TITLE: Continuity from below for Lebesgue outer measure
QUESTION [11 upvotes]: Let $\{E_n\}$ be an increasing sequence of subsets of $\mathbb{R}^n$, measurable or not. Then  $$m^* \bigg( \bigcup_{n=1}^{\infty}E_n \bigg)=\lim_{n\rightarrow\infty}m^*E_n$$

REPLY [2 votes]: Let $\mu$ be a measure on algebra $\mathcal A$ of subsets of $X$.
We will show that the outer measure $\mu^*$ defined as
$$
\mu^*(E) = \inf \left\{ \sum_{n=1}^\infty \mu(A_n) \middle| 
A_n \in \mathcal A, E \subset \bigcup_{n=1}^\infty A_n \right\}, \quad E\subset X,
$$
is continuous from below, i.e. $\mu^*(E_n) \uparrow \mu^*(E)$ if $E_n \uparrow E$, $E_n \subset X$.
Proof. If $m := \sup_n \mu^*(E_n) = +\infty$, then $\mu^*(E) = +\infty$ as well and there is nothing to prove, so we assume $m < +\infty$.
Fix any $\varepsilon > 0$.
For any $n$ we construct such $A_{nm} \in \mathcal A$, that $E_n \subset \bigcup_m A_{nm} =: B_n$ and $\mu^*(E_n) \ge \mu(B_n) - \varepsilon$, where $\mu$ denotes an extension of $\mu$ on $\sigma(\mathcal A)$.
Next, let $C_n := \bigcap_{k\ge n} B_n$. Obviously, $E_n = \bigcap_{k\ge n} E_k \subset C_n \subset B_n$ and thus,
$$
\mu(C_n) - \mu^*(E_n) \le \mu(B_n) - \mu^*(E_n) \le \varepsilon.
$$
Finally, $C_n \uparrow \bigcup_n C_n$ and $E = \bigcup_n E_n \subset \bigcup_n C_n$.
Therefore,
$$
0 \le \mu^*(E) - \lim\limits_{n\to\infty}   \mu^*(E_n) \le
\mu(\cup_n C_n) - \lim\limits_{n\to\infty}   \mu^*(E_n) =
\lim\limits_{n\to\infty} \mu(C_n) - \mu^*(E_n) \le \varepsilon.
$$
Since $\varepsilon$ is arbitrary, we get the desired property.