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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.

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