$\mu^*$ measurable set in the $\sigma$ ring

Question

If $\mu$ is a measure on a ring $R$, $A\subset E$ where $E\in \sigma(R)$ and $\mu^*(E)<\infty$, then $A$ is $\mu^*$ measurable if and only if $\mu^*(E) = \mu^*(A) + \mu^*(E-A)$ (1).

I would like to prove the $\Leftarrow$ implication, but I'm lost on what I have to consider. Specifically, if (1) holds I have to prove

$$ \mu^*(B) = \mu^*(B\cap A) + \mu^*(B\cap A^c) $$

for all $B\in\mathcal{H}(R)$ ($\supset \sigma(R)$). My problem is that I don't know how to relate $B$ with equation (1). For sure $B = (B\cap A)\cup (B\cap A^c)$, but more than that I wouldn't know how to continue.

EDIT:

So this is how far I am right now. I don't know if it's ok the reasoning though.

Let $\mathcal{H}$ be the hereditary class of sets that can be countably covered by elements of a ring $\mathcal{R}$. Let $\mathcal{M}$ be the class of $\mu^*$-measurable sets, i.e. $H\in \mathcal{H}$ is $\mu^*$-measurable in case

$$ \mu^*(B) = \mu^*(B\cap H) + \mu^*(B\cap H^c) $$

for all $B\in \mathcal{H}$. The following chain is known $\sigma(R)\subset \mathcal{M} \subset \mathcal{H}$. Thus, $E$ is $\mu^*$-measurable, i.e.

$$ \mu^*(B) = \mu^*(B\cap E) + \mu^*(B\cap E^c). $$

If we can prove $\mu^*(B) \geq \mu^*(B\cap A) + \mu^*(B\cap A^c)$ then we are done. Since $E = (E-A) \cup A$ we have

\begin{equation} \begin{split} \mu^*(B) &= \mu^*[B\cap((E-A) \cup A)] + \mu^*(B\cap E^c)\\ &\geq \mu^*(B\cap A) + \mu^*(B\cap E^c).\\ \end{split} \end{equation}

Up to here, I haven't used condition (1), so I guess one needs to use it in writting $E^c$ as a "function" of $A^c$ but I'm lost on how to write this relationship.

Answer 1

See here for a similar question on Lebesgue Measure .

I am going to cheat a little and use the notations of the question that I linked . So in your question I am going to replace $E$ with $M$ and $A$ with $E$ .

So reformulated your question becomes

If $\mu$ is a measure on a ring $R$, $E\subset M$ where $M\in \sigma(R)$ and $\mu^*(M)<\infty$, then $E$ is $\mu^*$ measurable if and only if $\mu^*(M) = \mu^*(E) + \mu^*(M-E)$ (1).

There exists a $R_{\sigma\delta}$ set $G$ such that $\mu^*(M-E)=\mu^*(G)$ . See Royden page $351$ for a proof. Otherwise you can try it yourself. It is almost exactly the same as showing for any set $E$ of finite Lebesgue outer measure there exists a $G_{\delta}$ set $G$ such that $\lambda^*(G) =\lambda^* (E)$ .

Then $M-E\subset M\cap G\subset G$ and hence $\mu^{*}(M-E)\leq \mu(M\cap G)\leq \mu(G) = \mu^{*}(M-E)\implies \mu^{*}(M-E)=\mu(M\cap G)$ .

As $M\cap G$ is measurable we have by Caratheodory cut condition on $M\cup G$ with $M$

$M\cap(M\cap G)= M\cup G$ and $M\cap(M\cap G)^{c}=M-G$

So $\mu(M)=\mu(M\cup G)+\mu(M-G)=\mu^{*}(M-E)+\mu(M-G)$ .

Now $\mu^*(M-E)=\mu(G)$ and we have the following claim that if $E$ is measurable and $F$ is any set then $\mu^{*}(E\cup F)+\mu^{*}(E\cap F)=\mu(E)+\mu^{*}(F)$ (You can prove this by using Cut condition on $E\cup F$ and $ F$ and is a standard result if you already know it. It is widely used in probability) . Now using this result for $E\subset F$ and $F-E$ to get $\mu^{*}(E\cup(F-E))+\mu^{*}(E\cap(F-E))=\mu^{*}(F)+\mu^{*}(\emptyset)=\mu(E)+\mu^{*}(F-E)$

Hence we have for $E$ measurable and $E\subset F$ that $\mu^{*}(F)=\mu(E)+\mu^{*}(F-E)$

Thus we have $\mu(M)-\mu^{*}(M-E)=\mu^{*}(E)$ and this would mean $\mu^{*}(E)=\mu(M-G)$

Thus we have a measurable set $M-G$ whose measure equals the outer measure of $E$ and $\mu^{*}(E)<\infty$ .

This means that $\mu^{*}(E-(M-G))=\mu^{*}(E)-\mu^{*}(M-G)=0$ (Using the claim) and hence $E-(M-G)$ is of measure $0$ and hence measurable. (Use the fact that the measure space obtained by restricting the outer measure to the sigma algebra of the $\mu^{*}$ measurable sets forms a complete measure space see Royden page 349 )

This means $E=(E-(M-G))\cup (M-G)$ is measurable.