Suppose that $\mathfrak A$ is an algebra on $X$ and that $\mu$ is a measure on $\mathfrak A$. It is given that $\pi^*(E\triangle F)=0, F \in M$, the set of all $\pi^*$ measurable sets. Then, it is to be shown that $E\in M$. $\pi^*$ is an outer measure on $X$ w.r.t. $\mu$. (Note: $E\triangle F:= (E\cap F^c)\cup (E^c\cup F)$, where little $c$ denotes complement in $X$.)
I'm having difficulty showing that.
I claim that $E\triangle F\in M$.
Proof: Let $E\triangle F=:K$. It is enough to show that for any $A\subset X$, $\pi^*(A)\ge \pi^*(A\cap K)+\pi^*(A\cap K^c).$
It's clear that $\overbrace{\pi^*(A\cap K)}^{\le \pi^*(K)=0}+\pi^*(A\cap K^c)=\pi^*(A\cap K^c)\le \pi^*(A).$ This proves the claim.
So we have $E\triangle F, F\in M$. But I am having difficulty showing that $E\in M$? I tried the following:
Let $A\subset X$ be chosen arbitrarily. It is enough to show that $\pi^*(A)\ge \pi^*(A\cap E)+\pi^*(A\cap E^c)$.
Since $F\in M$, $\pi^*(A\cap E)=\pi^*(A\cap E\cap F)+\overbrace{\pi^*(A\cap E\cap F^c)}^{\le \pi^*(E\triangle F)=0 }=\pi^*(A\cap E\cap F)$
and
$\pi^*(A\cap E^c)=\overbrace{\pi^*(A\cap E^c\cap F)}^{\le \pi^*(E\triangle F)=0}+\pi^*(A\cap E^c\cap F^c)=\pi^*(A\cap E^c\cap F^c)$.
I don't know how to proceed from here.
You already proved that $S = E \Delta F \in M$, and you know $F \in M$ by assumption,
so (I will just denote your $\pi^{\ast}$ by $\mu$), $\mu(A) \geq \mu(A \cap F) + \mu(A \cap F^c) = \mu(A \cap F \cap S) + \mu(A \cap F \cap S^c) + \mu(A \cap F^c \cap S) + \mu(A \cap F^c \cap S^c) = \mu(A \cap E \cap F) + \mu(A \cap E \cap F^c) + \mu(A \cap E^c \cap F) + \mu(A \cap E^c \cap F^c) = \mu(A \cap E \cap S) + \mu(A \cap E \cap S^c) + \mu(A \cap E^c \cap S) + \mu(A \cap E^c \cap S^c) = \mu(A \cap E) + \mu(A \cap E^c)$
which is what you need to show.