# Proving the Fubini-Tonelli Theorem

Okay, so the notes I'm using give the theorem as follows:

Let $(\Omega_1, \mathcal{A}_1, \mu_1)$ and $(\Omega_2, \mathcal{A}_2, \mu_2)$ be two $\sigma-$finite measure spaces.

$(1)$ If $f \colon \left(\Omega_1 \times \Omega_2 \right) \rightarrow \left(\overline{\mathbb{R}}, \mathcal{B} (\overline{\mathbb{R}}) \right)$ is a non-negative function, then $x \mapsto \int_{\Omega_2} f(x,y) \ d\mu_2 (y)$ is $\mathcal{A}_1 - \mathcal{B}(\overline{\mathbb{R}})$ measurable and $y \mapsto \int_{\Omega_1} f(x,y) \ d\mu_1 (x)$ is $\mathcal{A}_2 - \mathcal{B} (\overline{\mathbb{R}})$ measurable. Then we have

$\int_{\Omega_1 \times \Omega_2} f \ d (\mu_1 \otimes \mu_2) = \int_{\Omega_1} \left(\int_{\Omega_2} f(x,y) \ d\mu_2 (y) \right) \ d\mu_1 (x) = \int_{\Omega_2} \left(\int_{\Omega_1} f(x,y) \ d\mu_1 (x) \right) \ d \mu_2(y)$.

$(2)$ If $f \in \mathcal{L}^1 \left(\Omega_1 \times \Omega_2, \mathcal{A_1} \otimes \mathcal{A_2}, \mu_1 \otimes \mu_2 \right)$ then

• $f(x, \cdot) \in \mathcal{L}^1 \left(\Omega_2, \mathcal{A}_2, \mu_2 \right) \ \mu_1$-a.e.
• $f(\cdot, y) \in \mathcal{L}^1 \left(\Omega_1, \mathcal{A}_1, \mu_1 \right) \ \mu_2$-a.e.

I'm pretty sure I understand the proof of (1) pretty well as a consequence of Monotone Convergence and the properties of the product measure. What's confusing me is the notes' justification for (2); they claim that (2) follows from (1) and the fact that any non-negative measurable function is the pointwise limit of an isotone sequence of non-negative simple functions. I just don't see how. So first of all I want to know if that's just a typo (he refers to "Lemma 4.4"; Lemma 4.4 is in fact a theorem , which makes me suspicious), and what has been written makes no sense; secondly I want to check that my proof makes sense.

Obviously I'll only prove the first case; the proof of the second is no different. Wlog we can assume that $f$ is non-negative (then for arbitrary $f$ we know that the restrictions of $f^+$ and $f^-$ are integrable almost everywhere, say on $N_+^C, N_-^C$, then we have that the restriction of $f$ is integrable on their intersection, which is also the complement of the null set $N_+ \cup N_-$ ).

We have $\int_{\Omega_1 \times \Omega_2} f \ d (\mu_1 \otimes \mu_2) = \int_{\Omega_1} \left(\int_{\Omega_2} f(x,y) \ d\mu_2 (y) \right) \ d\mu_1 (x) < \infty$. In particular, the non-negative function $x \mapsto \int_{\Omega_2} f(x,y) \ d\mu_2 (y)$ is contained in $\mathcal{L}^1 \left(\Omega_1, \mathcal{A}_1, \mu_1 \right)$. Therefore, $\int_{\Omega_2} f(x,y) \ d\mu_2 (y)$ is finite outside of a null set, say $N$ (this was proved in the course; in general an integrable function is finite almost everywhere). So on $N^C$ the non-negative function $y \mapsto f(x,y)$ is integrable, i.e. $f(x, \cdot) \in \mathcal{L}^1 \left(\Omega_2, \mathcal{A}_2, \mu_2 \right) \mu_1$-a.e., as required.

Seems pretty basic, but I want to make sure I'm not missing anything, since I may be forced to prove this in an exam in the not too distant future.