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 $$\left\|f\right\|_{L^1(\mu_1\otimes\mu_2)}<\infty\;?$$

  • $\begingroup$ 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. $\mu_1$. $\endgroup$ Jan 30, 2016 at 14:21
  • $\begingroup$ 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. $\endgroup$ Jan 30, 2016 at 14:27
  • $\begingroup$ 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. $\endgroup$ Jan 30, 2016 at 14:29
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    $\begingroup$ @DavidC.Ullrich The product measure is usually denoted by $\mu_1\otimes\mu_2$. $\endgroup$
    – 0xbadf00d
    Jan 30, 2016 at 14:50
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    $\begingroup$ @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. $\endgroup$
    – 0xbadf00d
    Jan 30, 2016 at 15:15

1 Answer 1


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 $f \notin L^1(\mathbb{R}^2)$.


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