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:

$\mu^*(\phi) = 0$

$A \subseteq B \Rightarrow \mu^*(A) \leq \mu^*(B)$

$\mu^*(\cup A_j) \leq \sum \mu^*(A_j)$

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$\mu^*(E\cap(\cup (A_j))=\mu^*(\cup(A_j\cap E))$ 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 $\mu^*(\cup(A_j\cap E))=\sum\mu^*(A_j\cap E)$
• But an outer measure only has the property that the size of the union is $\leq$ the sum of the sizes. Do you not need $\mu^*$ to be a measure to deduce equality? I know it is, when restricted to measurable, but only via Caratheodony, is there an alternative? – T. Kiley Apr 3 '14 at 13:27
• Because the sets are $\mu^*$ measurable doesn't the axiom then hold? since $\mu^*$ is then a measure on them? – Ellya Apr 3 '14 at 13:32