Proof of Outer and Inner Regularity of Lebesgue Measure on $\mathbb{R}^n$

Notation and two preliminaries. For $k\in\mathbb{N}$ put

absolute values being those of Absolute Value in an Ordered Field. Each $W_{k}$ is open and each $\overline{W}_{k}$ is closed and bounded, hence compact by Heine-Borel Theorem in $\mathbb{R}^n$; both are Borel. The sequence $(W_{k})_{k\in\mathbb{N}}$ is increasing and $\bigcup_{k}W_{k}=\mathbb{R}^{n}$, because for $x\in\mathbb{R}^{n}$ the Archimedean property (The Archimedean Property of the Real Numbers) supplies $k\in\mathbb{N}$ with $\lVert x\rVert<k$, and $|x_{l}|\le\lVert x\rVert$ for every $l$ by Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$. Since $W_{k}$ is bounded, $\lambda_{n}(W_{k})<\infty$ by Balls Have Positive Lebesgue Measure and Bounded Sets Have Finite Lebesgue Measure.

(P1) The measures $\mu_{k}$. For $k\in\mathbb{N}$ define $\mu_{k}(A)=\lambda_{n}(A\cap W_{k})$ for $A\in\mathcal{B}(\mathbb{R}^{n})$. Then $\mu_{k}(\varnothing)=0$ and, since intersecting with $W_{k}$ preserves disjointness and commutes with countable unions, $\mu_{k}$ is countably additive; so $\mu_{k}$ is a measure on $\mathcal{B}(\mathbb{R}^{n})$, and it is finite because $\mu_{k}(\mathbb{R}^{n})=\lambda_{n}(W_{k})<\infty$. Note also that $\lambda_{n}(A)=\mu_{k}(A)$ whenever $A\subseteq W_{k}$.

(P2) Continuity from above. Let $\mu$ be a finite measure on $\mathcal{B}(\mathbb{R}^{n})$ and let $(A_{p})_{p\in\mathbb{N}}$ be a sequence in $\mathcal{B}(\mathbb{R}^{n})$ with $A_{p+1}\subseteq A_{p}$ for every $p$; put $A=\bigcap_{p}A_{p}$. Then the sequence $(\mu(A_{p}))_{p\in\mathbb{N}}$ converges to $\mu(A)$. Indeed, the sets $B_{p}=A_{1}\setminus A_{p}$ increase with union $A_{1}\setminus A$, and all values of $\mu$ are real and bounded above by $\mu(\mathbb{R}^{n})$, so claim 5 of Basic Properties of a Measure gives that $(\mu(B_{p}))_{p}$ converges to $\mu(A_{1}\setminus A)$. By claim 3 of the same lemma, $\mu(B_{p})=\mu(A_{1})-\mu(A_{p})$ and $\mu(A_{1}\setminus A)=\mu(A_{1})-\mu(A)$, and the assertion follows.

(P3) A geometric bound. For every $p\in\mathbb{N}$ one has $\sum_{m=1}^{p}2^{-m}=1-2^{-p}\le 1$, by induction on $p$ (Principle of Induction for the Natural Numbers). Let $c\in\mathbb{R}$ with $0\le c$. Every term of the sequence $(c\,2^{-m})_{m\in\mathbb{N}}$ is then real, and its partial sums $c(1-2^{-p})$ are bounded above by $c$; by the convention for sums of sequences in $[0,\infty]$ fixed in Measure, Measure Space, and Probability Measure, the sum of a sequence whose terms are real with partial sums bounded above is the least upper bound of those partial sums. Hence $\sum_{m\in\mathbb{N}}c\,2^{-m}\le c$.

Step 1: an approximation class. Let $\mathcal{A}$ be the set of all $B\in\mathcal{B}(\mathbb{R}^{n})$ with the property that for every $k\in\mathbb{N}$ and every real $\varepsilon>0$ there are a closed set $F$ and an open set $V$ with $F\subseteq B\subseteq V$ and $\mu_{k}(V\setminus F)\le\varepsilon$. We show $\mathcal{A}=\mathcal{B}(\mathbb{R}^{n})$.

(a) Every closed set lies in $\mathcal{A}$. Let $C\subseteq\mathbb{R}^{n}$ be closed and let $k\in\mathbb{N}$ and $\varepsilon>0$ be given. If $C=\varnothing$ take $F=V=\varnothing$. Otherwise take $F=C$ and, for $p\in\mathbb{N}$, put $V_{p}=\{x\in\mathbb{R}^{n}:d_{E}(x,C)<1/p\}$, where $d_{E}(\,\cdot\,,C)$ is the distance to $C$; it is continuous, being nonexpansive by The Distance to a Set is Nonexpansive, so $V_{p}$ is open, and clearly $C\subseteq V_{p}$. If $x\notin C$ then, $\mathbb{R}^{n}\setminus C$ being open, there is $r>0$ with $\lVert y-x\rVert\ge r$ for every $y\in C$, hence $d_{E}(x,C)\ge r$, and by the Archimedean property $x\notin V_{p}$ for large $p$. Therefore $\bigcap_{p}V_{p}=C$, so the decreasing sets $V_{p}\setminus C$ have empty intersection, and (P2) applied to the finite measure $\mu_{k}$ gives $p$ with $\mu_{k}(V_{p}\setminus C)\le\varepsilon$. Take $V=V_{p}$.

(b) $\mathcal{A}$ is closed under complements. If $F\subseteq B\subseteq V$ with $F$ closed and $V$ open, then $\mathbb{R}^{n}\setminus V$ is closed, $\mathbb{R}^{n}\setminus F$ is open,

so the same pair of estimates works for $\mathbb{R}^{n}\setminus B$.

(c) $\mathcal{A}$ is closed under countable unions. Let $(B_{m})_{m\in\mathbb{N}}$ be a sequence in $\mathcal{A}$ and $B=\bigcup_{m}B_{m}$; fix $k$ and $\varepsilon>0$. Choose closed $F_{m}$ and open $V_{m}$ with $F_{m}\subseteq B_{m}\subseteq V_{m}$ and $\mu_{k}(V_{m}\setminus F_{m})\le(\varepsilon/2)2^{-m}$. The set $V=\bigcup_{m}V_{m}$ is open and contains $B$. Put $G=\bigcup_{m}F_{m}$ and $G_{p}=\bigcup_{m\le p}F_{m}$; each $G_{p}$ is closed, because $\mathbb{R}^{n}\setminus G_{p}=\bigcap_{m\le p}(\mathbb{R}^{n}\setminus F_{m})$ is a finite intersection of open sets and hence open by Metric Open Sets Form a Topology, a set being closed exactly when its complement is open, and $G_{p}\subseteq G\subseteq B$. Since $V\setminus G\subseteq\bigcup_{m}(V_{m}\setminus F_{m})$, countable subadditivity (claim 4 of Basic Properties of a Measure) and (P3) give $\mu_{k}(V\setminus G)\le\varepsilon/2$. The sets $G\setminus G_{p}$ decrease with empty intersection, so (P2) gives $p$ with $\mu_{k}(G\setminus G_{p})\le\varepsilon/2$. Then $V\setminus G_{p}\subseteq(V\setminus G)\cup(G\setminus G_{p})$ yields $\mu_{k}(V\setminus G_{p})\le\varepsilon$, and $G_{p}\subseteq B\subseteq V$.

(d) Conclusion of Step 1. By (a) the set $\mathbb{R}^{n}$, being closed, lies in $\mathcal{A}$; with (b) and (c), $\mathcal{A}$ is a $\sigma$-algebra on $\mathbb{R}^{n}$ contained in $\mathcal{B}(\mathbb{R}^{n})$. By (a) and (b) it contains every closed and every open subset of $\mathbb{R}^{n}$. By The Borel $\sigma$-Algebras of Euclidean Space and of the Euclidean Metric Coincide, $\mathcal{B}(\mathbb{R}^{n})$ is the $\sigma$-algebra generated by the open subsets of $(\mathbb{R}^{n},d_{E})$, and by Intersections of Sigma-Algebras and Minimality of the Generated Sigma-Algebra that generated $\sigma$-algebra is contained in every $\sigma$-algebra containing the open sets. Hence $\mathcal{B}(\mathbb{R}^{n})\subseteq\mathcal{A}$, so $\mathcal{A}=\mathcal{B}(\mathbb{R}^{n})$.

Step 2: proof of claim 1. Let $B\in\mathcal{B}(\mathbb{R}^{n})$ and $\varepsilon>0$. Put $R_{1}=W_{1}$ and $R_{k}=W_{k}\setminus W_{k-1}$ for $k\ge2$; these are Borel, pairwise disjoint, $R_{k}\subseteq W_{k}$, and $\bigcup_{k}R_{k}=\bigcup_{k}W_{k}=\mathbb{R}^{n}$.

Fix $k$. By Step 1, $B\cap R_{k}\in\mathcal{A}$, so there is an open $V_{k}'$ with $B\cap R_{k}\subseteq V_{k}'$ and $\mu_{k+1}(V_{k}'\setminus(B\cap R_{k}))\le\varepsilon 2^{-k}$; we may even ignore the closed set produced there. Put $V_{k}=V_{k}'\cap W_{k+1}$, an open set which still contains $B\cap R_{k}$ because $R_{k}\subseteq W_{k}\subseteq W_{k+1}$. Then $V_{k}\setminus(B\cap R_{k})\subseteq W_{k+1}$, so by (P1)

Put $U=\bigcup_{k}V_{k}$, an open set containing $B=\bigcup_{k}(B\cap R_{k})$. If $x\in U\setminus B$ then $x\in V_{k}$ for some $k$ and $x\notin B\cap R_{k}$, so

and countable subadditivity together with (P3) gives $\lambda_{n}(U\setminus B)\le\varepsilon$.

Step 3: proof of claim 2. Apply claim 1 to $\mathbb{R}^{n}\setminus B\in\mathcal{B}(\mathbb{R}^{n})$: there is an open $U$ with $\mathbb{R}^{n}\setminus B\subseteq U$ and $\lambda_{n}(U\setminus(\mathbb{R}^{n}\setminus B))\le\varepsilon$. Put $F=\mathbb{R}^{n}\setminus U$, a closed set with $F\subseteq B$, and note $B\setminus F=B\cap U=U\setminus(\mathbb{R}^{n}\setminus B)$, so $\lambda_{n}(B\setminus F)\le\varepsilon$.

Step 4: proof of claim 3. Let $\lambda_{n}(B)<\infty$ and $\varepsilon>0$. By claim 2 there is a closed $F\subseteq B$ with $\lambda_{n}(B\setminus F)\le\varepsilon/2$. Put $K_{p}=F\cap\overline{W}_{p}$: it is closed and bounded, hence compact by Heine-Borel Theorem in $\mathbb{R}^n$, and $K_{p}\subseteq F\subseteq B$. The sets $K_{p}$ increase with union $F$, and $\lambda_{n}(F)\le\lambda_{n}(B)<\infty$, so all the values $\lambda_{n}(K_{p})$ are real and bounded above by $\lambda_{n}(F)$; claim 5 of Basic Properties of a Measure gives that $(\lambda_{n}(K_{p}))_{p}$ converges to $\lambda_{n}(F)$. Choose $p$ with $\lambda_{n}(F)-\lambda_{n}(K_{p})\le\varepsilon/2$; by claim 3 of that lemma this is $\lambda_{n}(F\setminus K_{p})$. Since $B\setminus K_{p}\subseteq(B\setminus F)\cup(F\setminus K_{p})$, subadditivity gives $\lambda_{n}(B\setminus K_{p})\le\varepsilon$, and $K=K_{p}$ is as required.

Step 5: proof of claim 4. Let $E\subseteq\mathbb{R}^{n}$. Suppose first $\lambda_{n}^{\ast}(E)=\infty$ and let $U$ be open with $E\subseteq U$. Then $U\in\mathcal{B}(\mathbb{R}^{n})$, so by monotonicity and agreement on Borel sets, $\infty=\lambda_{n}^{\ast}(E)\le\lambda_{n}^{\ast}(U)=\lambda_{n}(U)$, so $\lambda_{n}(U)=\infty$.

Suppose now $\lambda_{n}^{\ast}(E)<\infty$ and let $\varepsilon>0$. By the Borel hull claim there is $B\in\mathcal{B}(\mathbb{R}^{n})$ with $E\subseteq B$ and $\lambda_{n}(B)=\lambda_{n}^{\ast}(E)$. By claim 1 there is an open $U\supseteq B\supseteq E$ with $\lambda_{n}(U\setminus B)\le\varepsilon$. Since $U\subseteq B\cup(U\setminus B)$, subadditivity gives $\lambda_{n}(U)\le\lambda_{n}(B)+\varepsilon=\lambda_{n}^{\ast}(E)+\varepsilon$.