# 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.

#### Solutions Collecting From Web of "Prove that $\mu$(E$\Delta$F) = 0 implies $\mu$(E) = $\mu$(F)"

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 $0\leq \mu(E∩ [\sdiff]) \leq \mu(\sdiff) = 0$.

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 $\mu(E) = \mu(E \cap F) = \mu(F)$.

$\mu E\Delta F = \color{red}{\mu E-\mu (F\cap E)}+\color{blue}{\mu F-\mu (F\cap E )}=0$ 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,
$\mu (F\cap E )= \mu E=\mu F$