I'm struggle with the proof. How can I prove it?
Let $m^*$ be outer measure and $m$ be Lebesgue measure.
Let $A \subset \mathbb{R}$ be a set. Then, Prove that there is a measurable set $G$ such that $A \subset G$ and $m^{*}(A)=m(G)$.
Here is outer measure's definition:
Let $A \subseteq \mathbb{R} .$ Let $\mathcal{I}=\left\{I_{k} | k \in \mathbb{N}\right\}$ be a countable collection of open (closed, semiopen) intervals with $A \subset \bigcup_{k=1}^{\infty} I_{k} .$ We define an outer measure of $A,$ denoted by $m^{*}(A)$ as
$$
m^{*}(A):=\inf _{\mathcal{I}} \sum_{k=1}^{\infty} l\left(I_{k}\right).
$$
Here is Lebesgue measure's definition:
$E \subset \mathbb{R}$ is measurable if for every set $A \subset \mathbb{R}$ $m^{*}(A)=m^{*}(A \cap E)+m^{*}\left(A \cap E^{c}\right)$.
Let $m^{*}: \mathcal{P}(\mathbb{R}) \rightarrow[0, \infty]$ be the outer measurable. Let $\mathfrak{M} \triangleq\{E \subset \mathbb{R} | E \text { is measurable }\}$.
The restriction $\left.m^{*}\right|_{\mathfrak{M}}$ is called Lebesgue measure, denoted by $m$.
Thanks!
Best Answer
If $m^*(A)=\infty$: Let $G=\Bbb R.$
If $m^*(A)<\infty$: For $n\in \Bbb N$ let $A\subset G_n\subset \Bbb R$ where $G_n$ is open and $m^*(A)\le m(G_n)\le m^*(A)+1/n.$ Let $G=\cap_{n\in \Bbb N}G_n.$
Use this important general property:
$(\bullet)$ If $\{G_n: n\in \Bbb N\}$ is a countable family of measurable sets and each $G_n$ has finite measure then $m(\cap_{n\in \Bbb N}G_n)=\inf_{n\in \Bbb N}m(H_n)$ where $H_n=\cap_{j=1}^nG_j.$
Proof of $(\bullet)$: Let $G=\cap_{n\in \Bbb N}G_n.$ For $n\in \Bbb N$ let $J_n=H_n\setminus H_{n+1}.$ Then $\{G\}\cup \{J_n:n\in \Bbb N\}$ is a countable family of pair-wise disjoint measurable sets and for each $n\in \Bbb N$ we have $H_n=G\cup (\cup_{j\ge n}J_j)$ so $$ (*)\quad m(H_n)=m(G)+\sum_{j=n}^{\infty}m(J_j).$$ Now $\sum_{j\in \Bbb N} m(J_j)$ is a convergent series of non-negative reals... (It sums to $m(H_1\setminus G)=m(G_1\setminus G)$)... so $$(**)\quad \lim_{n\to \infty}\sum_{j=n}^{\infty}m(J_j)=0.$$ Apply $(**)$ to $(*)$ to see that $\langle m(H_n)\rangle_{n\in \Bbb N}$ is a decreasing sequence converging to $m(G).$
Remark: $(\bullet)$ also holds with the weaker condition that $m(G_{n_0})<\infty$ for at least one $n_0$: Apply $(\bullet)$ to $\{G'_n:n\in \Bbb N\}=\{G_n\cap G_{n_0}:n\in\Bbb N\}.$ It does not hold for all families, e.g. if $G_n=[n,\infty).$