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I have a simple homework question, but it's not so simple for me...

"Let $S$ be a $\sigma$-algebra on $X$, and $\mu$ a measure on $S$. Assume that $\mu(X)=1$.

Let $\mu^*$ be the outer measure induced by $\mu$. Assume there exists $E\subseteq X$ such that $\mu^*(E)=1$.

Prove that for every $A,B\in S$ such that $A\cap E=B\cap E$, then $\mu(A)=\mu(B)$".

I just can't see how to use $\mu^*(E)=1\ldots$

Thanks in advance!

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Contrast A,B$\in$S with $A,B\in S$. When you put all the math notation within $\TeX$ rather than just symbols you can't otherwise type, then the standard typestting conventions are followed. –  Michael Hardy Nov 19 '12 at 23:15
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If $A$ is measurable, then $\mu^* C = \mu^* (C \cap A) + \mu^* (C \setminus A)$ for all sets $C$. Choosing $C=E$ and $C=X$ gives $$\mu^* E = 1 = \mu^* (E \cap A) + \mu^* (E \setminus A) = \mu^* X = \mu^* A + \mu^* (X \setminus A)$$ Since $\mu^* (E \cap A) \leq \mu^* A$ and $\mu^* (E \setminus A) \leq \mu^* (X \setminus A)$, this implies that, in fact, $\mu^* (E \cap A) = \mu^* A = \mu A$.

Consequently, we have $\mu A = \mu^* (E \cap A) = \mu^* (E \cap B) = \mu B$.

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The key observation is $\mu(X)=\mu^{*}(E)$. You can show that $\mu^{*}(A\cap E)=\mu(A)$ and similarly $\mu^{*}(B\cap E)=\mu(B)$. Then you're done.

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