Proof of The Metric Subspace: Continuity, the Borel Sigma-Algebra, and the Restriction of a Borel Measure

Each result cited below is universally quantified over the data in its own statement, and is applied to the data named where it is used.

Claim 1(a). By Continuous Map Between Metric Spaces, read with the ambient metric space $(X,d)$, the subset $A$ and the map $f$, the first assertion says: for every $x\in A$ and every positive real $\varepsilon$ there is a positive real $\delta$ such that every $y\in A$ with $d(x,y)<\delta$ satisfies $d_{Y}(f(y),f(x))<\varepsilon$. Read instead with the ambient metric space $(A,d_{A})$, the subset $A$ of itself and the same map $f$, the same definition gives the second assertion: for every $x\in A$ and every positive real $\varepsilon$ there is a positive real $\delta$ such that every $y\in A$ with $d_{A}(x,y)<\delta$ satisfies $d_{Y}(f(y),f(x))<\varepsilon$. Since $d_{A}$ is the restriction of $d$, one has $d_{A}(x,y)=d(x,y)$ for all $x,y\in A$, so the two displayed conditions are the same condition on $f$, and each assertion holds exactly when the other does.

Claim 1(b). Likewise, by Continuous Map Between Metric Spaces the continuity of $u$ on $A$ as a map into $(Y,d_{Y})$ says: for every $x\in A$ and every positive real $\varepsilon$ there is a positive real $\delta$ such that every $y\in A$ with $d(x,y)<\delta$ satisfies $d_{Y}(u(y),u(x))<\varepsilon$; and the continuity of $u_{C}$ on $A$ as a map into $(C,d_{C})$ says the same with $d_{C}(u_{C}(y),u_{C}(x))$ in place of $d_{Y}(u(y),u(x))$. Since $u(x)\in C$ for every $x\in A$ and $d_{C}$ is the restriction of $d_{Y}$, one has $d_{C}(u_{C}(y),u_{C}(x))=d_{Y}(u(y),u(x))$ for all $x,y\in A$, so again the two conditions coincide.

Claim 2. Put $\mathcal{S}=\{B\cap A:B\in\mathcal{B}(X)\}$, a collection of subsets of $A$.

$\mathcal{S}$ is a $\sigma$-algebra on $A$. The requirements of σ-Algebra and Measurable Space hold: $A=X\cap A$ and $X\in\mathcal{B}(X)$; for $B\in\mathcal{B}(X)$ one has $A\setminus(B\cap A)=(X\setminus B)\cap A$ with $X\setminus B\in\mathcal{B}(X)$; and for a sequence $(B_{k})_{k\in\mathbb{N}}$ in $\mathcal{B}(X)$ one has $\bigcup_{k\in\mathbb{N}}(B_{k}\cap A)=\bigl(\bigcup_{k\in\mathbb{N}}B_{k}\bigr)\cap A$ with $\bigcup_{k\in\mathbb{N}}B_{k}\in\mathcal{B}(X)$, the three properties of $\mathcal{B}(X)$ used here being the same requirements for the $\sigma$-algebra $\mathcal{B}(X)$.

$\mathcal{B}(A)\subseteq\mathcal{S}$. Let $V\subseteq A$ be open in the metric space $(A,d_{A})$. By claim 2 of The Restriction of a Metric to a Subset Induces the Subspace Topology there is $U\in\mathcal{T}_{d}$ with $V=A\cap U$, and $U\in\mathcal{B}(X)$ by claim 1 of Borel Measurability and Bounded Integration on a Metric Space applied to $(X,d)$; hence $V\in\mathcal{S}$. By Borel Sigma-Algebra of a Metric Space, $\mathcal{B}(A)$ is the $\sigma$-algebra generated by the subsets of $A$ open in $(A,d_{A})$; since $\mathcal{S}$ is a $\sigma$-algebra on $A$ containing all of them, Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra gives $\mathcal{B}(A)\subseteq\mathcal{S}$.

$\mathcal{S}\subseteq\mathcal{B}(A)$. Put $\mathcal{D}=\{B\subseteq X:B\cap A\in\mathcal{B}(A)\}$. Then $\mathcal{D}$ is a $\sigma$-algebra on $X$: $X\cap A=A\in\mathcal{B}(A)$; if $B\in\mathcal{D}$ then $(X\setminus B)\cap A=A\setminus(B\cap A)\in\mathcal{B}(A)$; and if $(B_{k})_{k\in\mathbb{N}}$ is a sequence in $\mathcal{D}$ then $\bigl(\bigcup_{k\in\mathbb{N}}B_{k}\bigr)\cap A=\bigcup_{k\in\mathbb{N}}(B_{k}\cap A)\in\mathcal{B}(A)$, again by the requirements of σ-Algebra and Measurable Space for $\mathcal{B}(A)$. Moreover every $U\in\mathcal{T}_{d}$ belongs to $\mathcal{D}$: the set $U\cap A$ is open in $(A,d_{A})$ by claim 2 of The Restriction of a Metric to a Subset Induces the Subspace Topology, hence lies in $\mathcal{B}(A)$ by claim 1 of Borel Measurability and Bounded Integration on a Metric Space applied to the metric space $(A,d_{A})$. Since $\mathcal{B}(X)$ is the $\sigma$-algebra generated by $\mathcal{T}_{d}$, Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra gives $\mathcal{B}(X)\subseteq\mathcal{D}$, which says precisely that $B\cap A\in\mathcal{B}(A)$ for every $B\in\mathcal{B}(X)$, that is, $\mathcal{S}\subseteq\mathcal{B}(A)$.

The two inclusions give $\mathcal{B}(A)=\mathcal{S}$.

Claim 3. Assume $A\in\mathcal{B}(X)$.

If $B\in\mathcal{B}(X)$ and $B\subseteq A$, then $B=B\cap A\in\mathcal{B}(A)$ by claim 2. Conversely let $C\in\mathcal{B}(A)$. By claim 2 there is $B\in\mathcal{B}(X)$ with $C=B\cap A$; then $C\subseteq A$, and $C=X\setminus\bigl((X\setminus B)\cup(X\setminus A)\bigr)$ belongs to $\mathcal{B}(X)$ because $\mathcal{B}(X)$ contains $B$ and $A$ and is closed under complements and countable unions by σ-Algebra and Measurable Space, a union of two members being the union of a sequence in which they alternate. Hence $\mathcal{B}(A)=\{B\in\mathcal{B}(X):B\subseteq A\}$, which is the family written $\mathcal{B}(X)|_{A}$ in claim 1 of Assembly of Measure Spaces: Restriction, Transport, One-Point Spaces, and Countable Disjoint Unions, read for a measure space of the form $(X,\mathcal{B}(X),\mu)$ and the set $A\in\mathcal{B}(X)$; the description of that family in the cited claim involves only $X$, the $\sigma$-algebra and $A$, not the measure.

Let now $\mu$ be a Borel measure on $(X,d)$, so that $(X,\mathcal{B}(X),\mu)$ is a measure space by Borel Measure on a Metric Space. By claim 1 of Assembly of Measure Spaces: Restriction, Transport, One-Point Spaces, and Countable Disjoint Unions, setting $\mu|_{A}(B)=\mu(B)$ for $B\in\mathcal{B}(X)|_{A}$ defines a measure on $(A,\mathcal{B}(X)|_{A})$. Since $\mathcal{B}(X)|_{A}=\mathcal{B}(A)$, this is a measure on $(A,\mathcal{B}(A))$, that is, a Borel measure on $(A,d_{A})$ by Borel Measure on a Metric Space. Taking $B=A$, which lies in $\mathcal{B}(A)$, gives $\mu|_{A}(A)=\mu(A)$.

Claim 4. Assume $A\in\mathcal{B}(X)$ and let $\mu$ and $f$ be as stated.

Measurability of $f|_{A}$. For $B\in\mathcal{B}(\mathbb{R})$ one has $(f|_{A})^{-1}(B)=f^{-1}(B)\cap A$, and $f^{-1}(B)\in\mathcal{B}(X)$ by the measurability of $f$, so $(f|_{A})^{-1}(B)\in\mathcal{B}(A)$ by claim 2.

The positive and negative parts. Let $f^{+}$ and $f^{-}$ be the positive and negative parts of $f$ in the sense of Integrable Functions and Their Lebesgue Integral; by that item they are measurable with respect to $\mathcal{B}(X)$, they are nonnegative, and $f=f^{+}-f^{-}$. For $x\in X\setminus A$ we have $f(x)=0$, hence $f^{+}(x)=\max\{0,0\}=0$ and $f^{-}(x)=\max\{0,0\}=0$. Also, directly from the defining formulas of that item, $(f|_{A})^{+}=(f^{+})|_{A}$ and $(f|_{A})^{-}=(f^{-})|_{A}$ as functions on $A$.

Fix one of the two signs and write $u$ for the corresponding part, so that $u:X\to\mathbb{R}$ is nonnegative, measurable with respect to $\mathcal{B}(X)$, and vanishes on $X\setminus A$; read as a map into $[0,\infty]$ it is measurable in the sense of Measurable Functions and the Lebesgue Integral: Standing Notation (iii), the two readings agreeing there. Then $u$ is exactly the zero extension of $u|_{A}$ in the sense of claim 1 of Assembly of Measure Spaces: Restriction, Transport, One-Point Spaces, and Countable Disjoint Unions: the two functions agree on $A$ by definition and both vanish off $A$. Applying that claim to the measure space $(X,\mathcal{B}(X),\mu)$ and the set $A$, whose restriction is the measure space $(A,\mathcal{B}(A),\mu|_{A})$ by claim 3, the function $u|_{A}$ is measurable with respect to $\mathcal{B}(A)$ and

$$\int_{A}u|_{A}\,d(\mu|_{A})=\int_{X}u\,d\mu .$$

Conclusion. Since $f$ is integrable with respect to $\mu$, both $\int_{X}f^{+}\,d\mu$ and $\int_{X}f^{-}\,d\mu$ are finite by Integrable Functions and Their Lebesgue Integral. By the displayed identity applied to each sign, together with $(f|_{A})^{+}=(f^{+})|_{A}$ and $(f|_{A})^{-}=(f^{-})|_{A}$, both $\int_{A}(f|_{A})^{+}\,d(\mu|_{A})$ and $\int_{A}(f|_{A})^{-}\,d(\mu|_{A})$ are finite, so $f|_{A}$ is integrable with respect to $\mu|_{A}$ by Integrable Functions and Their Lebesgue Integral, and

$$\int_{A}f|_{A}\,d(\mu|_{A})=\int_{A}(f|_{A})^{+}\,d(\mu|_{A})-\int_{A}(f|_{A})^{-}\,d(\mu|_{A})=\int_{X}f^{+}\,d\mu-\int_{X}f^{-}\,d\mu=\int_{X}f\,d\mu ,$$

the first and the last equality being the definition of the integral of an integrable function in Integrable Functions and Their Lebesgue Integral.