Let $E=[0,1]$.
Here are the definitions I am using:
Let $A\subset E$, then we define the outer measure of $A$ as
$$\mu^*(A)=\inf \left\{\sum_k m(I_k): A\subset \cup_k I_k \right\}$$
where the infimum is taken over all any countable collection $\{I_k\}$ of intervals (open, closed or half open) whose union contains $A$, and we define the inner measure of $A$ as,
$$\mu_*(A)=1-\mu^*(E\setminus A)$$
and finally $A$ is said to be measurable if $\mu^*(A)=\mu_*(A)$.
As a note,
I have shown that
$$\mu^*(A)=\inf\{\mu(G): A\subset G, G \text{ is open relative to } E\}$$
so that we may use this characterization of the outer measure of a set or the one originally given above. We can also reformulate the definition of a measurable set: a set $A\subset E$ is measurable if and only if $\mu^*(E)=\mu^*(E\cap A)+\mu^*(E\setminus A)$ and this follows immediately from the fact that if $A$ is measurable according to the definition above, then $\mu^*(A)+\mu^*(E\setminus A)=1$ and since $A\subset E$, we have $A\cap E=A$.
Now, I want to show yet another equivalent characterization of measurability in the following,
$\textbf{Problem:}$
I am trying to prove if $A \subset E$ is measurable then for any $F\subset E$, we have
$$\mu^*(F)=\mu^*(F\cap A)+\mu^*(F\setminus A).$$
The hint in my book says to use $B\subset E$ is measurable if and only if for any $\epsilon >0$, $\exists G_1, G_2 \subset E$ and open relative to $E$ such that $B\subset G_1$, $E\setminus B\subset G_2$, and $\mu(G_1\cap G_2)<\epsilon$, so I also proved this (using the second characterization of $\mu^*$ provided here). For the current problem, this is my work so far:
Clearly,
$$\mu^*(F)\leq \mu^*(F\cap A)+\mu^*(F\setminus A)$$
by sub-additivity. For the other inequality, let $\epsilon >0$, then
by the above condition for measurability, we have (for some $G_1, G_2 \subset E$, etc),
$$F\cap A\subset A \subset G_1 \text{ and } F\setminus A\subset E\setminus A\subset G_2$$
and
$$\mu^*(F)+\mu^*(F\setminus A)\leq \mu(G_1)+\mu(G_2)=\mu(G_1 \cup G_2) + \mu(G_1\cap G_2)$$
but $G_1\cup G_2=E$, so the right hand side above is less than $1+\epsilon$. But I don't really see how this helps. We can conclude the left hand side is less than or equal to $1$ then since $\epsilon$ was arbitrary but I don't think I'm approaching this correctly. Any suggestions would be greatly appreciated, thanks.
Best Answer
Let $E=[0,1]$.
You already know that that $A\subset E$ is measurable (that is: $\mu^*(A)=\mu_*(A)$) if and only if
$$ \mu^*(E)=\mu^*(E\cap A)+\mu^*(E\setminus A)$$
You also know that:
if $A\subset E$ then $$\mu^*(A)=\inf\{\mu(G): A\subset G, G \text{ is open relative to } E\}$$
$A\subset E$ is measurable if and only if for any $\epsilon >0$, $\exists G_1, G_2 \subset E$ and open relative to $E$ such that $A\subset G_1$, $E\setminus A\subset G_2$, and $\mu(G_1\cap G_2)<\epsilon$ (The hint from your book is to use this property).
$\mu^*$ is subadditive.
In the proof below we will use properties 1, 2 and 3 and also the following four properties:
$\mu^*$ is monotone (that means, if $A\subset B$ then $\mu^*(A)\leqslant \mu^*(B)$).
if $G \subset E$ is open relative to $E$, then $\mu^*(G)=\mu(G)$.
if $G,H \subset E$ is open relative to $E$, then $\mu^*(G\setminus H)=\mu(G\setminus H)$.
$\mu$ is additive (in fact, it is $\sigma$-additive).
Proof: Let $A\subset E$ be measurable and let $F\subset E$. Let $\epsilon >0$.
Since $\mu^*(F)<\infty$, then, by property 1, there is $G_0$ open set relative to $E$, such that $F\subset G_0$ and $\mu^*(F)\leqslant\mu(G_0)<\mu^*(F)+\epsilon$.
Since $A$ is measurable, then, by property 2, $\exists G_1, G_2 \subset E$ and open relative to $E$ such that $A\subset G_1$, $E\setminus A\subset G_2$, and $\mu(G_1\cap G_2)<\epsilon$.
Let $D=E\setminus G_2$. Then $D\subset A$ and we have \begin{align} \mu^*(F\cap A)&+\mu^*(F\setminus A)\leqslant \mu^*(G_0\cap A)+\mu^*(G_0\setminus A) \leqslant & \textrm{ prop. 4} \\& \leqslant \mu^*(G_0\cap G_1)+\mu^*(G_0\setminus D) \leqslant & \textrm{ prop. 4} \\& \leqslant \mu^*(G_0\cap D)+\mu^*(G_0\cap (G_1\setminus D))+\mu^*(G_0\setminus D) = & \textrm{ prop. 3} \\& = \mu^*(G_0\setminus G_2)+\mu^*(G_0\cap (G_1\cap G_2))+\mu^*(G_0\cap G_2) = & \textrm{ definition of $D$} \\& = \mu(G_0\setminus G_2)+\mu(G_0\cap (G_1\cap G_2))+\mu(G_0\cap G_2) = & \textrm{ prop. 5 and 6} \\& = \mu(G_0)+\mu(G_0\cap (G_1\cap G_2)) \leqslant & \textrm{ prop. 7} \\& \leqslant \mu(G_0)+\mu(G_1\cap G_2) \leqslant & \textrm{ prop. 4} \\& \leqslant \mu^*(F)+\epsilon + \epsilon \end{align}
So, for any arbritary $\epsilon>0$, we have $\mu^*(F\cap A)+\mu^*(F\setminus A) \leqslant \mu^*(F)+2\epsilon$. So we have $$ \mu^*(F)\geqslant \mu^*(F\cap A)+\mu^*(F\setminus A)$$ Since $\mu^*$ is subadditive, we get $\mu^*(F)= \mu^*(F\cap A)+\mu^*(F\setminus A)$.