# Questions on Fubini's Theorem and $\sigma$-finite measure?

I asked a question about this a several days ago, but I think I have a better formulated question now. The reason I did not just edit the last question about this is that I feel the answers I got were helpful, and this time around I have more detail to add.

So, if $(X \times Y, \overline{\Sigma \times \tau}, \lambda)$ is a complete measure space, with $\lambda = \mu \times \nu$, then, if $f \in L^{1}(d\lambda)$, Fubini's theorem gives us: $\int \limits_{X \times Y} f \,d\lambda = \int \limits_{X} \left [ \int \limits_{Y} f \,d\nu \right ] \,d\mu$. We did not need to assume the space was $\sigma$-finite, because $f \geq 0$ being in $L^{1}(d\lambda)$ allowed us to construct a monotonically increasing sequence of simple functions supported on sets of finite measure which converge to $f$.

But, my professor said that $\sigma$-finiteness is hidden in the hypothesis of Fubini's theorem. Specifically, we proved that the following are equivalent:

$\exists f > 0$ measurable such that $f \in L^{1}(d\lambda)$ $\iff$ $\lambda$ is $\sigma$-finite

This statement was easy to prove. Then, my professor said that when we are looking at $f \in L^{1}(d\lambda)$, we have $\int \limits_{X \times Y} f \,d\lambda = \int \limits_{X \times Y} f \chi_{ \{x \mid f(x) \neq 0 \} } \,d\lambda$, and the measure given by $\chi_{ \{x \mid f(x) \neq 0 \} }\,d\lambda$ is $\sigma$-finite. This is because the positive part of $f$, $f^{+}$, is in $L^{1}( \chi_{ \{x \mid f(x) \neq 0 \} } \,d\lambda)$.

Question 1: Okay, clearly $f^{+}$ is strictly positive, and also measurable with respect to $\lambda$. How do I know that it is measurable with respect to $\chi_{ \{x \mid f(x) \neq 0 \} }\,d\lambda$? Also how would I show that $f^{+} \in L^{1}( \chi_{ \{x \mid f(x) \neq 0 \} }\,d\lambda)$? I need to show that $\int \limits_{X} f^{+} \,d(\chi_{ \{x \mid f(x) \neq 0 \} }\,d\lambda) < \infty$. I'm not sure how.

Question 2: Since we have the above equivalence for $\sigma$-finite measures, given any measure space and any measure $\lambda$, if $f\in L^{1}(d\lambda)$, then the map $f \chi_{ \{ x \mid f(x) > 0 \} }$ is a positive measurable function in $L^{1}(d\lambda)$. So any measure space is $\sigma$-finite by this argument... But that can't be true. What is wrong with the argument? Answer: This function actually takes the value $0$....

• Why do you say that $f^+$ is strictly positive? It's zero wherever f is zero or negative. – Nate Eldredge Aug 11 '14 at 22:16
• @NateEldredge Ok, then we can take the restriction of $f$ on the set where $f > 0$. That is, consider $f \chi_{ \{ x \mid f(x) > 0 \}}$. This is a strictly positive, measurable function in $L^{1}(d\lambda)$ since it is less than or equal to $f$. – layman Aug 11 '14 at 22:18
• @NateEldredge Wow, I just realized this function takes the value $0$..... – layman Aug 11 '14 at 22:36
• I don't understand the statement "Measurable with respect to $\lambda$" -- functions are measurable with respect to a $\sigma$-algebra, right? [and sometimes the $\sigma$-algebra is 'generated' by an outer measure; Carathéodory's theorem] The measure defined by $\eta(A) = \lambda(A \cap \{f > 0\})$ is a measure on the product $\sigma$-algebra because $\lambda$ is, and because the set $\{f > 0\}$ is measurable. – snar Aug 14 '14 at 3:02
• @snarski When I said measurable with respect to $\lambda$, I meant with respect to $\overline{\Sigma \times \tau}$, the $\sigma$-algebra of $\lambda^{*}$ measurable sets, where $\lambda^{*}$ is the product outermeasure. As you said, this $\sigma$-algebra is obtained via Carathéodory splitting. Is that more clear? Sorry for the confusion. – layman Aug 14 '14 at 3:27

I am answering my own question because I discovered the answer in my notes, and it may help someone else (it will definitely help me as I return to reference this page in the future).

Here is what is meant by $\sigma$-finiteness being "hidden" in the hypotheses of Fubini's theorem.

To prove Fubini's theorem, we assumed $f \in L^{1}(d\lambda)$. Notice that $f = f\chi_{ \{ x \mid f(x) \neq 0 \} }$, where $\chi_{A} = \begin{cases} 1 & x \in A \\ 0 & x \not \in A \end{cases}$.

Then that means $\int \limits_{X} f \,d\lambda = \int \limits_{X} f\chi_{ \{ x \mid f(x) \neq 0 \} } \,d\lambda$. And $\int \limits_{X} f\chi_{ \{ x \mid f(x) \neq 0 \} } \,d\lambda = \int \limits_{X} f \,d(\chi_{ \{ x \mid f(x) \neq 0 \} } \,d\lambda)$ (for a proof, see the answer and comments here). The measure given by $\chi_{ \{ x \mid f(x) \neq 0 \} } \,d\lambda$ is $\sigma$-finite, because:

$X = \bigcup \limits_{n = 1}^{\infty} \{ x \mid |f| \geq \frac{1}{n} \} \bigcup \{ x \mid f = 0 \}$.

Now we just need to show each of these sets in the countable union has finite measure with respect to the measure $\chi_{ \{ x \mid f(x) \neq 0 \} } \,d\lambda$.

For each $n$, we have $\chi_{ \{ x \mid f(x) \neq 0 \} } \,d\lambda( \{ x \mid |f| \geq \frac{1}{n} \}) = \int \limits_{ \{ x \mid |f| \geq \frac{1}{n} \} } 1 \chi_{ \{ x \mid f(x) \neq 0 \} } \,d\lambda = \int \limits_{ X \times Y } \chi_{\{ x \mid |f| \geq \frac{1}{n} \}} \chi_{ \{ x \mid f(x) \neq 0 \} } \,d\lambda = \int \limits_{ X \times Y } \chi_{\{ x \mid |f| \geq \frac{1}{n} \} \cap \{ x \mid f(x) \neq 0 \} } \,d\lambda = \int \limits_{ X \times Y } \chi_{\{ x \mid |f| \geq \frac{1}{n} \} } \,d\lambda \leq \int \limits_{ X \times Y } n |f| \,d\lambda < \infty$

since $f \in L^{1}(d\lambda)$. So for each $n$, $\{ x \mid |f| \geq \frac{1}{n} \}$ has finite measure with respect to the measure $\chi_{ \{ x \mid f(x) \neq 0 \} } \,d\lambda$. Also, it's clear that $\{ x \mid f(x) = 0 \}$ has measure $0$ with respect to this measure.

So, $(X, \Sigma, \chi_{ \{ x \mid f(x) \neq 0 \} } \,d\lambda)$ is $\sigma$-finite.