Fubini's theorem and $\sigma$-finiteness? 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.
 A: 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.
A: I'm answering my own question in the hopes that it may help someone else out there.  It will help me when I refer back to this page.
Here is the idea:  We can construct a sequence of simple functions $s_{n}$ which is monotonically increasing and converges to a non-negative measurable function $f$ (I discussed the usual construction in a comment on the other answer to this question).  This is true for any non-negative measurable function $f$.  The thing is, we don't necessarily know that the measure of the sets where each $s_{n}$ takes a non-zero, finite value is finite.  We could have that the measure of the set where $s_{n}$ takes the value $2$, for example, is $\infty$.  This could be true for all of the simple functions $s_{n}$, and so it would consequently be true for $f$.  (Here, $f$ has codomain $[0,\infty]$.)  
So, when we assume that $f \in L^{1}(d\lambda)$, that means $\int \limits_{Z} |f| \,d\lambda < \infty$, and since $|f| = f$ because $f$ is non-negative, then we have $f < \infty$ a.e. d$\lambda$.  Since $s_{n} \leq f$ for all $n$, then for each $n$, $s_{n}$ can only take the value $\infty$ on a set of measure $0$.  Furthermore, by the monotonicity of the integral ($\int \limits_{Z} s \,d\lambda \leq \int \limits_{Z} f \,d\lambda < \infty$), since the integral of the simple function is finite, the sets where it takes those non-zero values will have finite measure.  So if $f \in L^{1}(d\lambda)$ and it is non-negative, then our usual construction of a sequence of simple functions will work, because the simple functions we construct have the property that the measures of the sets where they take non-zero values are finite, since their integral is finite (and the simple functions are finite almost everywhere).
Now, if we are not in $L^{1}(d\lambda)$, then $f$ could have integral $\infty$.  This is a problem, because the usual construction of the sequence of simple functions does not necessarily have that the measure of sets where the function takes non-zero values is finite.  The sequence of simple functions could have integral $\infty$ by having the measure of a set where they take a non-zero value be infinite.  Then we can't run through the proof of Fubini's theorem, which depended on this property, to prove Tonelli's theorem.  But assuming the space is $\sigma$-finite helps us, because we can construct a sequence of simple functions converging up to $f$ which have the desired property.  Just use $\sigma$-finiteness.  Write the space $Z = \bigcup \limits_{n = 1}^{\infty} E_{n}$ with $\lambda(E_{n}) < \infty$ and $E_{n} \subseteq E_{n + 1}$.  Then $\chi_{E_{n}} \leq \chi_{E_{n + 1}}$, and since with our usual construction, $s_{n} \leq s_{n + 1}$, then $\{ s_{n} \chi_{E_{n}} \}$ is a sequence of simple functions that is increasing monotonically and converges up to $f$.  Then we can run through the regular proof of Fubini's theorem to prove Tonelli's theorem. 
