Let $X=Y=[0,1]$ and let $\mathcal{B}$ be the Borel $\sigma-$ algebra. Let $m$ be Lebesgue measure and $\mu$ counting measure on $[0,1]$.

(1) $D=\{(x,y): x=y\}$, show that $D$ is measurable w.r.t. $\mathcal{B}\times \mathcal{B}$.

I feel like that it is enough to show that $D\in \mathcal{B}([0,1]^2)=\mathcal{B}([0,1])\times \mathcal{B}([0,1])$. So we need to show $D$ is closed in $\mathbb{R}^2$. But how to show that?

(2) Show that $$\int_X\int_Y \chi_{D}\mu(dy)m(dx)\neq \int_Y\int_X\chi_{D}m(dx)\mu(dy)$$

I know that $\int_Y\int_X\chi_{D}m(dx)\mu(dy)=0$. But how to compute $$\int_X\int_Y \chi_{D}\mu(dy)m(dx)=\int_X\int_Y \chi_{D_x}(y)\mu(dy)m(dx)=\int_X\mu(\{x\})m(dx)=\int_Xm(dx)=?$$

(3) Why does this not contradict the Fubini-Tonelli theorem?

I feel like it does not satisfy $\sigma-$ finite measure?


1 Answer 1


The left-sided integral is $m([0,1])=1$. Remember that $\mu$ is not $\sigma$-finite on $[0,1]$, so Fubini Theorem fails in this case.

  • $\begingroup$ Why counting measure is not $\sigma-$ finite? I know the definition of $\sigma-$ finite but how to prove is not..? $\endgroup$
    – Hermi
    Dec 9, 2019 at 3:04
  • $\begingroup$ $[0,1]$ is uncountable, if it were the countable union of sets, some of the little set must be of uncountable as well, but the counting measure value on this set is $\infty$, not finite. $\endgroup$
    – user284331
    Dec 9, 2019 at 3:06
  • $\begingroup$ Okay, I see. How about the (1)? I remember there is a theorem: X is Hausdorff if, and only if, the diagonal $\{(x, x) | x ∈ X\}$ is a closed subspace of $X \times X$ $\endgroup$
    – Hermi
    Dec 9, 2019 at 3:09
  • $\begingroup$ Okay you can use that. $\endgroup$
    – user284331
    Dec 9, 2019 at 3:12
  • $\begingroup$ Check out my answer here: math.stackexchange.com/a/4587226/1060153 $\endgroup$ Nov 28, 2022 at 21:04

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .