Outermeasure and $\mu^{*}$-measurable sets proof If $\mu^{*}$ is an outer measure on $X$ and $\{A_j\}_{1}^{\infty}$ is a sequence of disjoint $\mu^{*}$-measurable sets, then for every $E\subset X$ $$\mu^{*}\left(E\cap\left(\bigcup_{1}^{\infty}A_j\right)\right) = \sum_{1}^{\infty}\mu^{*}(E\cap A_j)$$
proof: Let $\{A_j\}_{1}^{\infty}$ be a sequence of disjoint $\mu^{*}$-measurable sets and $E\subset X$. Set, $$A = \bigcup_{1}^{\infty} A_j$$ Since $A$ is $\mu^{*}$-measurable then we have $$\mu^{*}(E) \geq \mu^{*}(E\cap A) + \mu^{*}(E\cap A^{c}) \ \  \forall E\subset X, \  \mu(E) < \infty$$ $$ = \mu^{*}\left(E\cap\left(\bigcup_{1}^{\infty}A_j\right)\right) + \mu^{*}\left(E\cap\left(\bigcap_{1}^{\infty}A_j^{c}\right)\right)$$ I am not sure if my approach will yield the desired result, any suggestions is greatly appreciated. 
 A: By the $\sigma$-subadditivity of outer measures, we have
$$\mu^{\ast}\Biggl(E \cap \bigcup_{j = 1}^{\infty} A_j\Biggr) \leqslant \sum_{j = 1}^{\infty} \mu^{\ast}(E\cap A_j).$$
Since the $A_j$ are $\mu^{\ast}$-measurable and disjoint, we have
\begin{align}
\mu^{\ast}\Biggl(E \cap \bigcup_{j = n}^{\infty} A_j\Biggr) &= \mu^{\ast}\Biggl(\Biggl(E\cap \bigcup_{j = n}^{\infty} A_j\Biggr)\cap A_n\Biggr) + \mu^{\ast}\Biggl(\Biggl(E\cap \bigcup_{j = n}^{\infty} A_j\Biggr)\setminus A_n\Biggr)\\
&= \mu^{\ast}(E\cap A_n) + \mu^{\ast}\Biggl(E\cap \bigcup_{j = n+1}^{\infty} A_j\Biggr)
\end{align}
and hence
$$\mu^{\ast}\Biggl(E\cap \bigcup_{j = 1}^{\infty} A_j\Biggr) = \mu^{\ast}\Biggl(E\cap \bigcup_{j = n+1}^{\infty} A_j\Biggr) + \sum_{j = 1}^{n} \mu^{\ast} (E\cap A_j)$$
for all $n$. Since $\mu^{\ast}(B) \geqslant 0$ for all sets $B$, that implies
$$\sum_{j = 1}^{n} \mu^{\ast}(E\cap A_j) \leqslant \mu^{\ast}\Biggl(E\cap \bigcup_{j = 1}^{\infty} A_j\Biggr)$$
for all $n$. By definition of the sum of an infinite series,
$$\sum_{j = 1}^{\infty} \mu^{\ast}(E\cap A_j) \leqslant \mu^{\ast}\Biggl(E\cap \bigcup_{j = 1}^{\infty} A_j\Biggr).$$
