Show that for every $A \subseteq \mathbb{R}$ there exists a measurable set $A\subseteq E \subseteq \mathbb{R}$ such that $m^*(A) = m(E)$

Question

Some help doing this exercise?

Obviously if $A$ is measurable so let $ E = A$ then $m^*(E) = m(A)$. With that we can work in the case that A is not Lebesgue-measurable? some help in doing this exercise? I'm completely lost...

Answer 1

Hint: $E$ turns out to be a $G_\delta$ set.

Here is a complete answer.

For every $k\in \mathbb N$, there exists open set $E_k$ with $A\subset E_k$, such that $$m(E_k)\leq m^*(A)+\frac{1}{k}.$$ Now let $$E=\bigcap_{k=1}^\infty E_k,$$ then $E$ is a $G_\delta$ set, which is measure and contains $A$. Since $$m^*(A)\leq m(E)\leq m(E_k)\leq m^*(A)+\frac 1k,$$ for all $k\in\mathbb N$, it follows that $m(E)=m^*(A)$.