Proof of Borel Measurability and Bounded Integration on a Metric Space

Preliminary. Let $V,W$ be sets, let $h:V\to W$ be a map and let $\mathcal{A}$ be a $\sigma$-algebra on $V$. The family

is a $\sigma$-algebra on $W$: we have $h^{-1}(W)=V\in\mathcal{A}$; $h^{-1}(W\setminus B)=V\setminus h^{-1}(B)$, so $\mathcal{A}$ being closed under complements makes $h_{*}\mathcal{A}$ closed under complements; and $h^{-1}\bigl(\bigcup_{m}B_m\bigr)=\bigcup_{m}h^{-1}(B_m)$, so $h_{*}\mathcal{A}$ is closed under countable unions. Consequently, if $\mathcal{E}$ is a family of subsets of $W$ with $h^{-1}(E)\in\mathcal{A}$ for every $E\in\mathcal{E}$, then $\mathcal{E}\subseteq h_{*}\mathcal{A}$, and the minimality of the generated $\sigma$-algebra (claim 2 of Intersections of Sigma-Algebras and Minimality of the Generated Sigma-Algebra) gives $\sigma(\mathcal{E})\subseteq h_{*}\mathcal{A}$; that is, $h$ is measurable with respect to $\mathcal{A}$ and $\sigma(\mathcal{E})$. We refer to this as the generator criterion.

Claim 1. By claim 2 of Intersections of Sigma-Algebras and Minimality of the Generated Sigma-Algebra, $\mathcal{B}(X)$ contains the family $\mathcal{T}_d$ that generates it, so every open subset of $(X,d)$ is a Borel subset. If $F$ is closed then $X\setminus F$ is open by Closed Subset of a Topological Space, hence lies in $\mathcal{B}(X)$, and therefore so does $F=X\setminus(X\setminus F)$.

The family $\mathcal{C}$ is a $\pi$-system in the sense fixed in Dynkin's Pi-Lambda Theorem: it is nonempty, since $X$ is closed ($X\setminus X=\varnothing$ is open); and if $F_1,F_2\in\mathcal{C}$ then

is a union of two members of $\mathcal{T}_d$, hence a member of $\mathcal{T}_d$ because $\mathcal{T}_d$ is a topology, so $F_1\cap F_2\in\mathcal{C}$.

Finally, $\mathcal{C}\subseteq\mathcal{B}(X)$ gives $\sigma(\mathcal{C})\subseteq\mathcal{B}(X)$ by minimality. Conversely each $U\in\mathcal{T}_d$ satisfies $U=X\setminus(X\setminus U)$ with $X\setminus U\in\mathcal{C}$, so $U\in\sigma(\mathcal{C})$; thus $\mathcal{T}_d\subseteq\sigma(\mathcal{C})$ and minimality gives $\mathcal{B}(X)=\sigma(\mathcal{T}_d)\subseteq\sigma(\mathcal{C})$. Hence $\sigma(\mathcal{C})=\mathcal{B}(X)$.

Claim 2. Under the identification of $\mathbb{R}$ with $\mathbb{R}^{1}$ used in Borel Sigma-Algebra on the Real Line, the Euclidean distance on $\mathbb{R}^{1}$ is $d_{\mathbb{R}}$ by The Euclidean Distance on the Real Line is the Absolute Value Metric. By Euclidean Openness Agrees with Metric Openness on $\mathbb{R}^n$, a subset of $\mathbb{R}$ is open in the Euclidean sense if and only if it is open in the metric space $(\mathbb{R},d_{\mathbb{R}})$. The Borel $\sigma$-algebra of $(\mathbb{R},d_{\mathbb{R}})$ and $\mathcal{B}(\mathbb{R})$ are the $\sigma$-algebras generated by these two families, which coincide; hence the two $\sigma$-algebras are equal.

Claim 3. Let $V\subseteq Y$ be open in $(Y,d_Y)$. By Continuity of a Map Between Metric Spaces via Preimages of Open Sets, $f^{-1}(V)$ is open in $(X,d)$, hence lies in $\mathcal{B}(X)$ by claim 1. The open subsets of $(Y,d_Y)$ generate $\mathcal{B}(Y)$, so the generator criterion gives that $f$ is measurable with respect to $\mathcal{B}(X)$ and $\mathcal{B}(Y)$.

For the distance function, let $A\subseteq X$ be nonempty. By claim 5 of The Distance to a Set is Nonexpansive, the map $x\mapsto\operatorname{dist}_d(x,A)$ is continuous from $(X,d)$ to $(\mathbb{R},d_{\mathbb{R}})$, so by the first part it is measurable with respect to $\mathcal{B}(X)$ and the Borel $\sigma$-algebra of $(\mathbb{R},d_{\mathbb{R}})$, which is $\mathcal{B}(\mathbb{R})$ by claim 2.

Claim 4. Let $C\in\mathcal{G}$. Then

where $g^{-1}(C)\in\mathcal{B}(X)$ because $g$ is measurable, and therefore $Y^{-1}(g^{-1}(C))\in\mathcal{F}$ because $Y$ is measurable. Hence $g\circ Y$ is measurable with respect to $\mathcal{F}$ and $\mathcal{G}$.

Claim 5. Suppose first that $u$ is lower semicontinuous on $X$. Applying claim 2 of Semicontinuity via Sublevel and Superlevel Sets with the subset $A=X$, for which the subspace metric is $d$ itself and the subspace topology is $\mathcal{T}_d$, the set

is open in $(X,d)$, hence lies in $\mathcal{B}(X)$ by claim 1, for every real $c$. By claim 2 of Uniqueness of Finite Measures on a Generating Pi-System and the Density of the Exponential Law, the family $\{(c,\infty):c\in\mathbb{R}\}$ generates $\mathcal{B}(\mathbb{R})$, so the generator criterion shows that $u$ is measurable with respect to $\mathcal{B}(X)$ and $\mathcal{B}(\mathbb{R})$.

Suppose now that $u$ is upper semicontinuous on $X$. By claim 1 of Semicontinuity Under Negation and Characterization of Continuity, the pointwise negation $-u$ is lower semicontinuous on $X$, hence measurable with respect to $\mathcal{B}(X)$ and $\mathcal{B}(\mathbb{R})$ by the case just treated. The map $\varphi:\mathbb{R}\to\mathbb{R}$, $\varphi(t)=-t$, satisfies $d_{\mathbb{R}}(\varphi(s),\varphi(t))=|-s+t|=|s-t|=d_{\mathbb{R}}(s,t)$, so it is Lipschitz with constant $1$ from $(\mathbb{R},d_{\mathbb{R}})$ to itself and therefore continuous; by claims 2 and 3 applied to the metric space $(\mathbb{R},d_{\mathbb{R}})$, $\varphi$ is measurable with respect to $\mathcal{B}(\mathbb{R})$ and $\mathcal{B}(\mathbb{R})$. Since $u=\varphi\circ(-u)$, claim 4, applied with the measurable space $(X,\mathcal{B}(X))$ in the role of $(\Omega,\mathcal{F})$ and with the metric space $(\mathbb{R},d_{\mathbb{R}})$ in the role of $(X,d)$, shows that $u$ is measurable with respect to $\mathcal{B}(X)$ and $\mathcal{B}(\mathbb{R})$.

Claim 6. (a) Let $c$ be real with $0\le c$. The constant function on $X$ with value $c$ is the nonnegative simple function $c\,\mathbf{1}_X$, whose simple integral is $c\,\mu(X)$; by the agreement of the two notions of integral for nonnegative simple functions recorded in Lebesgue Integral of a Nonnegative Measurable Function, its integral as a nonnegative measurable function is also $c\,\mu(X)$, which is real because $\mu(X)<\infty$. Its positive part is itself and its negative part is the zero function, whose integral is $0$; hence by Integrable Function and the Lebesgue Integral it is integrable with $\int_X c\,d\mu=c\,\mu(X)$. If instead $c<0$, the positive part of the constant function is the zero function and its negative part is the constant function with value $-c>0$, so by the case just treated it is integrable with

(b) By the discussion in Integrable Function and the Lebesgue Integral, the positive part $f^{+}$, the negative part $f^{-}$ and $|f|=f^{+}+f^{-}$ are measurable, and $0\le f^{+}(x)\le M$, $0\le f^{-}(x)\le M$ and $0\le|f(x)|\le M$ for every $x\in X$. By the monotonicity of the integral of nonnegative measurable functions (claim 1 of Linearity and Monotonicity of the Lebesgue Integral) together with part (a),

so $f$ is integrable, and likewise $\int_X|f|\,d\mu\le M\,\mu(X)$. Claim 2 of Linearity and Monotonicity of the Lebesgue Integral gives $\bigl|\int_X f\,d\mu\bigr|\le\int_X|f|\,d\mu$, and combining the two estimates yields $\bigl|\int_X f\,d\mu\bigr|\le M\,\mu(X)$.

(c) If $0\le f(x)$ for every $x$, then $f^{+}=f$ and $f^{-}$ is the zero function, so by Integrable Function and the Lebesgue Integral,

and the right-hand side is exactly the integral of $f$ as a nonnegative measurable function in the sense of Lebesgue Integral of a Nonnegative Measurable Function. It is real because $f$ is integrable by part (b).