Outer measure induced by a measure Let $(X, \mathfrak{M}, \mu)$ be a measurable space.
Let $\mu^* \ : \ 2^X \ni Y \rightarrow \mu^*(Y)= \inf \{\mu(A) \ | \ A \in \mathfrak{M}, Y \subset A\}$.
Prove that $\forall Y \subset X  \ \ \exists A \in \mathfrak{M} \ : \ Y \subset A ,  \ \ \mu^*(Y)= \mu^*(A)$
and that all sets in $\mathfrak{M}$ are $\mu^* - $ measurable, this means that if $E \in \mathfrak{M}$, then $\forall T \subset X : \ \ \mu^*(T)= \mu^*(T \cap E) + \mu^*(T \setminus E)$.
The first thing would be obvious if the set $\{\mu(A) \ | \ A \in \mathfrak{M}, Y \subset A\}$ was closed.
Could you help me with this?
Thanks.
 A: For the first part, note that if $\mu^*(Y)=\infty$, then every set $A\in\mathcal{M}$ such that $Y\subset A$ satisfies $\mu(A)=\infty$. In particular, $\mu(X)=\infty$ and you can take $X$ as the set you're looking for.
If $\mu^*(Y)<\infty$, for each $n\in\mathbb{N}$, let $A_n$ be a set in $\mathcal{M}$ such that $\mu(A_n)<\mu^*(Y)+\frac{1}{n}$ and  $Y\subset A_n$. Let $A:=\bigcap_{n=1}^\infty A_n$.
Note that $Y\subset A$ and since $\mathcal{M}$ is a $\sigma$-algebra, $A\in\mathcal{M}$. By definition, $\mu^*(Y)\leq \mu(A)$. For $n\in\mathbb{N}$, $A\subset A_n$ hence $\mu(A)\leq \mu(A_n)<\mu^*(Y)+\frac{1}{n}$. Since this is valid for each positive integer $n$, we conclude that $\mu(A)\leq \mu^*(Y)$ and you can take $A$ as the set stated in the problem.
For the second part, note that, as defined, $\mu^*$ is subbaditive, which implies that for every pair of sets $E, T$, we have $\mu^*(T)\leq \mu^*(T\cap E)+\mu^*(T\setminus E)$ (since $T=(T\cap E)\cup (T\setminus E)$) so we only have to prove the reverse inequality when $E\in\mathcal{M}$.
If $\mu^*(T)=\infty$, by the previous remark, the equality is trivially satisfied, so assume that $\mu^*(T)$ is finite. Fix $\varepsilon>0$ and let $A\in\mathcal{M}$ such that $\mu(A)\leq \mu^*(T)+\varepsilon$.
Note that $T\cap E\subset A\cap E$, $T\setminus E\subset A\setminus E$ and $A\cap E, A\setminus E\in\mathcal{M}$. Hence, $\mu^*(T\cap E)\leq \mu(A\cap E)$ and similarly with $T\setminus E$. Then
 $$
\mu^*(T\cap E)+\mu^*(T\setminus E)\leq \mu(A\cap E)+\mu(A\setminus E)=\mu(A)\leq \mu^*(T)+\varepsilon
$$
Where the equalityequality holds since $\mu$ is a measure. Since $\varepsilon$ was arbitrary, we conclude that $ \mu^*(T\cap E)+\mu^*(T\setminus E)\leq \mu^*(T)$ and the result follows.
