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Let $\mu^* : \mathcal{P}(X) \rightarrow [0, \infty]$ be an outer measure, and let $M$ denote the set of $\mu^*$-measurable sets.

Let $A \subseteq X$ and let $E,F \in M$.

Why is the following statement true?

$\mu^*(A \cap E^c) = \mu^*(A \cap F \cap E^c) + \mu^*(A \cap E^c \cap F^c)$

EDIT: Is it because $F \in M$, with $(A \cap E^c)$ serving as our "test set"?

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Yes (and in this case, it would be better to write $\mu^*(A\cap E^c\cap F)$ in order make make this clearer) –  Davide Giraudo Sep 12 '12 at 9:23
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$A \cap E^c = (A \cap F \cap E^c) \cup (A \cap E^c \cap F^c)$. Is your question about this statement or about the theorem in the title? –  Karolis Juodelė Sep 12 '12 at 9:24
    
What I wrote is required to understand the proof of the theorem I wrote in the title. I understand everything now -- thanks for the help. –  George Sep 12 '12 at 10:23
    
You can answer your own question :-) (to remove this from the list of unanswered questions). –  leo Sep 13 '12 at 1:33
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1 Answer

Yes, it is a special case of the property defining measurable sets, applied to $F$ against $A\cap E^c$.

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