According to this question,
If $2$ r.v are equal a.s. can we write $\mathbb P((X\in B)\triangle (Y\in B))=0$
why is this so?
I get the equivalent statement $\mathbb P([(X\in B)\triangle (Y\in B)]^C)=1$ intuitively, but I don't see how to show rigorously that either follows from $P(\omega \in \Omega | X(\omega) - Y(\omega) \in \{ 0 \}) = 1$.
Note that the inclusion $\{X \in B\} \Delta \{Y \in B\} \subset \{X \ne Y\}$ holds, which immediately yields the result.
To see the inclusion: Assume $\omega$ is in the set on the left-hand side; Wlog $\omega \in \{X \in B\}$. Then by the definition of the symmetric difference $\omega \notin \{Y \in B\}$, i.e. we have $X(\omega) \in B$ and $Y(\omega) \notin B$. But this can only be if $X(\omega) \ne Y(\omega)$.