Show the outer measure of a union is the sum of the measures without Caratheodony I am attempting the following question:
Let $\mu^*$ denote an exterior measure, $\{A_j\}$ collection of disjoint, $\mu^*-measurable$  sets, show for any E:
$\mu^*(E \cap (\cup(A_j)) = \sum \mu^*(E\cap A_j)$
Using the fact that the collection of measurable sets is a $\sigma -algebra$ and hence closed under intersection, plus that an outer measure is a measure on measurable sets, the results follows immediately. However, given the next question is to state Caratheodony, it seems odd you would need it. 
However, I can't see how you can get an equality without something showing it is a measure on these sets. 
Update
Sorry, Carathédony states that given an outer measure on X, the set M of sets that are measurable wrt $\mu^*$ form a sigma algebra and $\mu^*$ restricted to M is a measure (e.g. we have the property that the size of the disjoint union of sets is equal to the sum of the measures). 
However, I would like to know if proving the above statement is possible without using the fact that $\mu^*$ is a measure. Instead, I would like only to use the definition of measureability:
A measurable if
$\forall E \in X, \mu^*(E) = \mu^*(E \cap A) + \mu^*(E \cap A^c)  $
and that $\mu^*$ is an outer measure:
$\mu^*(\phi) = 0$
$ A \subseteq B \Rightarrow \mu^*(A) \leq \mu^*(B)$
$\mu^*(\cup A_j) \leq \sum \mu^*(A_j)$
 A: To get equality you just need to show that:
$\mu^*(E\cap(\cup (A_j))=\mu^*(\cup(A_j\cap E))$ now show that $A_j\cap E$ are disjoint for all $j$, and you get equality. Since all $A_j$ are disjoint, it follows that $A_j\cap E$ are disjoint, since $A_j\cap E\subset A_j$.
Thus $\mu^*(\cup(A_j\cap E))=\sum\mu^*(A_j\cap E)$
A: We can establish equality for the finite case first. The following proof is from Royden's Real Analysis (4-th ed.), Ch. 17, p.348.
Lemma:
If $E \subset X$ and $\{ A_j \} _{j=1} ^n$ is a collection of disjoint, $\mu^*$ measurable sets, then
$$
  \mu ^* \left(E \cap \bigcup _{j=1} ^n A_j \right) = \sum _{k = 1} ^n \mu^*(E \cap A_j).
$$
Proof:
By induction. The equality holds for $n = 1$. Assume it holds for $n - 1$, then as $\{ A_j \}_{j = 1} ^n$ is disjoint,
$$
  E \cap \left( \bigcup _{j=1} ^n A_j \right) \cap A_n = E \cap A_n 
  \quad \text{and} \quad 
  E \cap \left( \bigcup _{j=1} ^n A_j \right) \cap A_n^c = E \cap \bigcup _{j=1} ^{n-1} A_j \tag{*}.
$$
By the measurability of all $A_n$'s and the inductive hypothesis (I.H.), taking $E \cap \bigcup _{j=1} ^n A_j$ as the set in the definition of measurability,
$$
\begin{align*}
  \mu^*\left( E \cap \bigcup _{j=1} ^n A_j \right) 
  &= \mu^*\left( \left[ E \cap \bigcup _{j=1} ^n A_j \right] \cap A_n \right) 
    + \mu^*\left( \left[ E \cap \bigcup _{j=1} ^n A_j \right] \cap A_n^c \right) \\
  &= \mu^*(E \cap A_n) + \mu^* \left( E \cap \bigcup _{j=1} ^{n-1} A_j \right)
  \tag{from (*)} \\
  &= \mu^*(E \cap A_n) + \sum _{j = 1} ^{n - 1} \mu^*(E \cap A_j)
  \tag{from I.H.} \\
  &= \sum _{j = 1} ^n \mu^*(E \cap A_j) \, .
\end{align*}
$$
You question:
As you mentioned,
$$
  \mu^*\left( E \cap \bigcup _{j = 1} ^\infty \right) \leq \sum _{j = 1}^\infty \mu^*(E \cap A_j)
$$
follows from the countable monotonicity of $\mu^*$. For the other direction, because $\mu^*$ is monotone,
$$
\begin{align*}
  \left( E \cap \bigcup _{j = 1} ^\infty A_j \right) 
  &\supset \left(E \cap \bigcup _{j = 1} ^n A_j \right)
  \quad \text{for all } n \\
  \implies \mu^* \left( E \cap \bigcup _{j = 1} ^\infty A_j \right)
  &\geq \mu^* \left( E \cap \bigcup _{j = 1} ^n A_j \right) \\
  &= \sum _{j = 1} ^n \mu^*(E \cap A_j) \, . \tag{from Lemma} 
\end{align*}
$$
The right hand side is independent of $n$, so we can take $n \to \infty$, giving
$$
  \mu^* \left( E \cap \bigcup _{j = 1} ^\infty A_j \right)
  \geq 
  \sum _{j = 1} ^\infty \mu^*(E \cap A_j) \, .
$$
