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

Question

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?

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.