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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 \tag*{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.

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    $\begingroup$ the diagonal is a borel set, thus it is automatically in the Caratheodory completion. This does not contradict Tonelli's theorem since $\mu$ is not $\sigma$ finite $\endgroup$ – Diesirae92 Dec 12 '17 at 10:48
  • $\begingroup$ @Dylan_Carlo_Beck Hey did you have any luck in proving this? I've been stuck on this. Thanks. I'd love it if you could provide a proof if you have one. $\endgroup$ – Darkdub Mar 14 '18 at 21:00
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@Diesirae92 explains the measurability and lack of contradiction; we confine ourselves to calculations.

In the first inner integral, $$ \int_Y \chi_D(x,y) \, \mu(dy) $$ $\chi_D(x,y) \, \mu(dy)$ basically means, "if you fix an $x\in X$, how many times does the slice at $x$ intersect with $D$?" The answer is clearly once, at a single point, so the counting measure returns $\mu(\{x\})=1$. Then $\int_Y \chi_D(x,y) \, \mu(dy) = 1$, and then $\int_X 1\,m(dx)=1$, yielding $$ \int_X \int_Y \chi_D(x,y) \, \mu(dy) \, m(dx) = 1 $$

For the second part, the inner integral $$ \int_X \chi_D(x,y) \, m(dx) $$asks, "if you fix a $y\in Y$, what is the Lebesgue measure of the intersection with $D$?" The intersection is still a single point, but now this has Lebesgue measure zero. So $\int_X \chi_D(x,y) \, m(dx) =0$, whence $\int _Y 0\,\mu(dy)=0$, and
$$ \int_Y \int_X \chi_D(x,y) \, m(dx) \, \mu(dy) = 0 $$

It is interesting to note that $$ \int_{X\times Y} \chi_D(x,y) \, d(m\times \mu)= \infty $$Very roughly, this amounts to asking, "how many points are in $D$," and of course the answer is the cardinality of $\mathbb{R}$.

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  • $\begingroup$ Very nicely done. I appreciate the thorough explanation of how to interpret each of the integrals. Thanks for solving this one! $\endgroup$ – Carlo Jul 6 at 17:08

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