Failure of Tonelli without $\sigma$-finiteness: the diagonal under Lebesgue times counting measure

Let $X=[0,1]$ with Lebesgue measure $\lambda$ and let $Y=[0,1]$ with counting measure $\kappa$, so that $\kappa (A)$ is the number of elements of $A$ when $A$ is finite and $+\mathrm{\infty }$ otherwise. Let

$f={\mathbf{1}}_{\mathrm{\Delta }},\mathrm{\Delta }=\{(x,y)\in [0,1{]}^2:x=y\},$

the indicator of the diagonal. Then $f$ is nonnegative and measurable for the product $\sigma$-algebra, and

${\int }_X({\int }_{Y}f(x,y)d\kappa (y))d\lambda (x)={\int }_0^{1}1d\lambda =1,$

${\int }_Y({\int }_{X}f(x,y)d\lambda (x))d\kappa (y)={\int }_{Y}0d\kappa =0.$

The two iterated integrals are $1$ and $0$. Tonelli's theorem does not apply because $\kappa$ is not $\sigma$-finite on the uncountable set $[0,1]$, and no hypothesis on $f$ can repair this: $f$ is an indicator of a closed set, as good as a function can be.

Not proved in this library. The computation is a two-line consequence of the definitions of the two integrals, but both integrals belong to the deferred measure track (Lebesgue measure and the Lebesgue integral ‡), so it is recorded here rather than proved.

What would prove it. Only the definitions: the inner integral against counting measure of ${\mathbf{1}}_{\{x\}}$ is $1$ for each fixed $x$, and the inner integral against Lebesgue measure of the indicator of a single point is $0$ for each fixed $y$. The measurability of $\mathrm{\Delta }$ in the product $\sigma$-algebra is the one point needing care, and it follows since $\mathrm{\Delta }$ is closed and the product $\sigma$-algebra here contains the Borel sets of the square.

Which page it serves. The Fubini and change of variables page, as the counterexample that shows why the Lebesgue statement carries a $\sigma$-finiteness hypothesis while the Riemann statement on a box does not need one. It is the cheapest possible witness: no choice, no pathological set, no continuum hypothesis, just a measure that is too large.

Contrast with the other failure recorded here. In this example $f$ is perfectly measurable and the measures are at fault. In Sierpinski's example under the continuum hypothesis ‡ the measures are the best possible and the function is at fault. Between them they show that both hypotheses of Fubini-Tonelli theorem and the $\sigma$-finiteness hypothesis ‡ are needed.