Continuity from below for regular outer measures. We define an outer measure in a set $X$ as a function $\mu: \mathcal{P}(X) \to [0,\infty] $ that satisfies:


*

*$\mu(\emptyset) = 0$;

*$A \subset B \Rightarrow \mu(A) \leq \mu(B)$;

*$\mu(\bigcup A_n) \leq \sum \mu(A_n)$ for $(A_n) \subset \mathcal{P}(X)$.


We say that a subset $A \subset X$ is $\mu$-measurable if 
$$ \mu(B) = \mu(B\cap A) + \mu(B \setminus A), \, \forall B \subset X, $$
and we denote the collection of all $\mu$-measurable sets by $\sigma(\mu)$.
Besides, if an outer measure satisfies
$$ \forall A \subset X, \, \exists E \in \sigma(\mu), \, A \subset E, \, \mu(A) = \mu(E), $$
we say that $\mu$ is regular.
I'm trying to proof the following problem

If $\mu$ is a regular outer measure and $(E_n)$ is an increasing sequence of subsets of $X$, then $\mu(\bigcup E_n) = \lim \mu (E_n)$.

The above problem is true if each $E_n$ is a $\mu$-measurable set. I'm trying to use this fact and the regularity of $\mu$ to proof that it holds. Well, the result is true if $\mu(E_n) = \infty$ of some $n$.
Then, take (E_n) an increasing sequence of subsets of $X$ with $\mu(E_n) < \infty.$ I've tried taking measurable sets $F_n$ such that $E_n \subset F_n$ and then take the unions of such $F_n$ to be an increasing sequence of measurable sets. However, I couldn't get anywhere.
Help?
 A: You have taken the $F_n$ as measurable sets with $E_n\subseteq F_n$ such that $\mu(E_n)=\mu(F_n)$ through regularity. Moreover without loss of generality you can assume that the $F_n$ are an increasing sequence. Indeed, you can increase $F_n$ by redefining it as its union with $F_{n-1}$. The regularity property holds for this union still since 
$$ \mu(F_n\cup F_{n-1}\setminus E_n)\leq \mu(F_n\setminus E_n)+ \mu(F_{n-1}\setminus E_n)\leq \mu(F_n\setminus E_n)+\mu(F_{n-1}\setminus E_{n-1})=0.$$
Then you have 
$$ \bigcup_{n=1}^\infty E_n \subseteq \bigcup_{n=1}^\infty F_n,$$
so you can use property 2 of your outer measure to get
$$ \mu\left(\bigcup_{n=1}^\infty E_n\right) \leq \mu\left(\bigcup_{n=1}^\infty F_n\right)=\lim_{n\to\infty}\mu(F_n)=\lim_{n\to\infty}\mu(E_n),$$
with the equalities following from your observation that the claim holds for measurable sets and the regularity condition. To get the lower inequality, note that for each $n$ you trivially have $E_n\subseteq \bigcup_{j=1}^\infty E_j$ and so by property 2 of your outer measure
$$\mu(E_n)\leq \mu\left(\bigcup_{j=1}^\infty E_j\right),$$
taking the limit $n\to\infty$ then gives the result.
