TITLE: Folland, Chapter 1 Problem 17
QUESTION [11 upvotes]: Problem 17: If $\mu^*$ is an outer measure on $X$ and $\{A_i\}_{i=1}^{\infty}$  is a sequence of disjoint $\mu^*$-measurable sets, then $\mu^*(E\cap \cup_{j=1}^{\infty} A_j)=\sum_{j=1}^{\infty}(E\cap A_j)$ for any $E\subset X$.
I've managed to show that the statement is true if $\mu(E)<+\infty$, by showing that
$$\mu^*(E)\ge \sum_{j=1}^{\infty}\mu^*(E\cap A_j)+\mu^*(E\cap B^C),$$
where $B=\cup_{j=1}^{\infty}A_j$, and in addition, that
$$\mu^*(E)=\mu^*(E\cap B)+\mu^*(E\cap B^C).$$
Now, if $\mu^*(E)<+\infty$, I can substitute the last result into the previous inequality, then subtract $\mu^*(E\cap B^C)$ from both sides of the result, since I am subtracting a finite number. The result is,
$$\mu^*(E\cap B)\ge \sum_{j=1}^{\infty} \mu^*(E\cap A_j),$$
which is equivalent to
$$\mu^*(E\cap \cup_{j=1}^{\infty} A_j)\ge \sum_{j=1}^{\infty}\mu^*(E\cap A_j).$$
Finally, because $\mu^*$ is an outer measure, I have easily that
$$\mu^*(E\cap \cup_{j=1}^{\infty} A_j)=\mu^*(\cup_{j=1}^{\infty}(E\cap A_j))\le \sum_{j=1}^{\infty}\mu^*(E\cap A_j).$$
However, I've failed to prove the result if $\mu^*(E)=\infty$. Perhaps it is not true? Counterexample? 
Any thoughts?

REPLY [2 votes]: Here's a cute alternative solution:
From the outer measure $\mu^*$, we get an outer measure $\mu^*|_{E}$ on $\mathcal{P}(E)$ defined as 
$$ \mu^*|_E(B) := \mu^*(E \cap B). $$
We claim then that each $E \cap A_j$ is $\mu^*|_E$-measurable, which is to say for any $B \subseteq E$ one has 
$$ \mu^*|_E(B) = \mu^*|_E(B \cap (E \cap A_j)) + \mu^*|_E(B \setminus (E \cap A_j)). $$
This of course follows from $\mu^*$-measurability of $A_j$, as 
\begin{align*}
\mu^*|_E(B) = \mu^*(B) &= \mu^*(B \cap A_j) + \mu^*(B \setminus A_j), \\
&= \mu^*(B \cap A_j \cap E) + \mu^*(B \setminus (A_j \cap E)), \\
&= \mu^*|_E(B \cap (E \cap A_j)) + \mu^*|_E(B \setminus (E \cap A_j)). 
\end{align*}
Using this, by Caratheodory's theorem, $\mu^*|_E$ is a measure on the collection of $\mu^*|_E$-measurable sets and therefore, 
\begin{align*}
\mu^*\bigg( E \cap \Big(\bigcup_{j=1}^\infty A_j\Big)\bigg) &= \mu^*|_E\bigg(\bigcup_{j=1}^\infty E \cap A_j\bigg), \\
&= \sum_{j=1}^\infty \mu^*|_E(E \cap A_j), \\
&= \sum_{j=1}^\infty \mu^*(E \cap A_j).
\end{align*}