I'm reviewing my analysis notes, and I am really confused about what is meant by $\sigma$-finiteness being a hidden hypothesis of Fubini's theorem.
Here is Fubini's theorem as was stated to me:
Suppose $(X, \Sigma, \mu)$ and $(Y, \tau, \nu)$ are complete measure spaces. Consider the complete product measure space $(X \times Y, \overline{\Sigma \times \tau}, \lambda)$. If $f \in L^{1}(d\lambda)$, then $\int \limits_{X\times Y} f d\lambda = \int \limits_{X} \left [ \int \limits_{Y} f d\nu \right ] d\mu$.
I was also told that a measure space $(X, \Sigma, \mu)$ being $\sigma$-finite is equivalent to there existing $f \in L^{1}(d\mu)$ such that $f > 0$. This equivalence was easy to prove.
I can't wrap my head around where $\sigma$-finiteness is "hiding" in this theorem (although my professor mentioned something about the ability to find a sequence of simple functions $s_{n}$ which is monotonic increasing and converges to our non-negative function $f$ — this was a necessary step in our proof). Thanks for any help you can offer.
Best Answer
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.