# Why does Fubini's Theorem fail?

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?

• can you even define product measure without sigma finite? Oct 12, 2019 at 19:53
• I think $\sigma$-finiteness is needed for uniqueness, but not for existence. Oct 12, 2019 at 20:06
• You can still obtain a Fubini theorem for products of $s$-finite measures but like Martin said it's not uniquely determined by what it does on the measurable rectangles. Oct 12, 2019 at 20:10