$A=B$, almost everywhere if and only if $\mu(A\Delta B)=0$

Question

Let $(S,\mathcal{S},µ)$ be a measure space.

We say that $A=B$, almost everywhere (a.e.), for $A,B∈S$ if $\chi_A=\chi_B$, a.e. How do I show that $A=B$ a.e., if and only if $\mu(A\Delta B)=0$

I figure $\Delta$ denotes the symmetric difference: $A\Delta B=(A\setminus B)\cup(B\setminus A)$.

what does that tells about $(A\setminus B)=(B\setminus A)$? How do I get to the wanted conclusion?

Answer 1

$\Leftarrow)$If $0=\mu(A\Delta B)=\int \chi_{A\Delta B}d\mu=\int(\chi_A+\chi_B-2\chi_{A\cap B})d\mu$. So $\mu(A)+\mu(B)=2\mu(A\cap B)$. From the fact that $\mu$ is monotonous, it follows that $\mu(A\cap B)\leq \min\{\mu(A),\mu(B)\}$. Therefore $\int\chi_Ad\mu=\mu(A)=\mu(A\cap B)=\mu(B)=\int\chi_Bd\mu$.