Folland, Chapter 1 Problem 17 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?
 A: Let $\{A_j\}_{1}^{\infty}$ be a sequence of disjoint $\mu^*$-measurable sets and $E\subset X$. By countable subadditivity, $$\mu^*\left(E\cap\left(\bigcup_{j=1}^{\infty}A_j\right)\right) = \mu^*\left(\bigcup_{1}^{\infty}E\cap A_j\right) \leq \sum_{1}^{\infty}\mu^*(E\cap A_j)$$
Let $B_n = \bigcup_{1}^{n}A_j$. For each $n\geq 2$, since $A_n$ is $\mu^*$-measurable we have 
\begin{align*}
\mu^*(E\cap B_n) &= \mu((E\cap B_n)\cap A_n) + \mu^*((E\cap B_n)\cap A_n^c)\\
&= \mu^*(E\cap A_n) + \mu^*(E\cap B_{n-1})
\end{align*}
By induction, 
\begin{align*}
\mu^*(E\cap B_n) &= \mu^*\left(E\cap \bigcup_{1}^{n}A_j\right) + \mu^*\left(\bigcup_{1}^{n}E\cap A_j\right)\\
&= \sum_{1}^{n}\mu^*(E\cap A_j) \ \forall n\geq 1
\end{align*}
So then by monotonicty,
$$\mu^*\left(E\cap \bigcup_{1}^{\infty}A_j\right) \geq \mu(E\cap B_n) = \sum_{1}^{n}\mu^*(E\cap A_j) \ \forall n\geq 1$$
thus as $n\rightarrow \infty$ we have that $$\mu^*\left(E\cap \bigcup_{1}^{\infty}A_j\right) \geq \sum_{1}^{\infty}\mu^*(E\cap A_j)$$
therefore $$\mu^*\left(E\cap \bigcup_{1}^{\infty}A_j\right) = \sum_{1}^{\infty}\mu^*(E\cap A_j)$$
A: Hint: The cases I would distinguish are $\mu^*(E\cap \bigcup_{i=1}^\infty A_i) <+\infty$ and $\mu^*(E\cap \bigcup_{i=1}^\infty A_i) =+\infty$, rather than $\mu^*(E) = +\infty$ and $\mu^*(E) < +\infty$.
Or said differently, you might just as well substitute $E \to E\cap  \bigcup_{i=1}^\infty A_i$, since the part of $E$ outside of $ \bigcup_{i=1}^\infty A_i$ has no relevance.
A: 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*}
