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I'm doing the problem 19 in Real Analysis of Folland like below:

Let $\mu^*$ be an outer measure on $X$ induced from a finite premeasure $\mu_0$. If $E \subset X$, define the inner measure of $E$ to be $\mu_*(E) = \mu_0(X) - \mu^*(E^c)$. Then $E$ is $\mu*$-measurable iff $\mu^*(E) = \mu_*(E)$

For the converse, I found the solution for it. For the $==>$ direction. I have a very simple solution compared to other solution I found on the Internet which I think may be wrong, but I can't find the hole in that solution. I post it here, so I hope someone can help me judge it. Thanks a lot.

Because $E$ is $\mu*$-measurable, we have $$\mu_0(X) = \mu^*(X) = \mu^*(X \cap E) + \mu^*(X \cap E^c) = \mu^*(E) + \mu^*(E^c)$$ Therefore $$\mu_*(E) = \mu_0(X) - \mu^*(E^c) = \mu^*(E)$$ Because $\mu_0$ is a finite premeasure, $\mu^*$ is a finite measure

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  • $\begingroup$ Problem 19 in what textbook? Also, what is your definition of measurable? There are many equivalent definitions, including the one that makes the problem trivial, so you want to establish all definitions clearly for this problem. $\endgroup$ – Gyu Eun Lee Jul 4 '15 at 2:07
  • $\begingroup$ I edited the question to make it clear where the source is. All in all, the measurable definition is based on the outer measure: $E$ is measurable iff for all $A \subset X$, we have $$\mu^*(A) = \mu^*(A \cap E) + \mu^*(A \cap E^c)$$ $\endgroup$ – le duc quang Jul 4 '15 at 3:04
  • $\begingroup$ Please share the proof of the other implication, if you will, because that's exactly where I'm stuck, and I assume more people will find this to be the harder implicatiln to prove $\endgroup$ – Akerbeltz Oct 29 '18 at 20:11
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Yes, your reasoning is correct.

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