Proof of Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps

Preimages are those of The Image and the Preimage of a Class under a Class (ii), and sequences are as in Sequences (i).

Proof of (i). By Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (iii), $\mathcal{C}\subseteq\sigma(\mathcal{C})=\mathcal{G}$, so if $f$ is $\mathcal{F}/\mathcal{G}$-measurable then $f^{-1}(C)\in\mathcal{F}$ for every $C\in\mathcal{C}$, by Measurable Map, and the Set of Measurable Maps (i). Conversely, suppose $f^{-1}(C)\in\mathcal{F}$ for every $C\in\mathcal{C}$, and let $\mathcal{D}=\{B\subseteq Y:f^{-1}(B)\in\mathcal{F}\}$, a set by Preimages under a Map: Membership, the Preimage Map, Sequences of Preimages, and the Set of Sets with Preimage in a Given Set (v). We check that $\mathcal{D}$ is a $\sigma$-algebra on $Y$. It is a subset of $\mathcal{P}(Y)$. $Y\in\mathcal{D}$, since $f^{-1}(Y)=X$ by Preimages of Unions, Intersections and Complements (iv) and $X\in\mathcal{F}$ by σ-Algebra and Measurable Space (i). If $B\in\mathcal{D}$, then $Y\setminus B\subseteq Y$ and $f^{-1}(Y\setminus B)=X\setminus f^{-1}(B)$ by Preimages of Unions, Intersections and Complements (iii), which lies in $\mathcal{F}$ by σ-Algebra and Measurable Space (i), so $Y\setminus B\in\mathcal{D}$. If $(B_{k})$ is a sequence in $\mathcal{D}$, then $k\mapsto f^{-1}(B_{k})$ is a sequence of subsets of $X$ by Preimages under a Map: Membership, the Preimage Map, Sequences of Preimages, and the Set of Sets with Preimage in a Given Set (iv), and $f^{-1}\bigl(\bigcup_{k}B_{k}\bigr)=\bigcup_{k}f^{-1}(B_{k})$ by Preimages of Unions, Intersections and Complements (i). The terms of this sequence lie in $\mathcal{F}$, so it is a sequence in $\mathcal{F}$ (Sequences and Maps of Sets: Complements, Traces, Two-Term Sequences, the Disjointing Sequence, and Sequences in a Subset (v)), and its union lies in $\mathcal{F}$ by σ-Algebra and Measurable Space (i). As $\bigcup_{k}B_{k}\subseteq Y$ (Indexed Families of Sets and Their Union, Intersection and Product (ii)), $\bigcup_{k}B_{k}\in\mathcal{D}$. So $\mathcal{D}$ is a $\sigma$-algebra on $Y$, and $\mathcal{C}\subseteq\mathcal{D}$ by the hypothesis. By Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (iv), $\mathcal{G}=\sigma(\mathcal{C})\subseteq\mathcal{D}$, that is, $f^{-1}(B)\in\mathcal{F}$ for every $B\in\mathcal{G}$, and $f$ is $\mathcal{F}/\mathcal{G}$-measurable by Measurable Map, and the Set of Measurable Maps (i).

Proof of (ii). Let $C\in\mathcal{H}$. For $x\in X$, $(g\circ f)(x)=g(f(x))$ (Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (ii)), so by Preimages under a Map: Membership, the Preimage Map, Sequences of Preimages, and the Set of Sets with Preimage in a Given Set (ii), applied to $g\circ f$, to $g$ and to $f$: $x\in(g\circ f)^{-1}(C)$ if and only if $g(f(x))\in C$, if and only if $f(x)\in g^{-1}(C)$, if and only if $x\in f^{-1}(g^{-1}(C))$. Both sets are subsets of $X$, so $(g\circ f)^{-1}(C)=f^{-1}(g^{-1}(C))$. Since $g$ is $\mathcal{G}/\mathcal{H}$-measurable, $g^{-1}(C)\in\mathcal{G}$, and since $f$ is $\mathcal{F}/\mathcal{G}$-measurable, $f^{-1}(g^{-1}(C))\in\mathcal{F}$. By Measurable Map, and the Set of Measurable Maps (i), $g\circ f$ is $\mathcal{F}/\mathcal{H}$-measurable.

Proof of (iii). Let $c:X\to Y$ be the map with value $y_{0}$ at every point, and let $B\in\mathcal{G}$. By Preimages under a Map: Membership, the Preimage Map, Sequences of Preimages, and the Set of Sets with Preimage in a Given Set (ii), $c^{-1}(B)=X$ if $y_{0}\in B$, and $c^{-1}(B)$ has no element, so equals $\emptyset$, if $y_{0}\notin B$. In the first case $c^{-1}(B)\in\mathcal{F}$ by σ-Algebra and Measurable Space (i), in the second by Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (vi). For the identity, $\mathrm{id}_{X}(x)=x$ for $x\in X$ by Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (iii), so by Preimages under a Map: Membership, the Preimage Map, Sequences of Preimages, and the Set of Sets with Preimage in a Given Set (ii) $\mathrm{id}_{X}^{-1}(B)=B$ for every $B\in\mathcal{F}$, which lies in $\mathcal{F}$. So $c$ is $\mathcal{F}/\mathcal{G}$-measurable and $\mathrm{id}_{X}$ is $\mathcal{F}/\mathcal{F}$-measurable, by Measurable Map, and the Set of Measurable Maps (i).

Proof of (iv). Let $\mathcal{T}_{d}$ and $\mathcal{T}_{e}$ be the sets of open subsets of $(M,d)$ and of $(N,e)$, so that $\mathcal{B}(M,d)=\sigma(\mathcal{T}_{d})$ and $\mathcal{B}(N,e)=\sigma(\mathcal{T}_{e})$ by Borel Sigma-Algebra of a Metric Space (i); $(M,\mathcal{B}(M,d))$ and $(N,\mathcal{B}(N,e))$ are measurable spaces, as stated in (iv). Let $V\in\mathcal{T}_{e}$. Since $h$ is continuous, $h^{-1}(V)$ is open in $(M,d)$ by Continuity of a Map Between Metric Spaces via Preimages of Open Sets (i), hence $h^{-1}(V)\in\mathcal{B}(M,d)$ by Borel Sets of a Metric Space: Open and Closed Sets, Points, Open Sets of Subspaces, and Subspaces (i). By (i), applied to $(M,\mathcal{B}(M,d))$, $(N,\mathcal{B}(N,e))$ and $\mathcal{C}=\mathcal{T}_{e}$, $h$ is $\mathcal{B}(M,d)/\mathcal{B}(N,e)$-measurable.

Proof of (v). For $a\in\mathbb{R}$ let the rays $(a,\infty)$, $[a,\infty)$, $(-\infty,a)$ and $(-\infty,a]$ be those of Intervals of the Real Line (iii); by Preimages under a Map: Membership, the Preimage Map, Sequences of Preimages, and the Set of Sets with Preimage in a Given Set (ii) and Intervals of the Real Line (iii),

$$f^{-1}((a,\infty))=\{x\in X:f(x)>a\},\quad f^{-1}([a,\infty))=\{x\in X:f(x)\ge a\},$$ $$f^{-1}((-\infty,a))=\{x\in X:f(x)<a\},\quad f^{-1}((-\infty,a])=\{x\in X:f(x)\le a\}.$$

Let $\mathcal{C}_{1}=\{(a,\infty):a\in\mathbb{R}\}$, $\mathcal{C}_{2}=\{[a,\infty):a\in\mathbb{R}\}$, $\mathcal{C}_{3}=\{(-\infty,a):a\in\mathbb{R}\}$ and $\mathcal{C}_{4}=\{(-\infty,a]:a\in\mathbb{R}\}$; each is a subset of the power set $\mathcal{P}(\mathbb{R})$ by The Families of Rays of the Real Line Are Sets of Subsets of the Real Line (i). Each of them generates $\mathcal{B}(\mathbb{R})$: $\mathcal{B}(\mathbb{R})=\sigma(\mathcal{C}_{1})=\sigma(\mathcal{C}_{4})$ by Intervals Are Borel, and Rays, Half-Open Intervals and Rational Intervals Generate the Borel Sigma-Algebra of the Real Line (iii), $\mathcal{B}(\mathbb{R})=\sigma(\mathcal{C}_{2})$ by The Borel σ-Algebra of the Real Line Is Generated by the Closed Rays, and by the Open Left Rays (i), and $\mathcal{B}(\mathbb{R})=\sigma(\mathcal{C}_{3})$ by The Borel σ-Algebra of the Real Line Is Generated by the Closed Rays, and by the Open Left Rays (ii); in particular $\mathcal{B}(\mathbb{R})$ is a $\sigma$-algebra on $\mathbb{R}$ by Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (iii), and $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ is a measurable space. By Measurable Map, and the Set of Measurable Maps (iii), (a) says that $f$ is $\mathcal{F}/\mathcal{B}(\mathbb{R})$-measurable. For each $r\in\{1,2,3,4\}$, (i), applied with $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ in place of $(Y,\mathcal{G})$ and $\mathcal{C}=\mathcal{C}_{r}$, shows that (a) holds if and only if $f^{-1}(C)\in\mathcal{F}$ for every $C\in\mathcal{C}_{r}$. By The Families of Rays of the Real Line Are Sets of Subsets of the Real Line (i), the members of $\mathcal{C}_{r}$ are exactly the rays of the corresponding form with endpoint $a\in\mathbb{R}$. So, by the displayed identities, this last condition is (b) for $r=1$, (c) for $r=2$, (d) for $r=3$ and (e) for $r=4$. So each of (b), (c), (d) and (e) is equivalent to (a), and all five conditions are equivalent.

Proof of (vi). For $c\in\mathbb{R}$ let $(c,\infty]=\{t\in[0,\infty]:c<t\}$ and let $\mathcal{R}$ be the set of these rays, a subset of $\mathcal{P}([0,\infty])$, as in The Borel σ-Algebra of [0,∞] (i); then $\mathcal{B}([0,\infty])=\sigma(\mathcal{R})$ by The Borel σ-Algebra of [0,∞] (ii), and $([0,\infty],\mathcal{B}([0,\infty]))$ is a measurable space by Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (iii). For $a\in\mathbb{R}$, Preimages under a Map: Membership, the Preimage Map, Sequences of Preimages, and the Set of Sets with Preimage in a Given Set (ii) gives $f^{-1}((a,\infty])=\{x\in X:f(x)>a\}$. If $f$ is measurable as a map into $[0,\infty]$, that is, $\mathcal{F}/\mathcal{B}([0,\infty])$-measurable by Measurable Map, and the Set of Measurable Maps (vi), then, as $(a,\infty]\in\mathcal{R}\subseteq\mathcal{B}([0,\infty])$ by Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (iii), $\{x\in X:f(x)>a\}=f^{-1}((a,\infty])\in\mathcal{F}$. Conversely, if $\{x\in X:f(x)>a\}\in\mathcal{F}$ for every $a\in\mathbb{R}$, then every member of $\mathcal{R}$ has its preimage in $\mathcal{F}$, so $f$ is $\mathcal{F}/\mathcal{B}([0,\infty])$-measurable by (i), that is, measurable as a map into $[0,\infty]$ by Measurable Map, and the Set of Measurable Maps (vi).

Proof of (vii). Let $a\in\mathbb{R}$ and $x\in X$. Since $a\in\mathbb{R}$ and $f(x)\in\mathbb{R}$, $f(x)>a$ holds in $[-\infty,\infty]$ if and only if it holds in $\mathbb{R}$, by Order and Suprema in the Extended Real Line (i). Hence the set $\{x\in X:f(x)>a\}$ is the same whether $>$ is read in $\mathbb{R}$ or in $[-\infty,\infty]$, for every $a\in\mathbb{R}$. By (v), $f$ is measurable as a map into $\mathbb{R}$ if and only if these sets lie in $\mathcal{F}$ for every $a\in\mathbb{R}$, and by (vi), applied to $f$ as a map into $[0,\infty]$, the same holds for $f$ measurable as a map into $[0,\infty]$.

Proof of (viii). By Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (ix), $f|_{A}$ is a map $A\to Y$ with $(f|_{A})(x)=f(x)$ for $x\in A$, and $(A,\mathcal{F}_{A})$ is a measurable space by Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (xi). Let $B\in\mathcal{G}$. By Preimages under a Map: Membership, the Preimage Map, Sequences of Preimages, and the Set of Sets with Preimage in a Given Set (ii), applied to $f|_{A}$ and to $f$, $(f|_{A})^{-1}(B)$ consists of the $x\in A$ with $f(x)\in B$, so $(f|_{A})^{-1}(B)=f^{-1}(B)\cap A$. Here $f^{-1}(B)\in\mathcal{F}$ as $f$ is $\mathcal{F}/\mathcal{G}$-measurable, so $f^{-1}(B)\cap A\in\mathcal{F}$ by Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (ix); it is a subset of $A$, so it lies in $\mathcal{F}_{A}$. By Measurable Map, and the Set of Measurable Maps (i), $f|_{A}$ is $\mathcal{F}_{A}/\mathcal{G}$-measurable.