Continuity from below for Lebesgue outer measure

I got it.

Since $\bigcup_{n=1}^{\infty}E_n \supset E_n$ for all $n$, We have $m^* \big( \bigcup_{n=1}^{\infty}E_n \big) \geqslant m^*E_n$. Therefore $m^* \big( \bigcup_{n=1}^{\infty}E_n \big)$ $\geqslant $ $\lim_{n\to\infty}m^*E_n$. So it's clear when $\lim_{n\to\infty}m^*E_n=\infty$. Then we assume $\lim_{n\to\infty}m^*E_n<\infty$. For all $\varepsilon>0$ and $n$, there exists $\{I_{n,i}\}_{i\in N_+}$, a sequence of open intervals in $\mathbb R^n$, covering $E_n$, s.t.

$$m\bigg(\bigcup_{i=1}^{\infty}I_{n,i}\bigg) \leqslant\sum_{i=1}^{\infty}m(I_{n,i}) =\sum_{i=1}^{\infty}|I_{n,i}| <m^*E_n+\frac{\varepsilon}{2^n}$$

by definition and properties of Lebesgue outer measure, and the L-measurability of open intervals. Then every $G_n:=\bigcup_{i=1}^{\infty}I_{n,i}\supset E_n$ is an open set, and $mG_1< m^*E_1+\varepsilon/2$. Assuming $m\big(\bigcup_{k\leqslant n}G_k\big)$ $<$ $m^*E_n+(1-1/2^n)\varepsilon$, we have

$$\begin{align*} m\Big(\bigcup_{k\leqslant n+1}G_k\Big) &=m\Big(\bigcup_{k\leqslant n}G_k\Big)+mG_{n+1} -m\bigg(\Big(\bigcup_{k\leqslant n}G_k\Big)\bigcap G_{n+1}\bigg) \\ &<\Big[m^*E_n+\Big(1-\frac{1}{2^n}\Big)\varepsilon\Big] +\Big(m^*E_{n+1}+\frac{\varepsilon}{2^{n+1}}\Big)-m^*E_{n} \\ &=m^*E_{n+1}+\Big(1-\frac{1}{2^{n+1}}\Big)\varepsilon. \end{align*}$$

So $$m\bigg(\bigcup_{k=1}^{n} G_k\bigg) <m^*E_n+\left(1-\frac{1}{2^n}\right)\varepsilon \quad \text{for}\ n=1,2,\cdots.$$

Finally,

$$ m^* \bigg( \bigcup_{n=1}^{\infty}E_n \bigg) \leqslant m\bigg( \bigcup_{n=1}^{\infty}G_n \bigg) =m\bigg( \bigcup_{n=1}^{\infty}\bigcup_{k=1}^n G_k \bigg) =\lim_{n\to\infty}m\bigg(\bigcup_{k=1}^n G_k \bigg) \leqslant \lim_{n\to\infty}m^*E_n +\varepsilon . $$

Since $\varepsilon>0$ was arbitrary, letting $\varepsilon\to 0$ yields the desired result. $\qquad \square$

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.