An outer measure is countable-additive on the measurable sets Given an outer measure $\mu^*$, i'm trying to prove that for any set $E$, and a countable number of distinct (pairwise) $\mu^*$-measurable sets $\{A_i\}_{i=1}^\infty$:
$$\mu^*\left(E \cap \left(\bigcup_{i=1}^\infty A_i\right)\right)= \sum_{i=1}^\infty \mu^*\left(E\cap A_i\right)$$
One inequality is obvious thanks to sub-additivity, but i'm having troubles with the other.
I figured out that i could maybe show
$$\mu^*\left(E \cap \left(\bigcup_{i=1}^\infty A_i\right)\right)\ge \sum_{i=1}^n \mu^*\left(E\cap A_i\right)$$
and then take the limit.
For that I would need to show only that
$$\sum_{i=1}^n\left( E\cap A_i\right)=\mu^*\left(\bigcup_{i=1}^n E\cap A_i\right)= \mu^*\left(E \cap \left(\bigcup_{i=1}^n A_i\right)\right)$$
If i'm right and all of this is really necessary, I guess that i'd need to prove that:

*

*for finite number of measurable subsets, the union is measurable

*finite-additivity for measurable sets.

I managed to prove the first for my outer measure $\mu^*$, but about the second - it is easy to show that for measurable subsets, but after doing all that I realised that $E \cap A_i $ does not have to be measurable itself.
Is anything of what I did useful to prove what I wanted?
How can I complete, or what steps should I take in order to prove it differently?
Thanks!
 A: You can show the finite case with finite induction on $n$. It works because the disjoint union is monotone increasing in terms of containment as $n$ increases. For the infinite case observe that for all $n$
$$
E\cap\bigcup_{k=1}^\infty A_k=\bigcup_{k=1}^\infty E\cap A_k\supseteq\bigcup_{k=1}^nE\cap A_k
$$
so that 
$$
\mu^*(E\cap\bigcup_{k=1}^\infty A_k)=\sum_{k=1}^\infty \mu^*(E\cap A_k)\geq\sum_{k=1}^n\mu^*(E\cap A_k).
$$
Since the LHS doesn't involve $n$, taking the limit finishes the proof!

More detail:
First I accidentally reversed the set names, see my edits above. Second, for the finite case, since each $A_k$ is measurable then for any set $E$ we have by induction:
$$\begin{align*}
\mu^*(E\cap\bigcup_{k=1}^nA_k)&=\mu^*((E\cap\bigcup_{k=1}^nA_k)\cap A_n)+\mu^*((E\cap\bigcup_{k=1}^nA_k)\cap A^C_n)\\
&=\mu^*(E\cap A_n)+\mu^*(E\cap\bigcup_{k=1}^{n-1}A_k)\\
&=\mu^*(E\cap A_n)+\sum_{k=1}^{n-1}\mu^*(E\cap A_k)\\
&=\sum_{k=1}^{n}\mu^*(E\cap A_k)\\
\end{align*}
$$
A: If $m^*\left(A \cap \bigcup_{k=1}^\infty E_k\right) = \infty$, the reserve inequality holds trivially.
If $m^*\left(A \cap \bigcup_{k=1}^\infty E_k\right) < \infty$, we can find a $G_\delta$ set $G$ that contains $A \cap \bigcup_{k=1}^\infty E_k$ and satisfies
$$
m^*(G) = m^*\left(A \cap \bigcup_{k=1}^\infty E_k\right)
$$
Since $G$ is measurable, we have
\begin{align*}
    m^*\left(A \cap \bigcup_{k=1}^\infty E_k\right) &= m^*(G) \\
    &\geq m^*\left(G \cap \bigcup_{k=1}^\infty E_k\right) \\
    &= m^*\left(\bigcup_{k=1}^\infty (G \cap E_k)\right) \\
    &= \sum_{k=1}^\infty m^*(G\cap E_k) \\
    &\geq \sum_{k=1}^\infty m^*(A\cap E_k)
\end{align*}
where the fourth line follows because $G \cap E_k$ is measurable for all $k$ and the inequalities follow from the monotonicity of outer measure.
