Meaning $P(X=Y)$. I am working on this measure theory problem. We have two RV's $X$ and $Y$ defined on the space $(\Omega, \mathcal{A}, \mathbb{P})$. We have some conditions, and need to prove as a result that that $\mathbb{P}(X=Y) = 0$. (I do not want to put the whole question on the internet, therefore the vague description).
However, my question is mostly, how do I interpret the $\mathbb{P}(X=Y)$? 
 A: This is an elaboration of what has already been written in the comments.
In measure-theoretic probability, events are the same thing as measurable subsets of the probability space $\Omega$, and the probability of an event is the same thing as the $\mathbb P$-measure of that measurable set. In this case, the event "$X=Y$" means "the subset of $\Omega$ consisting of $\omega\in\Omega$ such that $X(\omega)=Y(\omega)$". Since $X$ and $Y$ are random variables (i.e., measurable functions from $\Omega$ to $\mathbb R$) we also have that $X-Y$ is a measurable function, and thus its zero set is a measurable subset of $\Omega$. The $\mathbb P$-measure of this set is what you are after:
$$
P(X=Y):=\mathbb P\Bigl(\bigl\{\omega\in\Omega\colon X(\omega)=Y(\omega)\bigr\}\Bigr).
$$
For example, if $\Omega=[0,1]^2$ and $\mathbb P$ is the Lebesgue measure, and $X,Y$ are the corresponding coordinate functions, then the event $X=Y$ consists of the diagonal subset $\{(t,t)\colon t\in[0,1]\}$ which has Lebesgue measure zero, hence $P(X=Y)=0$ in this case. It's not a bad picture to have in mind either, even when $X,Y$ are arbitrary non-atomic independent random variables.
