Maps that are a.e. equal have almost the same graphs

Question

Let $X$ and $Y$ be two measurable spaces, and let $p$ be a probability measure on $X\times Y$. Denote by $p_X$ the marginal of $p$ on $X$, that is an image of $p$ under projection on $X$. Consider two measurable functions $f, g:X\to Y$ such that $f = g$ holds $p_X$-a.e. Is that true that $$ p\left(\mathrm{Gr}[f]\,\Delta\, \mathrm{Gr}[g]\right) = 0 \tag{1} $$ where $\Delta$ is the symmetric difference of sets and $$ \mathrm{Gr}[f]:=\{(x,f(x)):x\in X\} $$ is the graph of $f$ in $X\times Y$. Actually, I am mostly interested in the case when both $X$ and $Y$ are Borel spaces, and $f$ and $g$ are universally measurable maps, so in case $(1)$ does not hold in general, I would be still happy to know whether it holds true under some the latter assumptions.

I guess, one of the sufficient conditions would be that $p$ admits a regular kernel $\mu$ w.r.t. $p_X$.

Answer 1

The answer is easily yes, because we have $$ \operatorname{Gr}(f)\Delta \operatorname{Gr}(g)\subset N\times Y$$ where $N:=\{x\in X\, :\, f(x)\neq g(x) \}$ by assumption has null measure $$p_X(N):= p(\operatorname{Pr}_X^{-1}(N))=p( N\times Y )=0.$$