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. $\qquad \square$