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

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?

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.
• Why counting measure is not $\sigma-$ finite? I know the definition of $\sigma-$ finite but how to prove is not..? Dec 9, 2019 at 3:04
• $[0,1]$ 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. Dec 9, 2019 at 3:06
• 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 $X \times X$ Dec 9, 2019 at 3:09