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 $m$.


  • 1
    $\begingroup$ I can help. The first thing you need to do is determine the exact definition of $m^*(A)$. What is it? $\endgroup$
    – Umberto P.
    Apr 16, 2020 at 23:56
  • 2
    $\begingroup$ It's perhaps worth noting that an outer measure is a rather general concept, though clearly here you are talking the one that induces Lebesgue measure. Anyway, you're going to need to provide more context for how everything is defined, as some people define Lebesgue measure as the restriction of a certain outer measure to the outer-measurable sets. $\endgroup$ Apr 16, 2020 at 23:57
  • $\begingroup$ I'm sorry to confuse you. I provided some definitions! Thanks! $\endgroup$
    – himath
    Apr 17, 2020 at 1:18
  • $\begingroup$ Do you know that open sets are measurable? $\endgroup$
    – Umberto P.
    Apr 17, 2020 at 1:25
  • $\begingroup$ Yes, I know that. In fact, I deal with just real line. So, I understand that the open sets are $(a,b)\subset \mathbb{R}$. $\endgroup$
    – himath
    Apr 17, 2020 at 1:34

1 Answer 1


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 $G_n=[n,\infty).$


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