Lebesgue Outer Measure on $\mathbb{R}^n$

We work in the setting of Product Measures and Lebesgue Measure on ℝⁿ: Standing Notation and Background, whose notation is fixed for every dimension and is used here with a natural number $n$ satisfying $1\le n$: thus $\mathcal{B}(\mathbb{R}^{n})$ is the Borel $\sigma$-algebra of $\mathbb{R}^{n}$ and $\lambda_{n}$ is Lebesgue measure on it, with the conventions for $[0,\infty]$ fixed there.

Let $E$ be an arbitrary subset of $\mathbb{R}^{n}$ and put

$$\mathcal{C}(E)=\{\lambda_{n}(B):B\in\mathcal{B}(\mathbb{R}^{n})\ \text{ and }\ E\subseteq B\}\subseteq[0,\infty].$$

Since $\mathbb{R}^{n}\in\mathcal{B}(\mathbb{R}^{n})$ and $E\subseteq\mathbb{R}^{n}$, the value $\lambda_{n}(\mathbb{R}^{n})$ belongs to $\mathcal{C}(E)$, so $\mathcal{C}(E)$ is nonempty.

Definition. (i) The Lebesgue outer measure of $E$ is the element $\lambda_{n}^{\ast}(E)$ of $[0,\infty]$ defined as follows. If $\lambda_{n}(B)=\infty$ for every $B\in\mathcal{B}(\mathbb{R}^{n})$ with $E\subseteq B$, then $\lambda_{n}^{\ast}(E)=\infty$. Otherwise the set

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

is a nonempty subset of $\mathbb{R}$ which is bounded below by $0$, because $\lambda_{n}$ takes values in $[0,\infty]$; by Existence of the Infimum of a Nonempty Subset of $\mathbb{R}$ Bounded Below it therefore has a greatest lower bound in $\mathbb{R}$, and $\lambda_{n}^{\ast}(E)$ is defined to be that greatest lower bound. In both cases we write

$$\lambda_{n}^{\ast}(E)=\inf\{\lambda_{n}(B):B\in\mathcal{B}(\mathbb{R}^{n}),\ E\subseteq B\},$$

the infimum being understood in $[0,\infty]$ in the sense just described.