Here is Prob. 22, Chap. 2, in the book Real Analysis by H. L. Royden & P. M. Fitzpatrick, 4th edition:
For any set $A$, define $m^{**}(A) \in [0, \infty]$ by $$ m^{**}(A) = \inf \left\{ m^*(O) \, | \, O \supseteq A, O \mbox{ open } \right\}. $$ How is this set function $m^{**}$ related to outer measure $m^*$?
My Attempt:
If O is any set (and in particular any open set) containing $A$, then by the monotonicity of the outer measure, we have $m^*(O) \geq m^*(A)$, that is, $m^*(A) \leq m^*(O)$, and thus $m^*(A)$ is a lower bound for the set $$ \left\{ m^*(O) \, | \, O \supseteq A, O \mbox{ open } \right\}, $$ and since $m^{**}(A)$ is the infimum of this same set, we have the inequality $$ m^*(A) \leq m^{**}(A). \tag{1} $$
Next, we show that $$ m^{**}(A) \leq m^*(A) $$ also holds.
If $m^*(A) = \infty$, then we trivially have $$ m^{**}(A) \leq m^*(A). $$ Therefore we assume that $m^*(A) < \infty$.
Let $\epsilon$ be any real number such that $\epsilon > 0$. As $m^{*}(A)$ is the infimum of the set of all the sums $\sum_{k=1}^\infty l \left( I_k \right)$, where $\left\{ I_k \right\}_{k=1}^\infty$ is any countable collection of non-empty bounded open intervals for which $A \subseteq \bigcup_{k = 1}^\infty I_k$, so for our particular $\epsilon > 0$ there exists a countable collection $\left\{ I_{\epsilon, k} \right\}_{k =1}^\infty$ of non-empty bounded open intervals such that $A \subseteq \bigcup_{k=1}^\infty I_{\epsilon, k}$ for which we have the inequalities $$ m^*(A) \leq \sum_{k=1}^\infty l \left( I_{\epsilon, k} \right) < m^*(A) + \epsilon. \tag{2} $$
Let us put $O_\epsilon := \bigcup_{k = 1}^\infty I_{\epsilon, k}$. This set $O_\epsilon$ is an open set containing $A$, and we have $$ m^*\left(O_\epsilon \right) = m^* \left( \bigcup_{k = 1}^\infty I_{\epsilon, k} \right) \leq \sum_{k = 1}^\infty m^* \left( I_{\epsilon, k} \right) = \sum_{k=1}^\infty l \left( I_{\epsilon, k} \right), $$ which together with (2) yields $$ m^* \left(O_\epsilon \right) < m^*(A) + \epsilon. $$ But as $O_\epsilon$ is an open set containing $A$, we also have $m^{**}(A) \leq m^*\left( O_\epsilon \right)$. Refer to the definition of $m^{**}(A)$ in the statement of this problem above. We therefore obtain $$ m^{**}(A) \leq m^* \left( O_\epsilon \right) < m^*(A) + \epsilon, $$ which implies $$ m^{**}(A) < m^*(A) + \epsilon. \tag{3} $$
Since (3) holds for every real number $\epsilon > 0$, we can conclude that $$ m^{**}(A) \leq m^*(A), $$ which together with (1) above yields $$ m^{**}(A) = m^*(A). $$
Is my finding correct? If so, is my derivation of this identity correct and clear enough too? Or, are there any lacunas?