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?

  • 4
    $\begingroup$ can you even define product measure without sigma finite? $\endgroup$ Oct 12, 2019 at 19:53
  • 1
    $\begingroup$ I think $\sigma$-finiteness is needed for uniqueness, but not for existence. $\endgroup$ Oct 12, 2019 at 20:06
  • $\begingroup$ 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. $\endgroup$
    – Matt Carr
    Oct 12, 2019 at 20:10


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