Outer measure induced by measure, equality of subsets Let $(X,\mathcal{M},\mu)$ be a measure space such that $\mu(X)=1$, and let $\mu^{*}$ be the outer measure induced by $\mu$. Suppose $E\subset X$ satisfies $\mu^{*}(E)=1$.
If $A,B\in \mathcal{M}$ and $A\cap E = B \cap E$, then $\mu(A)=\mu(B)$?
 A: Clearly $A$ and $B$ are $\mu^*$- measurable sets, therefore $\mu(A) = \mu^*(A)$ and $\mu(B) = \mu^*(B)$. Also note that $\mu^*(E^c) = 0$, because $\mu^*(E) = 1$.
$$\mu(A) = \mu^*(A) = \mu^*((A\cap E) \cup (A\cap E^c)) \leq \mu^*(A\cap E) + \mu^*(A\cap E^c) = \mu^*(A\cap E)$$
We know $\mu^*(A\cap E^c) = 0$ because $A\cap E^c\subset E^c$. Now we do the same reasoning for $B$.
$$\mu(B) = \mu^*(B) = \mu^*((B\cap E) \cup (B\cap E^c)) \leq \mu^*(B\cap E) + \mu^*(B\cap E^c) = \mu^*(B\cap E)$$
Finally, we know that $A\cap E = B\cap E$, therefore $\mu^*(A\cap E) = \mu^*(B\cap E) \implies$ $\implies\mu(A) = \mu(B)$.
A: I just encountered this question and found an answer.
First, I proved every measurable set $F \subset E^c$ has measure zero:
$E \subset F^c \implies \mu^*(E) \leq \mu(F^c)$
and since $\mu^*(E) = 1$, $\mu(F^c) = 1$ and $\mu(F) = 0$
Now just write $A$ and $B$ as disjoint unions
$A = (A\cap B)\cup(A\setminus B)$
$B = (A\cap B)\cup(B\setminus A)$
and since $A\setminus B$ and $B\setminus A$ are in $E^c$, they have measure zero and $\mu(A)=\mu(B)$.
