The property of Lebesgue-measurable set.

Question

Let $E\subset \mathbb{R^n}$ be Lebesgue-measurable.

Then, prove that there exists $G_{\delta}$ set $H$ s.t. $E\subset H$ and $m^*(H\backslash E)=0.$

My proof is here.

For $E$, there exists $G_{\delta}$ set $H$ s.t. $E\subset H$ and $m^*(E)=m^*(H)$. (This is already proven.)

Then, $m(H)=m(E)+m(H\backslash E)$ holds since $H=E\cup H\backslash E$, $E$ and $H$ are Lebesgue-measurable, $E \cap H\backslash E=\emptyset$ and $m$ is measure.

Then, if $m(H)$, $m(E)<\infty,$ I get $m(H\backslash E)=0$ since $m(H)= m(E)$.

But it is not clear whether or not $m(H)$, $m(E)<\infty.$

Doesn't my proof work?

Answer 1

Choose a $G_{\delta}$ set $H_N$ such that $E_N \subset H_N$ and $m^{*}(H_n\setminus E_N)=0$ where $E_N=E\cap R_N$ and $R_N=\{(x_1,x_2,...,x_n): -N <x_i <N \, \forall i\}$. This possible by your argument since $m(H_n)$ and $m(E_N)$ are finite (when we choose $H_N$ with $m^{*}(H_N)=m^{*}(E_N)$).

Now take $H=\bigcap_N H_N$. Can you finish?

[Note that $H\setminus E \subset \bigcap_n (H_N\setminus E_N$].