Here is Prob. 7, Chap. 2, in the book Real Analysis by H. L. Royden and P. M. Fitzpatrick, 4th edition:
A set of real numbers is said to be a $G_\delta$ set provided it is the intersection of a countable collection of open sets. Show that for any bounded set $E$, there is a $G_\delta$ set $G$ for which $$ E \subseteq G \qquad \mbox{ and } \qquad m^*(G) = m^*(E). $$
My Attempt:
Note that, for any subset $E$ of $\mathbb{R}$, the outer measure $m^*(E)$ by definition is the infimum (or greatest lower bound) of the set of all the sums of the form $\sum_{k = 1}^\infty l \left( I_k \right)$, where $\left\{ I_k \right\}_{k=1}^\infty$ is a countable collection of non-empty bounded, open intervals such that $E \subseteq \bigcup_{k=1}^\infty I_k$.
Let $E$ be a (non-empty) bounded set of the set $\mathbb{R}$ of real numbers. Then $E$ has an infimum $\alpha$ and a supremum $\beta$, where $\alpha, \beta \in \mathbb{R}$, and we have the inclusion $$ E \subseteq [\alpha, \beta ], $$ which implies $$ m^*(E) \leq m^* \big( [\alpha, \beta] \big) = l \big( [\alpha, \beta] \big) = \beta - \alpha < +\infty, $$ by virtue of Proposition 1 in Royden. In short $m^*(E) < +\infty$, that is, $m^*(E)$ is a non-negative real number.
Now for each positive integer $n$, as $m^*(E)$ is the infinum (or greatest lower bound) of a particular sum and as $$ m^*(E) < m^*(E) + \frac{1}{n}, $$ so there exists a countable colletion $\left\{ I_{nk} \right\}_{k=1}^\infty$ such that $$ E \subseteq \bigcup_{k=1}^\infty I_{nk} \tag{1} $$ and $$ m^*(E) \leq \sum_{k=1}^\infty l \left( I_{nk} \right) < m^*(E) + \frac{1}{n}; \tag{2} $$ let us put $$ U_n := \bigcup_{k=1}^\infty I_{nk}. \tag{Definition 1} $$ Then each set $U_n$ being a union of bounded open intervals on the real line is an open set of $\mathbb{R}$ and we also have $$ E \subseteq U_n $$ by (1) above.
Now let us put $$ G := \bigcap_{n=1}^\infty U_n. \tag{Definition 2} $$ This set $G$ is a $G_\delta$ set of real numbers, and we also have $E \subseteq G$, which implies $$ m^*(E) \leq m^*(G), \tag{3} $$ by the monotonicity of the outer measure.
For each positive integer $n$, since $G \subset U_n$ [Refer to (Definition 2) above.], therefore we also have $$ m^*(G) \leq m^* \left( U_n \right). \tag{4} $$ Therefore for each positive integer $n$ we have \begin{align} & \ \ \ m^*(G) \\ &\leq m^* \left( U_n \right) \\ &= m^* \left( \bigcup_{k=1}^\infty I_{nk} \right) \qquad [\mbox{ refer to (Definition 1) above }] \\ &\leq \sum_{k=1}^\infty m^* \left( I_{nk} \right) \qquad [ \mbox{ by Proposition 3 in Royden } ] \\ &= \sum_{k=1}^\infty l \left( I_{nk} \right) \qquad [ \mbox{ by Proposition 1 in Royden } ] \\ &< m^*(E) + \frac{1}{n}, \qquad [ \mbox{ by virtue of (2) above } ] \end{align} which implies $$ m^*(G) < m^*(E) + \frac{1}{n}, $$ and upon taking the limits of both sides of this last inequality as $n \to \infty$, we obtain $$ m^*(G) \leq m^*(E), $$ which together with (3) above gives $$ m^*(G) = m^*(E), $$ as required.
Is my proof correct and clear enough in each and every detail? Or, are there any issues in any portion of it?