Measurable cover For any bounded set $E$ there is a $G_{\delta}$ set $G$ s.t. 
$$E \subseteq G \text{ and } m^*(G)=m^*(E)$$
This is a problem from Royden's Real Analysis. I have finished the proof for this but I was wondering:


*

*Does $E$ need to be bounded? 

*Does $E$ need to be measurable?


Since in my proof, I didn't use the condition that $E$ is bounded or measurable. So can I conclude that for $any$ set $E$, the result still holds?
Thanks!
 A: I don't think $E$ needs to be bounded or measurable. Outer measure is defined for any subset of $\mathbb R$. If $m^*(E) = \infty$ we can take $G = \mathbb R$. Otherwise, by definition of outer measure, for every positive integer $n$ we can find a set $G_n$ which is the union of a sequence of open intervals (hence an open set) such that $E \subseteq G_n$ and 
$$m^*(E) \leq m^*(G_n) \leq m^*(E) + 1/n$$
Then the intersection $G = \bigcap_{n=1}^{\infty}G_n$ is a $G_{\delta}$ set containing $E$, and by monotonicity of outer measure we have
$$m^*(E) \leq m^*(G) \leq m^*(G_n) \leq m^*(E) + 1/n$$
As this holds for every $n$, we conclude that $m^*(G) = m^*(E)$.
A: I think here is another proof when $E$ is a bounded set of real numbers.

As $E$ is bounded, so $E \subset [\alpha, \beta]$, where $\alpha := \inf E$ and $\beta := \sup E$. Let us put $G := [\alpha, \beta]$. Then we have $E \subset G$, which implies that
$$
m^*(E) \leq m^*(G); \tag{1}
$$
moreover, for any $\epsilon > 0$, $E \not\subset [ \alpha + \epsilon/2, \beta - \epsilon / 2 ]$, and so
$$ m^*(E) \not\leq m^*\big( [ \alpha + \epsilon/2, \beta - \epsilon / 2 ] \big). $$
But since both $m^*(E)$ and $m^*\big( [ \alpha + \epsilon/2, \beta - \epsilon / 2 ] \big)$ are real numbers, we must have
$$ m^*(E) >  m^*\big( [ \alpha + \epsilon/2, \beta - \epsilon / 2 ] \big) = (\beta - \epsilon/2) - (\alpha + \epsilon / 2) = \beta - \alpha - \epsilon = m^*(G) - \epsilon, $$
which in turn implies
$$
m^*(G) < m^*(E) + \epsilon. 
$$
Since the last inequality holds for any $\epsilon > 0$, we must have
$$
m^*(G) \leq m^*(E). \tag{2} 
$$
From (1) and (2) we have
$$
m^*(E) = m^*(G).
$$


Moreover, since we cna write this set $G$ as
$$
G = [\alpha, \beta] = \bigcap_{n=1}^\infty \left( \alpha - \frac{1}{n}, \beta + \frac{1}{n} \right),
$$
therefore $G$ is indeed a $G_\delta$-set.

