MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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. $p_X$.

share|cite|improve this question
up vote 3 down vote accepted

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

share|cite|improve this answer
Indeed, easier than I thought! – Ilya Jul 11 '13 at 13:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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