Let $X = Y = [0,1].$ Let $\mathcal{B}$ denote the Borel $\sigma$-algebra. Let $m$ denote the Lebesgue measure on $[0,1]$, and let $\mu$ denote the counting measure on $[0,1].$ Prove that $D = \{(x,y) : x = y \}$ is measurable with respect to $\mathcal{B} \times \mathcal{B}.$ Furthermore, prove that $$\int_X \int_Y \chi_D(x,y) \, \mu(dy) \, m(dx) \neq \int_Y \int_X \chi_D(x,y) \, m(dx) \, \mu(dy).$$ Explain why this does not contradict the Fubini Theorem.
Given that $D$ is measurable with respect to $\mathcal{B} \times \mathcal{B},$ we believe that we can prove the statement about the integrals. We claim that we have \begin{align*} \int_X \int_Y \chi_D(x,y) \, \mu(dy) \, m(dx) &= 1, \text{ but} \\ \\ \int_Y \int_X \chi_D(x,y) \, m(dx) \, \mu(dy) &= \infty; \end{align*}
however, we are not certain about this, and we cannot prove that $D$ is measurable. We would appreciate any hints or tips on how to approach this problem.