If X and Y are equal almost surely, then they have the same distribution 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 $P(\omega \in \Omega | X(\omega) - Y(\omega) \in \{ 0 \}) = 1$.
 A: 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 $X(\omega) \ne Y(\omega)$.
A: $$\{X=Y\} \subseteq [(X\in B)\triangle (Y\in B)]^C$$
Suppose $\omega \in  \{X=Y\}$ but on the contrary that $\omega \in (X\in B)\triangle (Y\in B)$.
Then we have two cases by $X$: $X(\omega) \in B$ or $X(\omega) \notin B$
If $X(\omega) \in B$, then $\omega$ belongs to $X \in B \cap Y \notin B$. Hence $Y(\omega) \notin B$. However, $Y(\omega)=X(\omega) \in B$. ↯
If $X(\omega) \notin B$, then $\omega$ belongs to $X \notin B \cap Y \in B$. Hence $Y(\omega) \in B$. However, $Y(\omega)=X(\omega) \notin B$. ↯
A: $$\{X=Y\} \subseteq [(X\in B)\triangle (Y\in B)]^C$$
Suppose $\omega \in  \{X=Y\}$. To show that $\omega \in [(X\in B)\triangle (Y\in B)]^C$.
We have to show that $\omega$ belongs to both of the following sets
$X \in B \cup Y \notin B$
$X \notin B \cup Y \in B$
For $X \in B \cup Y \notin B$, we have two cases by $X$: $X(\omega) \in B$ or $X(\omega) \notin B$


*

*If $X(\omega) \in B$, then $\omega$ belongs to $X \in B \cap Y \notin B$.

*If $X(\omega) \notin B$, then $Y(\omega) \notin B$ and hence $\omega$ belongs to $X \in B \cap Y \notin B$.
For $X \notin B \cup Y \in B$, we again have two cases by $X$: $X(\omega) \in B$ or $X(\omega) \notin B$


*

*If $X(\omega) \in B$, then $Y(\omega) \in B$ and hence $\omega$ belongs to $X \notin B \cup Y \in B$.

*If $X(\omega) \notin B$, then $\omega$ belongs to $X \notin B \cup Y \in B$.
