# A Non-Counterexample to the Fubini Theorem with Counting and Lebesgue Measures

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.

• 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 – Diesirae92 Dec 12 '17 at 10:48
• @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. – Darkdub Mar 14 '18 at 21:00

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 $$\mathbb{R}$$.