Continuity from below for Lebesgue outer measure Let $\{E_n\}$ be an increasing sequence of subsets of $\mathbb{R}^n$, measurable or not. Then  $$m^* \bigg( \bigcup_{n=1}^{\infty}E_n \bigg)=\lim_{n\rightarrow\infty}m^*E_n$$ 
 A: 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.
A: 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. $\qquad \square$
A: Perhaps I am stating the trivial thing, but the Borel sets should be closed under countable union, etc. And if I remember correctly the popular definition of Lesbegue measure allows us to use Borel sets to approach measurable sets. To prove this for Borel set should be more or less automatic, like making standard set manipulations, etc. 
You may be interested to read about the limit superior and limit inferior. 
Edit: sorry I ignored your condition that $E_{i}$ are not necessarily measurable. Under such condition you may be interested in cutting $\bigcup E_{i}$ into $\bigcup (E_{i}-E_{j})$s. Since the two sets are the same and the second set is a union of separated subsets, you should be able to use subaddivity to prove this. 
