$G_\delta$ set with the same Lebesgue outer measure Statement: If $E$ is a bounded set of real numbers, there exists a $G_\delta$ set $G$ such that $G$ contains E and has the same Lebesgue outer measure with $E$.
I completed the proof for the case that $E$ is countable, but how should I prove this for the case that $E$ is uncountable?
 A: How are you defining the Lebesgue outer measure $\tau$ of a set $E$? It should be the infimum of the set $M$ consisting of all those numbers $r$ that are the measures of certain open sets that cover $E$, right?
Note that this infimum is finite because $E$ is bounded: To be bounded means that for some $l$, $E\subseteq(-l,l)$, so $2l\in M$ and $\tau=\inf M\le 2l$.
Take a sequence $r_n$ of elements of $M$ that is decreasing and converges to $\tau$. This sequence exists because $\tau$ is an infimum: For each $n$ there must be an $r\in M$ with $\tau\le r<\tau+(1/n)$.
For each $r_n$ pick a witnessing open set $E_n$ covering $E$ with measure $r_n$. We may as well assume that $E_n$ is bounded. If it is not, replace it with $E_n'=E_n\cap(-l,l)$. Of course, this may replace $r_n$ with a smaller number. That's fine.
The point, of course, is that $F=\bigcap_nE_n$ is a $G_\delta$ set that contains $E$ (since each $E_n$ is open and contains $E$). To check that $F$ has the same outer measure as $E$, note that any open set covering $F$ also covers $E$, so the outer measure $\rho$ of $F$ satisfies $\rho\ge\tau$. On the other hand, $E_n$ covers $F$, so $\rho\le\tau+(1/n)$. Therefore, $\rho\le\tau$.
