3
$\begingroup$

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. $p_X$.

$\endgroup$

1 Answer 1

3
$\begingroup$

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 $$p_X(N):= p(\operatorname{Pr}_X^{-1}(N))=p( N\times Y )=0.$$

$\endgroup$
1
  • $\begingroup$ Indeed, easier than I thought! $\endgroup$
    – SBF
    Jul 11, 2013 at 13:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.