Outer measure "distributive" property Let $X$ be a set, $\mathcal{A}\subseteq 2^X$ is a ring of sets of $X$,
$\mu: \mathcal{A}\to [0, \infty]$ is a countably additive function.
And let $\mu^*$ is outer measure induced by $\mu$ on $2^X$.
Let $\{X_n\}$ be a sequence of $\mu^*$-measurable disjoint sets and
$A$ is arbitrary subset. It is necessary to prove the following property:
$$\mu^*\left(A\cap \bigg(\bigcup_{n}X_n\bigg)\right)=\sum_{n}\mu^*(A\cap X_n).$$
Let us remind the deficnition of outer measure $\mu^*:$
$$\mu^* \left({E}\right) = \inf \ \left\{{\sum_{n=1}^\infty \mu\left({A_n}\right) : A_n \in \mathcal{A}, \ E \subseteq \bigcup_{n=1}^\infty A_n}\right\}.$$
And we say that set $F\subseteq X$ is $\mu^*$-measurable if for every set $E\subseteq X$
$$\mu^*(E)=\mu^*(E\cap F)+\mu^*(E\setminus F).$$
I will be glad for any idea, comment, hint or advice.
 A: As Umberto suggested, I'm going to go ahead and assume that you meant to type $A \cap (\bigcup_{n = 1}^{\infty} X_n)$ as opposed to $A \cap (\bigcap_{n = 1}^{\infty} X_n)$.
First note that $$A \cap \left(\bigcup_{n = 1}^{\infty} X_n\right) = \bigcup_{n = 1}^{\infty} (A \cap X_n)$$ so we are interested in the outer measure of a disjoint union of sets, where the $\mu^*$-measurability of the sets $X_n$ is going to play an important role, naturally.
Since $X_1$ is $\mu^*$-measurable, I can write
\begin{align*}
\mu^* \left( \bigcup_{n = 1}^{\infty} (A \cap X_n) \right) &= \mu^* \left( \bigcup_{n = 1}^{\infty} (A \cap X_n) \, \, \, \cap \, \, \, X_1 \right) + \mu^* \left( \bigcup_{n = 1}^{\infty} (A \cap X_n) \, \, \, \setminus \, \, \, X_1 \right).
\end{align*}
Noticing that $$\bigcup_{n = 1}^{\infty} (A \cap X_n) \, \, \, \cap \, \, \, X_1 = A \cap X_1$$ because the $\{ X_n \}_{n = 1}^{\infty}$ are disjoint, we can rewrite the outer sum decomposition as
\begin{align*}
\mu^* \left( \bigcup_{n = 1}^{\infty} (A \cap X_n) \right) &= \mu^* \left( A \cap X_1 \right) + \mu^* \left( \bigcup_{n = 1}^{\infty} (A \cap X_n) \, \, \, \setminus \, \, \, X_1 \right).
\end{align*}
Now notice that $$\left( \bigcup_{n = 1}^{\infty} (A \cap X_n) \, \, \, \setminus \, \, \, X_1\right) = \bigcup_{n = 2}^{\infty} (A \cap X_n)$$ and apply the same trick with $X_2$ that was used with $X_1$ above, and you will see the desired sum gradually unfold before your eyes.
