Why does from this follow that $\mathcal{M}(\mu^*)$ is a $\sigma$-algebra and $\mu^*$ a measure on it? Let $\mu^*$ be an outer measure on $\Omega$. A set $A\subset\Omega$ is called $\mu^*$-measurable, if
$$
\forall E\subset\Omega:~~~\mu^*(A\cap E)+\mu^*(A^C\cap E)=\mu^*(E).
$$
Then
$$
\mathcal{M}(\mu^*):=\left\{A\subset\Omega:~\forall E\subset\Omega:~\mu^*(A\cap E)+\mu^*(A^C\cap E)=\mu^*(E)\right\}
$$
is the set of all $\mu^*$-measurable sets.
Now we have a theorem that says:


Let $\mu^*$ be an outer measure on $\Omega$. Then $\mathcal{M}(\mu^*)$ is a $\sigma$-algebra and the restriction from $\mu^*$ on $\mathcal{M}(\mu^*)$ is a measure.


This theorem, says the script, follows from the following three lemmas:
(1) If $\mu^*$ is an outer measure, then $\mathcal{M}(\mu^*)$ is an algebra.
(2) An outer measure $\mu^*$ is $\sigma$-additive on $\mathcal{M}(\mu^*)$.
(3) If $\mu^*$ is an outer mesaure, then $\mathcal{M}(\mu^*)$ is a $\sigma$-algebra.
Maybe I am only slow-witted, but it is not totally clear to me why the theorem above follows from (1) - (3). Can you explain that to me, please?
Greetings
#
Miro
 A: By the definition of a measure (i.e. here), given a set $\Omega$ and a $\sigma$-algebra $\Sigma$ on it, a function $\mu:\Sigma\to \overline{\mathbb{R}}$ is a measure if it satisfies:


*

*for all $A\in\Sigma$, $\mu(A)\geq 0$;

*$\mu(\emptyset)=0$;

*$\mu$ is $\sigma$-additive.


Denote $\tilde{\mu}$ the restriction of $\mu^*$ to $\mathcal{M}(\mu^*)$ (this means that $\tilde{\mu}$ is only defined on $\mathcal{M}(\mu^*)$ and for all $A\in\mathcal{M}(\mu^*)$ we have $\tilde{\mu}(A)=\mu^*(A)$). Now you have by (3) that $\mathcal{M}(\mu^*)$ is a $\sigma$-algebra and by (2) you know that $\tilde{\mu}$ is $\sigma$-additive on it. Thus $\tilde{\mu}$ verifies 3. of the definition above. By the definition of outer measure (see here) it follows that for all $A\subset\Omega$, $\mu^*(A)\geq 0$ and $\mu^*(\emptyset)=0$. Let $A\in\mathcal{M}(\mu^*)\subset \mathcal{P}(\Omega)$ (where $\mathcal{P}(\Omega)$ is the power set of $\Omega$). Then $\tilde{\mu}(A)=\mu^*(A)\geq 0$. Since $\mathcal{M}(\mu^*)$ is a $\sigma$-algebra, we have $\emptyset\in\mathcal{M}(\mu^*)$ and $\tilde{\mu}(\emptyset)=\mu^*(\emptyset)=0$. Therefore 1. and 2. of the definition above are also satisfied for $\tilde{\mu}$ and it is a measure.
