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Let $(\Omega_i,\mathcal A_i,\mu_i)$ be a $\sigma$-finite measure space and $f:\Omega_1\times\Omega_2\to\mathbb R$ be measurable with respect to $\mathcal A_1\otimes\mathcal A_2$. Can we conclude, that if $$\left\|f(\;\cdot\;,\omega_2)\right\|_{L^1(\mu_1)}<\infty\;\;\;\text{for }\mu_2\text{-almost every }\omega_2\in\Omega_2$$ and $$\left\|f(\omega_1,\;\cdot\;)\right\|_{L^1(\mu_2)}<\infty\;\;\;\text{for }\mu_1\text{-almost every }\omega_1\in\Omega_1\;,$$ then $$\left\|f\right\|_{L^1(\mu_1\otimes\mu_2)}<\infty\;?$$

0xbadf00d
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  • How do you propose to compute $|f|_{L^1(\mu_1)}$ if e.g. $\Omega_1 = \Omega_2 = [0,1]$ and the $\mu_i$ are both the Lebesgue measure? A general function $f \in L^1([0,1]^2)$ may not even be measurable wrt. $\mu_1$. – Hans Engler Jan 30 '16 at 14:21
  • Of course there's no such thing as $||f||_{L^p(\mu_1)}$ here; you must really be talking about norms of "slices". Tonelli's theorem seems like a bound of the sort you mean to be asking about. – David C. Ullrich Jan 30 '16 at 14:27
  • It's $\mu_1\times\mu_2$ btw, not $\mu_1\otimes\mu_2$. Yes, it's $\mathcal A_1\otimes\mathcal A_2$. The product $\sigma$-algebra is (analogous to) a tensor product; the product measure is not. – David C. Ullrich Jan 30 '16 at 14:29
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    @DavidC.Ullrich The product measure is usually denoted by $\mu_1\otimes\mu_2$. – 0xbadf00d Jan 30 '16 at 14:50
  • Really? For example? It's $\mu_1\times\mu_2$ in every reference I know - for example Rudin, Folland, Wikipedia. The reason it that it's not a tensor product. You're confusing the product measure with the (tensor) product $\sigma$-algebra, which is written using $\otimes$. Or maybe not - reference? – David C. Ullrich Jan 30 '16 at 14:57
  • @DavidC.Ullrich See, for example, https://books.google.de/books?id=xDy3BAAAQBAJ&pg=PA277&lpg=PA277&dq=klenke+finite+product+measure&source=bl&ots=u6WfkufwpH&sig=S_Mejor1mejFBJ71OQ4QvT6XHOQ&hl=de&sa=X&ved=0ahUKEwjnmtSM5tHKAhUFDQ4KHe0QCdQQ6AEIHDAA#v=onepage&q=klenke%20finite%20product%20measure&f=false. Theorem.14.14. – 0xbadf00d Jan 30 '16 at 14:59
  • @DavidC.Ullrich Concerning your first comment: I had something else in mind and updated the question accordingly. – 0xbadf00d Jan 30 '16 at 15:01
  • @HansEngler I've updated the question, since I had something else in mind. Now $f(;\cdot;,\omega_2)$ is $\mathcal A_1$-measurable and $f(\omega_1,;\cdot;)$ is $\mathcal A_2$-measurable, for all $\omega_i\in\Omega_i$. – 0xbadf00d Jan 30 '16 at 15:03
  • That's one reference that uses non-standard (and bad) notation. To settle the question of whether it's "usually" written that way we'd need to look in a lot of places. Given that I learned about product measures 35 years ago and this is the first time I've seen that $\otimes$ it's going to take a lot more than one book to convince me that it's usually written that way. (Especially since the $\otimes$ simply makes no sense - there's nothing tensorish about $\mu_1\times\mu_2$.) – David C. Ullrich Jan 30 '16 at 15:07
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    @DavidC.Ullrich There are other references. However, thinking about that it seems like $\mu_1\otimes\mu_2$ is often used by European authors while I've never seen this notation used by an American author. Maybe that's why you've never seen it before. – 0xbadf00d Jan 30 '16 at 15:15
  • @DavidC.Ullrich: I've seen this notation in several (German) standard text books on functional analysis (Werner) as well as on measure theory (Elstrodt). – Roland Jan 30 '16 at 15:58

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The answer is no. Take $\Omega_1 = \Omega_2 = \mathbb{R}$ with the usual Lebesgue measure and set $f(x_1,x_2) = e^{-(x_1-x_2)^2}$. Then $\|f(\cdot, x_2)\|_{L^1(\Omega_1)} = \|f(x_1, \cdot)\|_{L^1(\Omega_2)} = const. < \infty$ for all $x_1, x_2$ but $f \notin L^1(\mathbb{R}^2)$.

Hans Engler
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