The following version of Fubini's theorem is from Aliprantis and Border:
Fubini's Theorem: Assume that $\mu: \mathcal{S}_{1} \rightarrow[0, \infty]$ and $v: \mathcal{S}_{2} \rightarrow[0, \infty]$ are measures on two semirings of subsets of the sets $X$ and $Y,$ and $Y,$ respectively. If $f: X \times Y \rightarrow \mathbb{R}$ is a $\mu \times v$ -integrable function, then both iterated integ rals $\iint f d v d \mu$ and $\iint f d \mu d v$ exist and $$ \int f d(\mu \times v)=\iint f d v d \mu=\iint f d \mu d v $$
A classical example that fails Fubini's theorem is the following from wikipedia:
Suppose that X is the unit interval with the Lebesgue measurable sets and Lebesgue measure, and Y is the unit interval with all subsets measurable and the counting measure, so that Y is not σ-finite. If f is the characteristic function of the diagonal of X×Y, then integrating f along X gives the 0 function on Y, but integrating f along Y gives the function 1 on X. So the two iterated integrals are different.
I understand why the example fails Tonelli's version of the theorem because one measure is not $\sigma$-finite. However, why does it fail the cited theorem?
