For two stochastic processes if $\Bbb E[|X(t)-Y(t)|^2]=0$ why does that imply $\Bbb P[X(t)=Y(t)]=1\ \forall t\in[0,T]$?

Question

Let $(X(t))_{t\in[0,T]},\ (Y(t))_{t\in[0,T]}$ be two continuous time stochastic processes.

If $\Bbb E[|X(t)-Y(t)|^2]=0\ \forall t\in[0,T]$ why does that imply $\Bbb P[X(t)=Y(t)]=1\ \forall t\in[0,T]$ ?

I think this question is equivalent to the slightly simpler

Let $X,Y$ be two random variables. If $\Bbb E[|X-Y|^2]=0$ why does that imply $\Bbb P[X=Y]=1$ ?

(This is from the proof of the uniqueness theorem for SDEs where $X$ and $Y$ are two solutions of $dX=f(t,X)dt+g(t,X)dW$ under some assumptions on $f,g$)

Answer 1

If $X\neq Y$ on a set with positive probability, then $|X-Y|^2>0$ on that set, and then $E|X-Y|^2 > 0$.

Or, put somewhat differently, if $Z$ is a non-negative random variable, then $E(Z)=0$ iff $Z = 0$ almost surely. Apply that to $Z=|X-Y|^2$ and note that $|X-Y|^2 = 0$ iff $X=Y$.

Answer 2

I guess it can be showed via the integral expression of the expectation

$\Bbb E[|X-Y|^2]=\int\limits_{\Omega}|X(\omega)-Y(\omega)|^2d\Bbb P=0\implies X=Y \ a.e$