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
Best Answer
Yes, your reasoning is correct.