M$.

If $G$ is a $\sigma$-algebra, then $G$ is closed under complements and countable unions.

The triple $(\Omega, \mathcal{F}, \mu)$ is a measure space.

If $A$ is a subset of the power set of $\Omega$, then $(\Omega, A, \mu)$ is a measure space, where $\mu$ is the zero measure.

The trivial sigma algebra is a sigma algebra.

The empty set and the whole space are a measure space.

If $(\Omega, M, \mu)$ is a measure space, then $M \subseteq \mathcal{P}(\Omega)$.

If two measures agree on a ring of sets, then they agree on the positive sets of that ring.

If two measures agree on a $\sigma$-algebra, then they are equal.

If two measures agree on all measurable sets, then they are equal.

If two measures agree on all sets in the sigma algebra generated by a collection of sets, then they agree on the measure of the collection.

The space of the measure of $\Omega$ is $\Omega$, the sets of the measure of $\Omega$ are the sigma-algebra generated by $A$ if $A \subseteq \mathcal{P}(\Omega)$ and $\{\emptyset, \Omega\}$ otherwise, and the measure of the measure of $\Omega$ is $\mu$ if $A \subseteq \mathcal{P}(\Omega)$ and $\mu$ is a measure on the sigma-algebra generated by $A$, and $0$ otherwise.

If $A$ is a subset of the power set of $\Omega$, then the sets of the measure space $(\Omega, A, \mu)$ are the $\sigma$-algebras generated by $A$, and the space of the measure space is $\Omega$.

The space $\Omega$ is in the $\sigma$-algebra of the measure space $(\Omega, M, \mu)$.

The sets of a measure space are the same as the sets of the measure.

The space of a measure space is the same as the original space.

If $M$ is a $\sigma$-algebra on $\Omega$ and $M'$ is a sub-$\sigma$-algebra of $M$, then the $\sigma$-algebra generated by $M'$ is a sub-$\sigma$-algebra of the $\sigma$-algebra generated by $M$.

The measure of a $\sigma$-algebra is zero.

If $A$ is a $\sigma$-algebra on $C$ and $B$ is a subset of $D$, then the $\sigma$-algebra generated by $A$ and $B$ is a subset of the $\sigma$-algebra generated by $C$ and $D$.

If $M$ is a $\sigma$-algebra of subsets of $\Omega$ and $\mu$ is a measure on $M$, then $A \in M$ implies $A \in \mathcal{M}$.

A $\sigma$-algebra is empty if and only if it contains only the empty set.

If $M$ is the measure on $\Omega$ induced by $\mu$, then the outer measure of $X$ with respect to $M$ is equal to $\mu(X)$.

If $\mu$ is a measure on a $\sigma$-algebra $M$, then the measure of any set $A \in M$ is equal to the outer measure of $A$.

Every measure can be written as a triple $(\Omega, A, \<mu>)$ where $\Omega$ is a set, $A$ is a $\sigma$-algebra on $\Omega$, and $\mu$ is a measure on $A$.

If the sets of a topological space $A$ are a subset of the sets of a topological space $B$, then the space of $A$ is a subset of the space of $B$.

If two measure spaces have the same measurable sets, then they are equal.

If a set $A$ is not in the $\sigma$-algebra of a measure space $(X, \mathcal{A}, \mu)$, then $\mu(A) = 0$.

If the measure of a set $A$ is not zero, then $A$ is a measurable set.

If $A$ is not a measurable set, then its measure is $0$.

If two measures agree on all measurable sets, then they are equal.

If two $\sigma$-algebras are equal, then the corresponding $\sigma$-fields are equal.

If $f$ is a function from a measurable space $(X, \mathcal{A})$ to a measurable space $(Y, \mathcal{B})$, then $f$ is measurable if $f^{-1}(B) \in \mathcal{A}$ for all $B \in \mathcal{B}$.

If $f$ is a measurable function from a measurable space $M$ to a measurable space $A$, then for any $x \in M$, $f(x) \in A$.

If $f$ is a measurable function from a measure space $(M, \mathcal{M})$ to a measurable space $(A, \mathcal{A})$, then the preimage of any measurable set in $A$ is measurable in $M$.

If $f$ is a measurable function from $M$ to $N$, and $P$ is a measurable subset of $N$, then the preimage of $P$ under $f$ is a measurable subset of $M$.

If $f$ is a function from a measurable space $(X, \mathcal{M})$ to a measurable space $(Y, \mathcal{N})$, and if $\mathcal{N}$ is generated by a collection of subsets of $Y$, then $f$ is measurable if and only if $f^{-1}(A) \in \mathcal{M}$ for every $A \in \mathcal{N}$.

If $f$ is a function from a measurable space $(X, \mathcal{M})$ to a set $\Omega$ and $N$ is a $\sigma$-algebra on $\Omega$, then $f$ is measurable with respect to the measure space $(\Omega, N, \mu)$ if and only if $f^{-1}(y) \in \mathcal{M}$ for all $y \in N$.

A function $f$ is measurable with respect to the measure space $(\Omega, \mathcal{N}, \mu)$ if and only if for every $A \in \mathcal{N}$, the preimage $f^{-1}(A)$ is measurable with respect to the original measure space $(\Omega, \mathcal{M}, \mu)$.

If two $\sigma$-algebras are equal, then a function is measurable with respect to one if and only if it is measurable with respect to the other.

If two functions are equal almost everywhere, then they are measurable if and only if the other is measurable.

If two functions are equal almost everywhere, then they are measurable if and only if the other is measurable.

If two measurable spaces are equal, and two functions are equal on the space of the first measurable space, then the two functions are measurable with respect to the two measurable spaces if and only if they are measurable with respect to the two measurable spaces.

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 $L$.

If $f$ is a measurable function from $M$ to $N$ and $g$ is a measurable function from $N$ to $L$, then $g \circ f$ is a measurable function from $M$ to $L$.

If $c$ is in the space of the measure $M'$, then the constant function $x \mapsto c$ is measurable with respect to $M$.

The identity function is measurable.

The identity function is measurable.

If two $\sigma$-algebras are equal, then the identity function is measurable.

If $P_i$ is a sequence of measurable sets, then the set $\{x \in X : \forall i, P_i(x)\}$ is measurable.

If $M'$ is a subset of the power set of $\Omega$ and $M$ is a subset of $M'$, then the measurable sets of the measure space $(\Omega, M, \mu)$ are a subset of the measurable sets of the measure space $(\Omega, M', \mu')$.

The space of a countable space is the set of points, and the sets of a countable space are the powerset of the set of points.

A function $f$ is measurable from a countable space $A$ to a measurable space $M$ if and only if $f$ is a function from $A$ to $M$.

If $f_i$ is a measurable function for each $i \in I$ and $g$ is a measurable function from $M$ to the countable space $I$, then the composition $f_i \circ g$ is measurable.

If $A$ is a countable set, then a function $f$ is measurable from a measurable space $M$ to the countable space $A$ if and only if $f$ is a function from $M$ to $A$ and the preimage of every element of $A$ is measurable.

If $A$ is a finite set, then a function $f$ is measurable with respect to the counting measure on $A$ if and only if $f$ is a function from the underlying space to $A$ and the preimage of each element of $A$ is measurable.

A function $f$ is measurable if and only if it is a function from the space $M$ to a countable set $A$ and the preimage of every element of $A$ is measurable.

If $f_i$ is a measurable function from $M$ to $N$ for each $i \in \mathbb{N}$, and $g$ is a measurable function from $M$ to $\mathbb{N}$, then the function $x \mapsto f_{g(x)}(x)$ is measurable.

The constant function $x \mapsto c$ is measurable.

The identity function on a countable set is measurable.

If $f$ is a measurable function from $L$ to $N$ and $g$ is a measurable function from $M$ to $L$, then the composition $f \circ g$ is a measurable function from $M$ to $N$.

If the null set is measurable, then the null set is measurable if and only if the null set is measurable.

If $G$ is a function from $I$ to the power set of $\Omega$, then the space of the measure $\mu$ extended by $G$ is $\Omega$.

If $G$ is a function from $I$ to the power set of $\Omega$, then the sets of the measure space $(\Omega, \mathcal{F}, \mu)$ are the same as the sets of the measure space $(\Omega, \mathcal{F}, \mu)$ extended by $G$.

If $\mu'$ is a measure on a $\sigma$-algebra $\mathcal{M}$ that extends a measure $\mu$ on a sub-$\sigma$-algebra $\mathcal{I}$, then $\mu'$ and $\mu$ agree on $\mathcal{I}$.

If $\mu'$ is a measure on a set $\Omega$ and $\mu$ is a measure on a set $G$ such that $\mu'$ and $\mu$ agree on $G$, then the measure $\mu$ can be extended to a measure $\mu'$ on $\Omega$.

The space of a vimage algebra is the space of the original algebra.

The sets of a vimage algebra are the preimages of the sets of the original algebra.

If $f$ is a function from $X$ to the measurable space $M$, then the sets of the vimage algebra of $f$ are the preimages of the sets of $M$ under $f$.

If two measure spaces have the same sets, then the vimage algebras of the two measure spaces have the same sets.

If $X = Y$, $f = g$, and $M = N$, then $f^{-1}(M) = g^{-1}(N)$.

If $A$ is a measurable set, then $f^{-1}(A) \cap X$ is a measurable set in the vimage algebra of $X$ under $f$.

If $f$ is a measurable function from $N$ to $M$, then the image of $N$ under $f$ is a measurable subset of $N$.

If $f$ is a function from $X$ to the measurable space $M$, then $f$ is measurable with respect to the vimage algebra of $f$ on $M$.

If $g$ is a function from a measurable space $N$ to a measurable space $X$ and $f$ is a function from $X$ to a measurable space $M$, then $g$ is measurable with respect to the algebra generated by $f$.

If $f$ is a function from $\Omega$ to $\Omega'$, then the preimage of a $\sigma$-algebra on $\Omega'$ is a $\sigma$-algebra on $\Omega$.

If $f: X \to Y$ and $g: Y \to M$ are functions, then the algebra generated by the image of $f$ in the algebra generated by the image of $g$ is the same as the algebra generated by the image of the composition $g \circ f$.

The space of a subspace is the intersection of the original space with the subspace.

If $\Omega$ is a measurable set, then the space of the restriction of $M$ to $\Omega$ is $\Omega$.

The sets of a subspace are the intersections of the sets of the original space with the subspace.