TITLE: If X and Y are equal almost surely, then they have the same distribution, but the reverse direction is not correct
QUESTION [7 upvotes]: Show that if two random variables X and Y are equal almost surely, then they
  have the same distribution. Show that the reverse direction is not correct.

If $2$ r.v are equal a.s. can we write $\mathbb P((X\in B)\triangle (Y\in B))=0$ (How to write this better ?)
then 
$\mathbb P(X\in B)-\mathbb P(Y\in B)=\mathbb P(X\in B \setminus Y\in B)\le \mathbb P((X\in B)\triangle (Y\in B))=0$
$\Longrightarrow P(X\in B)=\mathbb P(Y\in B)$
but the other direction makes no sense for me, i don't know how this can be true.

REPLY [2 votes]: If $X$ is a random variable following uniform $\mathcal U(-1,1)$ distribution, then $X$ and $-X$ are identically distributed, but obviously $X$ and $-X$  are not almost surely equal, in fact $P(X=-X)=0$