Proof of Inner and Outer Regularity of a Finite Borel Measure on a Metric Space, and Lipschitz Approximation of Indicators

Throughout, each result cited is universally quantified over the data appearing in its own statement and is applied to the data named here. Write $\mathcal{C}$ for the family of subsets of $X$ closed in $(X,\mathcal{T}_{d})$ and $\mathcal{U}$ for the family of subsets open in $(X,d)$; by claim 1 of Borel Measurability and Bounded Integration on a Metric Space both are contained in $\mathcal{B}(X)$, and $\mathcal{B}(X)$ is the $\sigma$-algebra generated by $\mathcal{C}$. Since $\mu(X)<\infty$, claim 2 of Basic Properties of a Measure makes $\mu(A)$ a real number with $0\le\mu(A)\le\mu(X)$ for every $A\in\mathcal{B}(X)$, and for $A\subseteq A'$ in $\mathcal{B}(X)$ claim 3 of that lemma gives $\mu(A'\setminus A)=\mu(A')-\mu(A)$. Let $\iota$ be the canonical map from $\mathbb{N}$ to $\mathbb{R}$; by claim 3 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field each $\iota(m)$ is positive, as is its inverse $\iota(m)^{-1}$. Open balls $B_{d}(x,r)$ are those of Open Ball in a Metric Space.

Step 1 (Continuity from above). Let $(A_{m})_{m\in\mathbb{N}}$ be a sequence in $\mathcal{B}(X)$ with $A_{m+1}\subseteq A_{m}$ for every $m$, and let $A=\bigcap_{m\in\mathbb{N}}A_{m}$. Writing $B_{m}=X\setminus A_{m}$, one has $A=X\setminus\bigcup_{m\in\mathbb{N}}B_{m}$, so $A\in\mathcal{B}(X)$ because a $\sigma$-algebra is closed under complements and countable unions. The sets $B_{m}$ increase, $\bigcup_{m}B_{m}=X\setminus A$, and every $\mu(B_{m})$ is real and at most $\mu(X)$, so claim 5 of Basic Properties of a Measure applies and $(\mu(B_{m}))_{m\in\mathbb{N}}$ converges to $\mu(X\setminus A)$. By claim 3 of Basic Properties of a Measure, $\mu(A_{m})=\mu(X)-\mu(B_{m})$ for every $m$ and $\mu(A)=\mu(X)-\mu(X\setminus A)$; hence, by claim 3 of Arithmetic of Limits of Real Sequences applied to the constant sequence with value $\mu(X)$ and to $(\mu(B_{m}))_{m}$, the sequence $(\mu(A_{m}))_{m\in\mathbb{N}}$ converges to $\mu(A)$.

Step 2 (A closed set contains its closure points). Let $F\in\mathcal{C}$ and let $x$ belong to the closure $\operatorname{cl}_{X}(F)$. If $x\notin F$ then $x\in X\setminus F$, which is open because $F$ is closed, so there is a positive real $\varepsilon$ with $B_{d}(x,\varepsilon)\subseteq X\setminus F$, that is, $B_{d}(x,\varepsilon)\cap F=\varnothing$; this contradicts claim 2 of Characterization of the Closure in a Metric Space by Open Balls. Hence $x\in F$.

Step 3 (Shrinking neighbourhoods of a nonempty closed set). Let $F\in\mathcal{C}$ be nonempty, let $\operatorname{dist}_{d}(\cdot,F)$ be the distance to $F$, and for $m\in\mathbb{N}$ put

Each $V_{m}$ is open: for $x\in V_{m}$ the real number $r=\iota(m)^{-1}-\operatorname{dist}_{d}(x,F)$ is positive, and every $y\in B_{d}(x,r)$ satisfies $\operatorname{dist}_{d}(y,F)\le\operatorname{dist}_{d}(x,F)+d(x,y)<\iota(m)^{-1}$ by claim 4 of The Distance to a Set is Nonexpansive together with claim 3 of Properties of the Absolute Value in an Ordered Field, so $B_{d}(x,r)\subseteq V_{m}$. Each $V_{m}$ is Borel, being open. Moreover $F\subseteq V_{m}$, since $\operatorname{dist}_{d}(x,F)=0$ for $x\in F$ by Vanishing of the Distance to a Set Characterizes the Closure; the sets decrease, since $\iota(m)\le\iota(m+1)$ by claim 6 of Properties of the Canonical Map from the Natural Numbers to an Ordered Field, and multiplying that inequality by $\iota(m)^{-1}\iota(m+1)^{-1}$, which is nonnegative by claims 7 and 5 of Elementary Order Arithmetic in an Ordered Field, gives $\iota(m+1)^{-1}\le\iota(m)^{-1}$ by claim 5 of Elementary Arithmetic in an Ordered Field; and $\bigcap_{m\in\mathbb{N}}V_{m}=F$, because a point $x$ of the intersection has $\operatorname{dist}_{d}(x,F)<\iota(m)^{-1}$ for every $m$, hence $\operatorname{dist}_{d}(x,F)=0$ by claim 3 of The Archimedean Property of the Real Numbers and the nonnegativity of the distance, hence $x\in\operatorname{cl}_{X}(F)$ by Vanishing of the Distance to a Set Characterizes the Closure and so $x\in F$ by Step 2. By Step 1, $(\mu(V_{m}))_{m\in\mathbb{N}}$ converges to $\mu(F)$.

Step 4 (Closed sets are approximable). Let $\mathcal{A}$ be the family of those $B\in\mathcal{B}(X)$ such that for every positive real $\varepsilon$ there are $F\in\mathcal{C}$ and $U\in\mathcal{U}$ with $F\subseteq B\subseteq U$ and $\mu(U\setminus F)\le\varepsilon$. Then $\mathcal{C}\subseteq\mathcal{A}$. Indeed, let $F\in\mathcal{C}$ and let $\varepsilon$ be a positive real number. If $F=\varnothing$, take $F$ itself and $U=\varnothing$, which is both closed and open by claim 1 of Complements, Unions and Intersections of Closed Sets in a Topological Space and Metric Open Sets Form a Topology, so that $\mu(U\setminus F)=\mu(\varnothing)=0\le\varepsilon$. If $F\ne\varnothing$, let $V_{m}$ be as in Step 3; since $(\mu(V_{m}))_{m}$ converges to $\mu(F)$ there is $m$ with $\mu(V_{m})-\mu(F)\le\varepsilon$, and $F\subseteq F\subseteq V_{m}$ with $\mu(V_{m}\setminus F)=\mu(V_{m})-\mu(F)\le\varepsilon$.

Step 5 (The approximable sets form a $\sigma$-algebra); claim 1. We show $\mathcal{A}$ is a $\sigma$-algebra on $X$. It contains $X$, which is closed by claim 1 of Complements, Unions and Intersections of Closed Sets in a Topological Space, by Step 4.

Complements. Let $B\in\mathcal{A}$ and let $\varepsilon$ be a positive real number; choose $F\subseteq B\subseteq U$ as in the definition of $\mathcal{A}$. Then $X\setminus U\subseteq X\setminus B\subseteq X\setminus F$, the set $X\setminus U$ is closed and $X\setminus F$ is open by Closed Subset of a Topological Space, and $(X\setminus F)\setminus(X\setminus U)=U\setminus F$, so $\mu\bigl((X\setminus F)\setminus(X\setminus U)\bigr)\le\varepsilon$. Hence $X\setminus B\in\mathcal{A}$.

Countable unions. Let $(B_{m})_{m\in\mathbb{N}}$ be a sequence in $\mathcal{A}$ and $B=\bigcup_{m}B_{m}$. Fix a positive real $\varepsilon$ first; then, for each $m$, choose $F_{m}\in\mathcal{C}$ and $U_{m}\in\mathcal{U}$ with $F_{m}\subseteq B_{m}\subseteq U_{m}$ and $\mu(U_{m}\setminus F_{m})\le\varepsilon\,(\tfrac{1}{2})^{m+1}$. Put $U=\bigcup_{m}U_{m}$, which is open by Metric Open Sets Form a Topology, and $G=\bigcup_{m}F_{m}\in\mathcal{B}(X)$. Then $F_{m}\subseteq B\subseteq U$ for every $m$, so $G\subseteq B\subseteq U$. Since $U\setminus G\subseteq\bigcup_{m}(U_{m}\setminus F_{m})$, claims 2 and 4 of Basic Properties of a Measure and claims 3 and 4 of Series of Nonnegative Real Numbers, Comparison, and the Geometric Series give

For $N\in\mathbb{N}$ let $H_{N}=\bigcup_{m=1}^{N}F_{m}$, closed by claim 2 of Complements, Unions and Intersections of Closed Sets in a Topological Space. The $H_{N}$ increase with union $G$, so $(\mu(H_{N}))_{N}$ converges to $\mu(G)$ by claim 5 of Basic Properties of a Measure, and there is $N$ with $\mu(G)-\mu(H_{N})\le\tfrac{\varepsilon}{2}$, that is $\mu(G\setminus H_{N})\le\tfrac{\varepsilon}{2}$. Now $H_{N}\subseteq B\subseteq U$ and $U\setminus H_{N}=(U\setminus G)\cup(G\setminus H_{N})$, so $\mu(U\setminus H_{N})\le\varepsilon$ by claim 4 of Basic Properties of a Measure. Hence $B\in\mathcal{A}$.

Thus $\mathcal{A}$ is a $\sigma$-algebra on $X$ containing $\mathcal{C}$, so $\mathcal{B}(X)\subseteq\mathcal{A}$ by claim 2 of Intersections of Sigma-Algebras and Minimality of the Generated Sigma-Algebra; as $\mathcal{A}\subseteq\mathcal{B}(X)$ by definition, $\mathcal{A}=\mathcal{B}(X)$. This is claim 1.

Step 6 (Claims 2 and 3). Let $B\in\mathcal{B}(X)$ and let $S=\{\mu(F):F\subseteq B,\ F\in\mathcal{C}\}$. It is nonempty, since $\varnothing\in\mathcal{C}$ and $\varnothing\subseteq B$, and $\mu(B)$ is an upper bound for it by claim 2 of Basic Properties of a Measure; so $S$ is bounded above. Let $b$ be an upper bound for $S$ and suppose, for a contradiction, that $b<\mu(B)$. Then $\varepsilon=\mu(B)-b$ is positive, and claim 1, applied with the positive real number $\tfrac{\varepsilon}{2}$, gives $F\in\mathcal{C}$ and $U\in\mathcal{U}$ with $F\subseteq B\subseteq U$ and $\mu(U\setminus F)\le\tfrac{\varepsilon}{2}$. Since $B\setminus F\subseteq U\setminus F$, claim 2 of Basic Properties of a Measure gives $\mu(B)-\mu(F)=\mu(B\setminus F)\le\tfrac{\varepsilon}{2}$, so $\mu(F)\ge\mu(B)-\tfrac{\varepsilon}{2}=b+\tfrac{\varepsilon}{2}$; since $0<\tfrac{\varepsilon}{2}$ by claim 8 of Elementary Order Arithmetic in an Ordered Field, claim 1 of that lemma gives $b<b+\tfrac{\varepsilon}{2}$ and claim 2 then gives $b<\mu(F)$. As $\mu(F)\in S$, this contradicts the assumption that $b$ is an upper bound for $S$. Hence $\mu(B)\le b$ for every upper bound $b$ of $S$, and $\mu(B)$ is the least upper bound of $S$. This is claim 2.

For claim 3 let $S'=\{\mu(U):B\subseteq U,\ U\in\mathcal{U}\}$. It is nonempty, since $X$ is open and $B\subseteq X$, is bounded below by $0$, and $\mu(B)$ is a lower bound for it by claim 2 of Basic Properties of a Measure. Let $b'$ be a lower bound for $S'$ and suppose $\mu(B)<b'$. Then $\varepsilon'=b'-\mu(B)$ is positive, and claim 1, applied with $\tfrac{\varepsilon'}{2}$, gives $F\in\mathcal{C}$ and $U\in\mathcal{U}$ with $F\subseteq B\subseteq U$ and $\mu(U\setminus F)\le\tfrac{\varepsilon'}{2}$. Since $U\setminus B\subseteq U\setminus F$, claim 2 of Basic Properties of a Measure gives $\mu(U)-\mu(B)=\mu(U\setminus B)\le\tfrac{\varepsilon'}{2}$, so $\mu(U)\le\mu(B)+\tfrac{\varepsilon'}{2}=b'-\tfrac{\varepsilon'}{2}$; since $0<\tfrac{\varepsilon'}{2}$ by claim 8 of Elementary Order Arithmetic in an Ordered Field, claims 1 and 2 of that lemma give $\mu(U)<b'$, contradicting that $b'$ is a lower bound for $S'$. Hence $b'\le\mu(B)$ for every lower bound $b'$ of $S'$, and $\mu(B)$ is the greatest lower bound of $S'$.

Step 7 (Claim 4). Let $B\in\mathcal{B}(X)$ and let $\varepsilon$ be a positive real number. By claim 1, applied with the positive real number $\tfrac{\varepsilon}{2}$, there are $F\in\mathcal{C}$ and $U\in\mathcal{U}$ with $F\subseteq B\subseteq U$ and $\mu(U\setminus F)\le\tfrac{\varepsilon}{2}$; in particular $\mu(B\setminus F)\le\tfrac{\varepsilon}{2}$, since $B\setminus F\subseteq U\setminus F$.

If $F=\varnothing$, take $h$ to be the constant function with value $0$, which is Lipschitz with constant $0$ and takes values in $\{0\}$, and take $N=B$; then $\mu(N)=\mu(B\setminus F)\le\varepsilon$ and $h(x)=0=\mathbf{1}_{B}(x)$ for every $x\in X\setminus N$.

Suppose $F\ne\varnothing$ and let $V_{m}$ be as in Step 3. Since $(\mu(V_{m}))_{m}$ converges to $\mu(F)$, choose $k\in\mathbb{N}$ with $\mu(V_{k})-\mu(F)\le\tfrac{\varepsilon}{2}$, that is $\mu(V_{k}\setminus F)\le\tfrac{\varepsilon}{2}$; the order of choice is $\varepsilon$, then $F$ and $U$, then $k$. Define

the maximum of two real numbers. Then $0\le h(x)$, and $h(x)\le1$ because $1-\iota(k)\operatorname{dist}_{d}(x,F)\le1$, the distance being nonnegative and $\iota(k)$ positive; so $h$ takes values in $[0,1]$, the lower bound by claim 1 of Elementary Properties of the Maximum of Two Elements and the upper bound by claim 3 of that lemma, applied with the upper bound $1$ for both $0$ and $1-\iota(k)\operatorname{dist}_{d}(x,F)$.

The map $h$ is Lipschitz with constant $\iota(k)$. To see this, note first that $|\max(0,a)-\max(0,b)|\le|a-b|$ for all real $a,b$: by symmetry we may assume $\max(0,b)\le\max(0,a)$; if $\max(0,a)=0$ then $\max(0,b)=0$ by claim 1 of Elementary Properties of the Maximum of Two Elements and the difference is $0$; otherwise $\max(0,a)=a$ and $b\le\max(0,b)$, so $\max(0,a)-\max(0,b)\le a-b\le|a-b|$ by claim 3 of Properties of the Absolute Value in an Ordered Field. Applying this with $a=1-\iota(k)\operatorname{dist}_{d}(x,F)$ and $b=1-\iota(k)\operatorname{dist}_{d}(y,F)$ and using claim 4 of The Distance to a Set is Nonexpansive together with claim 4 of Properties of the Absolute Value in an Ordered Field,

Put $N=(B\setminus F)\cup(V_{k}\setminus F)$, a member of $\mathcal{B}(X)$ with $\mu(N)\le\varepsilon$ by claim 4 of Basic Properties of a Measure. Let $x\in X\setminus N$. If $x\in F$ then $\operatorname{dist}_{d}(x,F)=0$ by Vanishing of the Distance to a Set Characterizes the Closure, so $h(x)=\max(0,1)=1$, and $\mathbf{1}_{B}(x)=1$ because $F\subseteq B$. If $x\notin F$ then $x\notin V_{k}$, since $x\notin V_{k}\setminus F$, so $\iota(k)^{-1}\le\operatorname{dist}_{d}(x,F)$, and multiplying by the nonnegative number $\iota(k)$ gives $1\le\iota(k)\operatorname{dist}_{d}(x,F)$ by claim 5 of Elementary Arithmetic in an Ordered Field, giving $h(x)=\max(0,c)$ with $c\le0$, that is $h(x)=0$; and $x\notin B$, since $x\notin B\setminus F$ and $x\notin F$, so $\mathbf{1}_{B}(x)=0$. In both cases $h(x)=\mathbf{1}_{B}(x)$, which proves claim 4.