Proof of Elementary Properties of Lebesgue Outer Measure on $\mathbb{R}^n$

Throughout, for $E\subseteq\mathbb{R}^{n}$ write

$$\mathcal{C}(E)=\{\lambda_{n}(B):B\in\mathcal{B}(\mathbb{R}^{n}),\ E\subseteq B\},\qquad \mathcal{C}_{\mathbb{R}}(E)=\{a\in\mathcal{C}(E):a\in\mathbb{R}\},$$

as in the definition of Lebesgue outer measure: $\lambda_{n}^{\ast}(E)=\infty$ if $\mathcal{C}_{\mathbb{R}}(E)=\varnothing$, and otherwise $\lambda_{n}^{\ast}(E)$ is the greatest lower bound of $\mathcal{C}_{\mathbb{R}}(E)$. We use throughout the monotonicity and the countable subadditivity of $\lambda_{n}$, that is claims 2 and 4 of Basic Properties of a Measure, and the fact that $\mathcal{B}(\mathbb{R}^{n})$ is closed under countable unions, complements, and hence countable intersections, by σ-Algebra and Measurable Space.

Claim 2 (monotonicity). Let $E\subseteq F\subseteq\mathbb{R}^{n}$. Every $B\in\mathcal{B}(\mathbb{R}^{n})$ with $F\subseteq B$ also satisfies $E\subseteq B$, so $\mathcal{C}(F)\subseteq\mathcal{C}(E)$ and therefore $\mathcal{C}_{\mathbb{R}}(F)\subseteq\mathcal{C}_{\mathbb{R}}(E)$. If $\mathcal{C}_{\mathbb{R}}(F)=\varnothing$ then $\lambda_{n}^{\ast}(F)=\infty$ and the asserted inequality holds. Otherwise $\mathcal{C}_{\mathbb{R}}(E)$ is nonempty as well, and both outer measures are greatest lower bounds; since $\lambda_{n}^{\ast}(E)$ is a lower bound of $\mathcal{C}_{\mathbb{R}}(E)$ it is a lower bound of the subset $\mathcal{C}_{\mathbb{R}}(F)$, whence $\lambda_{n}^{\ast}(E)\le\lambda_{n}^{\ast}(F)$ because $\lambda_{n}^{\ast}(F)$ is the greatest lower bound of $\mathcal{C}_{\mathbb{R}}(F)$.

Claim 1 (agreement on Borel sets). Let $B\in\mathcal{B}(\mathbb{R}^{n})$. Since $B\subseteq B$ we have $\lambda_{n}(B)\in\mathcal{C}(B)$, and by monotonicity of $\lambda_{n}$ every $B'\in\mathcal{B}(\mathbb{R}^{n})$ with $B\subseteq B'$ satisfies $\lambda_{n}(B)\le\lambda_{n}(B')$. If $\lambda_{n}(B)=\infty$ this forces $\lambda_{n}(B')=\infty$ for every such $B'$, so $\mathcal{C}_{\mathbb{R}}(B)=\varnothing$ and $\lambda_{n}^{\ast}(B)=\infty=\lambda_{n}(B)$. If $\lambda_{n}(B)$ is real, then $\lambda_{n}(B)\in\mathcal{C}_{\mathbb{R}}(B)$, so $\lambda_{n}^{\ast}(B)\le\lambda_{n}(B)$; and $\lambda_{n}(B)$ is a lower bound of $\mathcal{C}_{\mathbb{R}}(B)$ by the displayed monotonicity, so $\lambda_{n}(B)\le\lambda_{n}^{\ast}(B)$. Hence $\lambda_{n}^{\ast}(B)=\lambda_{n}(B)$.

Claim 3 (Borel hull). Let $E\subseteq\mathbb{R}^{n}$. If $\lambda_{n}^{\ast}(E)=\infty$, take $B=\mathbb{R}^{n}$; then $E\subseteq B$, $B\in\mathcal{B}(\mathbb{R}^{n})$, and $\lambda_{n}(B)$ cannot be real (otherwise $\mathcal{C}_{\mathbb{R}}(E)\ne\varnothing$), so $\lambda_{n}(B)=\infty=\lambda_{n}^{\ast}(E)$. Now suppose $a=\lambda_{n}^{\ast}(E)$ is real. For each $m\in\mathbb{N}$ the number $a+\tfrac1m$ is strictly larger than the greatest lower bound $a$ of $\mathcal{C}_{\mathbb{R}}(E)$, hence is not a lower bound of it, so there is $B_{m}\in\mathcal{B}(\mathbb{R}^{n})$ with $E\subseteq B_{m}$ and $\lambda_{n}(B_{m})\le a+\tfrac1m$. Put $B=\bigcap_{m\in\mathbb{N}}B_{m}\in\mathcal{B}(\mathbb{R}^{n})$. Then $E\subseteq B$, and $B\subseteq B_{m}$ gives $\lambda_{n}(B)\le a+\tfrac1m$ for every $m$. If we had $a<\lambda_{n}(B)$, then by The Archimedean Property of the Real Numbers there would be $m\in\mathbb{N}$ with $\tfrac1m<\lambda_{n}(B)-a$, a contradiction; hence $\lambda_{n}(B)\le a$. Conversely $\lambda_{n}(B)\in\mathcal{C}_{\mathbb{R}}(E)$ (it is real, being at most $a+1$), so $a\le\lambda_{n}(B)$. Thus $\lambda_{n}(B)=a=\lambda_{n}^{\ast}(E)$.

Claim 4 (countable subadditivity). Let $(E_{m})_{m\in\mathbb{N}}$ be a sequence of subsets of $\mathbb{R}^{n}$ and put $E=\bigcup_{m\in\mathbb{N}}E_{m}$. If $\lambda_{n}^{\ast}(E_{m_{0}})=\infty$ for some $m_{0}$, the sum on the right-hand side is $\infty$ and there is nothing to prove. Otherwise, by claim 3 choose for each $m$ a set $B_{m}\in\mathcal{B}(\mathbb{R}^{n})$ with $E_{m}\subseteq B_{m}$ and $\lambda_{n}(B_{m})=\lambda_{n}^{\ast}(E_{m})$. Then $B=\bigcup_{m\in\mathbb{N}}B_{m}\in\mathcal{B}(\mathbb{R}^{n})$ and $E\subseteq B$, so by claims 2 and 1,

$$\lambda_{n}^{\ast}(E)\le\lambda_{n}^{\ast}(B)=\lambda_{n}(B)\le\sum_{m\in\mathbb{N}}\lambda_{n}(B_{m})=\sum_{m\in\mathbb{N}}\lambda_{n}^{\ast}(E_{m}),$$

the middle inequality being countable subadditivity of $\lambda_{n}$.

Claim 5 (null sets). By Null Set of a Measure, a subset $E$ of $\mathbb{R}^{n}$ is $\lambda_{n}$-null exactly when there is $B\in\mathcal{B}(\mathbb{R}^{n})$ with $E\subseteq B$ and $\lambda_{n}(B)=0$. If such a $B$ exists then $0\in\mathcal{C}_{\mathbb{R}}(E)$, and since $0$ is a lower bound of $\mathcal{C}_{\mathbb{R}}(E)$ (the values of $\lambda_{n}$ are nonnegative) we get $\lambda_{n}^{\ast}(E)=0$. Conversely, if $\lambda_{n}^{\ast}(E)=0$ then the Borel hull $B$ produced by claim 3 satisfies $E\subseteq B$ and $\lambda_{n}(B)=0$, so $E$ is $\lambda_{n}$-null.

If $E'\subseteq E$ with $E$ $\lambda_{n}$-null, then $0\le\lambda_{n}^{\ast}(E')\le\lambda_{n}^{\ast}(E)=0$ by claim 2, so $E'$ is $\lambda_{n}$-null. If $(E_{m})_{m\in\mathbb{N}}$ are $\lambda_{n}$-null, then by claim 4 and the first part, $\lambda_{n}^{\ast}(\bigcup_{m}E_{m})\le\sum_{m}0=0$, so $\bigcup_{m}E_{m}$ is $\lambda_{n}$-null.