Let $(X,\mathcal{M},\mu)$ be a measure space and $\mu^*:\mathbb{P}(X)\to [0,+\infty] $ given by $\mu^*(B)=\inf\{\mu(E): E\in\mathcal{M}, B\subset E\} $.
a) Prove that $\mu^*$ is an outer measure
b) Let $\mu(X)\lt\infty$, prove that for every $ B\subset X$, there exists $E \in \mathcal{M}$ such that $ B\subset E$ and $\mu^*(B)=\mu(E)$.
So I think that I have completed the first part, but I would like someone to check my proof and also help me with the part b).
For the first part I have this:
Let $H_B=\{\mu(E): E\in\mathcal{M}, B\subset E\}$, then $\mu^*(B)=\inf H_B $
and $\mu^*(B)\le\lambda$, for every $\lambda\in H_B$
We prove the 3 conditions of the definition of outer measure :
- $\mu^*(\emptyset)=0$
It's clear that $0\in H_\emptyset $, since $E=\emptyset\in\mathcal{M}$ and $\mu(\emptyset)=0.$ Then $0\le\mu^*(\emptyset)\le0 \implies\mu^*(\emptyset) = 0$
- $A\subset B \implies\mu^*(A) \le\mu^*(B)$
Let's see that $H_B\subset H_A$, because then $\mu^*(A)=\inf H_A \le \inf H_B =\mu^*(B) $
Let $\lambda\in H_B$, then there exist E such that $B\subset E$ and $\mu(E)= \lambda$. Since $A\subset B\implies A\subset E$ and $\lambda\in H_A$
- $\mu^*(\bigcup_{i\in I}A_i)\le \sum_{i\in I} \mu^*(A_i), A_i\in X,\forall i$
We have that for every i there exist a $\lambda_i\in H_{A_i} $ and an $E_i\in \mathcal{M}$ with $A_i\subset E_i$ and $\lambda_i=\mu(E_i)$ . This implies that $\bigcup_{i\in I} A_i \subset \bigcup_{i\in I}E_i$ and $\mu^*(\bigcup_{i\in I}A_i)\le \mu(\bigcup_{i\in I}E_i)$, since $\mu(\bigcup_{i\in I}E_i)\in H_{\bigcup_{A_i}}$. Now, because $\mu$ is a measure:
$\mu^*(\bigcup_{i\in I}A_i)\le \mu(\bigcup_{i\in I}E_i) \le \sum_{i\in I}\mu(E_i)$, so if $\mu^*(\bigcup_{i\in I}A_i)\le \sum_{i\in I}\mu(E_i)$ for any arbitrary $E_i$ that we have taken,then in particular it is less or equal than the infimum of them, so $\mu^*(\bigcup_{i\in I}A_i)\le \sum_{i\in I} \mu^*(A_i)$
I am not really sure that the last part of the proof of the condition 3 is correct, and for the part b) I don't really even know where to start, so I would be very thankful for the help!! :)
I think in your last argument you showed that $\mu^*({\bigcup}_{i\in I}A_i)\leq inf\{ \sum_{i \in I} \mu(E_i):A_i\subset E_i\}$ which is different than what you wanted.
Another way to approach this (standard technique):
$I$ is countable so i can assume that $I=\Bbb N$, i can also assume that $\mu^*(A_n)<\infty$ $\forall n \in \Bbb N$ , otherwise the desired inequality holds.
Now let $\epsilon >0$ then by the characterization of infimum i can find $\forall n \in \Bbb N$ sets $A_n\subset B_n$ with $B_n \in \mathcal M$ and $\mu^*(A_n)\leq \mu(B_n) < \mu^*(A_n)+\frac{\epsilon}{2^n}$.
Now $\bigcup A_n \subset \bigcup B_n$ so $\mu^*(\bigcup A_n)\leq \mu(\bigcup B_n)\leq \sum \mu (B_n)\leq \sum \mu^*(A_n)+\epsilon$ thus
$\mu^*(\bigcup A_n)<\sum \mu^*(A_n) + \epsilon $.Now , $\epsilon>0$ was arbitrary so we have the desired inequality.
Hint for b : use the characterization of infimum for $\epsilon = 1/n$ then you can find sets $B\subset B_n$ such that $B_n \in \mathcal M$ and $\mu^*(B)\leq \mu(B_n) <\mu^*(B)+1/n$, note that we can do that since $\mu^*(B)<\infty$.
Now take $E=\bigcap B_n$.