Proving that $(\mu^{*})^{*}=\mu^*$ for a measure $\mu$

Question

I am not sure I am using the standard definitions so I will open by defining what I need:

1. Let $X$ be a set, $\nu:\, \mathscr{P}(X)\to[0,\infty]$ will be called an external measure if $\nu(\emptyset)=0$ and for any $\{A_{i}\}_{i=1}^{\infty}\subseteq\mathscr{P}(X)$ (not neccesarily disjoint) it holds that $\nu(\cup_{i=1}^{\infty}A_{i})\leq\sum_{i=1}^{\infty}\nu(A_{i})$

2. Let $\nu$ be an external measure on a set $X$ then we say that a set $A$ is $\nu$ measurable if for any $E\subseteq X$: $\nu(E)=\nu(E\cap A)+\nu(E\cap A^{c})$

Let $\mu$ be a $\sigma$-additive measure on an algebra $A\subseteq X$, let $\mu^*$ be the outer measure on $X$ that comes from $\mu$.

It was proven in another exercise that the set of $\nu:=\mu^{*}$ measurable sets, $M$, is a $\sigma$ algebra and that $\nu$ is $\sigma$-additive on $M$.

Since $\nu$ is also a measure on the algebra $M$ there is an outer measure $\nu^*=(\mu^*)^{*}$.

The exercise wishes to prove that $(\mu^{*})^{*}=\mu^*$.

I don't really know how to even start here, what I first wanted to figure out is what is $M$. my intuition is that $A=M$ but I tried to prove it and couldn't start (I wrote the definitions but I couldn't see why if $B\in A$ it is $\nu$-measurable, not to mention the other direction which seems even harder).

Can someone please help me get started on this problem ? maybe a hint or an observation that might help me figure out what to do?

ADDED: by an outer measure that comes from a measure I mean $$\forall E\subseteq X:\,\nu(E):=Inf\{\sum_{i=1}^{\infty}\mu(A_{i})\,|\, A_{i}\in A,E\subseteq\cup A_{i}\}$$

Answer 1

Let $\scr A$ be the $\sigma-$algebra on which $\mu$ is defined, and let $\scr M$ be the set of measurable sets of $\mu^*$.

Claim: $ \forall E\subseteq X[\mu^{**}(E)\leq\mu^*(E)]$

Proof: Note that $\scr A\subseteq\scr M$. Thus: $\{\sum_{i=1}^{\infty}\mu(A_{i})\,|\, E\subseteq\cup A_{i}, A_{i}\in \scr{A}\}\subseteq$$\{\sum_{i=1}^{\infty}\mu(A_{i})\,|\, E\subseteq\cup A_{i}, A_{i}\in \scr{M}\}$,

Hence:

$\inf\{\sum_{i=1}^{\infty}\mu(A_{i})\,|\, E\subseteq\cup A_{i}, A_{i}\in \scr{A}\}\geq$$\inf\{\sum_{i=1}^{\infty}\mu(A_{i})\,|\, E\subseteq\cup A_{i}, A_{i}\in \scr{M}\}$

Now we proceed with the rest of the proof. Let $\epsilon>0$. Let $(A_i)_{i\in Z^+}$ be a sequence of sets in $\scr{M}$, such that $E\subseteq\cup_{i\in Z^+}A_i$ and $\sum_{i=1}^{\infty}\mu^*(A_i)<\mu^{**}(E)+\epsilon$. For each $A_i$, we choose a sequence of sets $(A_{i,j})_{j\in Z^+}$ in $\scr A$ such that $ A_i\subseteq \cup_{j\in Z^+}A_{i,j}$ and $\sum_{j\in Z^+}\mu(A_{i,j})<\mu^*(A_i)+\frac{\epsilon}{2^{i}}$. It follows that: $$\sum_{i\in Z^+}\sum_{j\in Z^+}\mu(A_{i,j})\leq\sum_{i\in Z^+}[\mu^*(A_i)+\frac{\epsilon}{2^{i}}]=\sum_{i\in Z^+}\mu^*(A_i)+\epsilon\leq (\mu^{**}(E)+\epsilon)+\epsilon$$

Note that $(A_{i,j})_{i,j\in Z^+}$ is a countable sequence of elements of $\scr A$ whose union contains $E$. Therefore, $\mu^*(E)\leq\sum_{i\in Z^+}\sum_{j\in Z^+}\mu(A_{i,j})$. Hence: $$\mu^*(E)\leq\mu^{**}(E)+2\epsilon$$ Since $\epsilon$ is an arbitrary positive number, we get $\mu^*(E)\leq\mu^{**}(E)$. We combine this with the previous claim to get $\mu^*(E)=\mu^{**}(E)$.

Answer 2

I seem to recall that given a premeasure $\mu$ on an algebra $\scr{A}$, each $A\in \scr{A}$ is a $\mu^*$-measurable subset, i.e. $\scr{A\subseteq M}$, where $\scr{M}$ is the $\sigma$-algebra of $\mu^*$-measurable subsets. It can also be shown (I think) that $\nu$ agrees with $\mu$ on $\scr{A}$. (the proofs of these statements can be found in Folland's Real Analysis book)

Now one can apply the same statements from above to the measure $\nu$ on $\scr{M}$ to show that the $\nu^*$ agrees with $\nu$ on $\scr{M}$. But as $\nu$ agrees with $\mu$ on $\scr{A}$ it follows that $\nu^*$ and $\mu$ agree on $\scr{A}$.

I think this is a solution, but of course I may be wrong. Unfortunately, I do not have any books with me at the moment, so if I were you I would double check a text for the validity of the statements I gave in the second paragraph.