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**tttcomrader** If $\displaystyle \mu $ is an outer measure on X and $\displaystyle \{ A_n \} ^ \infty _{n=1} $ is a sequence of disjointed $\displaystyle \mu $ measurable sets, prove that $\displaystyle \mu ( E \cap ( \bigcup ^ \infty _{n=1} A_n )) = \sum ^ \infty _{n=1} \mu (E \cap A_n ) \ \ \ \forall E \subset X $

Proof so far.

Since $\displaystyle \mu $ is an outer measure on $\displaystyle X $, we know that:

1. $\displaystyle \mu (A) \leq \mu (B) $ if $\displaystyle A \subset B $

2. $\displaystyle \mu ( \bigcup _{n=1} ^ \infty A_n ) \leq \sum ^ \infty _{n=1} \mu (A_n) $

And since $\displaystyle \{ A_n \} ^ \infty _{n=1} $ is $\displaystyle \mu $-measurable, then for each n, we have $\displaystyle \mu (E) = \mu (E \cap A_n )+ \mu (E \cap A_n ^c ) \ \ \ \forall E \subset X $

Now, $\displaystyle \mu (E \cap ( \bigcup _{n=1} ^ \infty A_n)) \leq \sum _{n=1} ^ \infty \mu (E \cap A_n ) $ since $\displaystyle \mu $ is an outer measure.

So what is left to prove is that $\displaystyle \mu (E \cap ( \bigcup _{n=1} ^ \infty A_n)) \geq \sum _{n=1} ^ \infty \mu (E \cap A_n ) $

So I have $\displaystyle \sum _{n=1} ^ \infty \mu (E \cap A_n ) = \mu (E \cap A_1)+ \mu (E \cap A_2) + . . . $

And I know I will have to use the measurable property, but I don't seem to be making any progress here...

Any hints? Thank you.