Prove that $\mu$(E$\Delta$F) = 0 implies $\mu$(E) = $\mu$(F) So here is what we were given. Let (X, $\mathscr{M}$, $\mu$) be a measure space and let E, F $\in$ $\mathscr{M}$. Prove that $\mu$(E$\Delta$F) = 0 implies $\mu$(E)=$\mu$(F). Know that E$\Delta$F = E$\setminus$F $\cup$ F$\setminus$E.
Here is what I have so far. $$(E \cup F) = (E\setminus F)\cup (E \cap F) \cup (F\setminus E)$$ $$\Rightarrow (E \cap F) \cup (E \Delta F)$$ 
It's from here where I'm not sure where to go. Would I say that going on the steps that I gave above $$\mu(E \cup F) = \mu(E\setminus F)\cup \mu(E \cap F) \cup \mu(F\setminus E)$$ $$\Rightarrow \mu(E)-\mu(F) + \mu(0) + \mu(F)-\mu(E)$$
Am I on the right track with this one? Again and as per usual any help is always appreciated.
 A: Hint: we can rewrite $E$, $\newcommand{\sdiff}{\color{darkgreen}{EΔ F}}$
$$\mu(E) = \mu(E∩[\sdiff]) + \mu(E∩ F) = \mu(\sdiff) + \mu(E∩ F) = 0+\mu(E∩ F) $$
Since $0\leq \mu(E∩ [\sdiff]) \leq \mu(\sdiff) = 0$. 
A: $\mu E\Delta F = \color{red}{\mu E-\mu (F\cap E)}+\color{blue}{\mu F-\mu (F\cap E )}=0$ consider with $\mu (F\cap E )\leq \mu E$ & $\mu (F\cap E )\leq \mu F$(since $F\cap E\subseteq E$..)
thus we have sum of two positive number equal zero which give both equal zero, 
$\mu (F\cap E )= \mu E=\mu F$
A: What you have so far is correct, and I'm sure you realize that $E\cap F$ and 
$E \Delta F$ are disjoint.  
Now observe that $E = (E \cap F) \cup (E \cap F^C)$ and that these two sets are pairwise disjoint.  Similarly, $F = (F \cap E) \cup (F \cap E^C)$.
Thus by countable additivity, we have that $\mu(E) = \mu(E \cap F) = \mu(F)$.
A: Recall that $$E_{1}\triangle E_{2}=\big(E_{1}\setminus E_{2}\big)\cup\big(E_{2}\setminus E_{1}\big)=\big(E_{1}\cap E_{2}^{C})\cup\big(E_{2}\cap E_{1}^{C})$$
Since $(E_1 \setminus E_2)$ and $(E_2 \setminus E_1)$ are pair-wise disjoint and the finite additivity of $\mu$ we get that:
$$ \mu\bigg[\big(E_{1}\cap E_{2}^{C})\cup\big(E_{1}\cap E_{2}^{C})\bigg]=\mu\bigg[\big(E_{1}\cap E_{2}^{C})\bigg]+\mu\bigg[\big(E_{2}\cap E_{1}^{C})\bigg]$$


*Since $E_{1},E_{2}$ are $\mu$-measurable we get that 
$$ \mu\big[E_{1}]=\mu\big[E_{1}\cap E_{2}]+\mu\big[E_{1}\cap E_{2}^{C}]\implies \mu\big[E_{1}\cap E_{2}^{C}]=\mu\big[E_{1}]-\mu\big[E_{1}\cap E_{2}]$$
$$ \mu\big[E_{2}]=\mu\big[E_{2}\cap E_{1}]+\mu\big[E_{2}\cap E_{1}^{C}]\implies \mu\big[E_{2}\cap E_{1}^{C}]=\mu\big[E_{2}]-\mu\big[E_{2}\cap E_{1}] $$
Recall that $\mu$ is non-negative and monotone and $\mu\big[E_{1}\cap E_{2}]$ is smaller or equal than either $\mu\big[E_{1}]$ or $\mu\big[E_{2}]$ we get that:
$$
\mu\big[E_{1}]-\mu\big[E_{1}\cap E_{2}]=-\big(\mu\big[E_{2}]-\mu\big[E_{1}\cap E_{2}]\big)
$$
$$\iff $$
$$
\begin{array}{c}
\mu\big[E_{1}]-\mu\big[E_{1}\cap E_{2}]=0\\
\mu\big[E_{2}]-\mu\big[E_{1}\cap E_{2}]=0
\end{array}\implies m[E_{1}]-\mu[E_{2}]=0\implies \mu[E_{1}]=\mu[E_{2}]
$$
Hope this helps anyone studying measure theory right now :)
