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$\mu^*$ is an outer measure on $X$ and $\{A_j\}_{1}^\infty$ is a sequence of disjoint $\mu^*$-measurable sets. Prove that $\mu^*(E\cap(\bigcup_1^\infty A_j))=\sum_1^\infty \mu^*(E\cap A_j)$ for any $E\subset X$.

This is what I have shown so far:

$\mu^*(E \cap (\bigcup_1^\infty A_j))=\mu^*(\bigcup_1^\infty (E \cap A_j))\leq \sum_1^\infty \mu^*(E\cap A_j) ;$ since $\mu^*$ is an outer measure.

If I can show the reverse inequality I will win. Can anyone help me?

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With induction it can be shown that $\mu^{*}\left(E\cap\left(\bigcup_{j=1}^{n}A_{j}\right)\right)=\sum_{j=1}^{n}\mu^{*}\left(E\cap A_{j}\right)$ for each $n$.

The inductionstep is: $$\mu^{*}\left(E\cap\left(\bigcup_{j=1}^{n}A_{j}\right)\right)=\mu^{*}\left(E\cap\left(\bigcup_{j=1}^{n}A_{j}\right)\cap A_{n}^{c}\right)+\mu^{*}\left(E\cap\left(\bigcup_{j=1}^{n}A_{j}\right)\cap A_{n}\right)=$$$$\mu^{*}\left(E\cap\left(\bigcup_{j=1}^{n-1}A_{j}\right)\right)+\mu^{*}\left(E\cap A_{n}\right)=\sum_{j=1}^{n-1}\mu^{*}\left(E\cap A_{j}\right)+\mu^{*}\left(E\cap A_{n}\right)$$

Then: $$\mu^{*}\left(E\cap\left(\bigcup_{j=1}^{\infty}A_{j}\right)\right)\geq\mu^{*}\left(E\cap\left(\bigcup_{j=1}^{n}A_{j}\right)\right)=\sum_{j=1}^{n}\mu^{*}\left(E\cap A_{j}\right)$$ This for each $n$, so: $$\mu^{*}\left(E\cap\left(\bigcup_{j=1}^{\infty}A_{j}\right)\right)\geq\sum_{j=1}^{\infty}\mu^{*}\left(E\cap A_{j}\right)$$

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