Fubini's theorem and $\sigma$-finiteness?

Solution 1:

Your Professor was probably talking about Tonelli's Theorem in regard to $\sigma$-finiteness.

If $f \in L^{1}(\mu\times\nu)$, then Fubini's theorem holds, regardless of $\sigma$-finiteness of $\mu$, $\nu$ or not. Of course all of the measures must be complete, including the product measure. The way this is proved is by reducing to the case of positive $f$ because the positive $f_{+}$ and negative parts have finite integrals. That allows you to approximate $f_{+}$, for example, by a non-decreasing sequence of non-negative simple functions $\{\varphi_{n}\}$ converging upward to $f_{+}$ with the property that each is supported on a set of finite measure. This approximation is a critical part of the standard proof.

Tonelli's Theorem is a generalization of Fubini's Theorem for the case of positive functions $f$, where the assumption of integrability of $f$ is dropped; that is, you allow for the possibility that $\int f\,d(\mu\times\nu) = \infty$. You still get the same conclusion as Fubini's Theorem for such a case, provided you assume that the measures $\mu$ and $\nu$ are $\sigma$-finite. By adding this assumption of $\sigma$-finiteness, you are still able to get the existence of $\{\varphi_{n}\}$ as above which are once again supported on sets of finite measure. So the proof of Fubini's Theorem goes through, even without assuming $f$ has a finite integral. However, in this case, I think you can see the need for $\sigma$-finiteness, whereas in Fubini's Theorem, it was only necessary to assume $\int |f|\,d(\mu\times \nu) < \infty$ in order to get the desired approximation.