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I'm reading Theorem 6.19 in textbook Real Analysis: Modern Techniques and Their Applications by Gerald B. Folland.

Suppose that $(X, \mathcal{M}, \mu)$ and $(Y, \mathcal{N}, \nu)$ are $\sigma$-finite measure spaces, and let $f$ be an $(\mathcal{M} \otimes \mathcal{N})$-measurable function on $X \times Y$.

a. If $f \geq 0$ and $1 \leq p<\infty$, then $$ \left[\int\left(\int f(x, y) d \nu(y)\right)^{p} d \mu(x)\right]^{1 / p} \leq \int\left[\int f(x, y)^{p} d \mu(x)\right]^{1 / p} d \nu(y) $$ b. If $1 \leq p \leq \infty, f(\cdot, y) \in L^{p}(\mu)$ for a.e. $y$, and the function $y \mapsto\|f(\cdot, y)\|_{p}$ is in $L^{1}(\nu)$, then $f(x, \cdot) \in L^{1}(\nu)$ for a.e. $x$, the function $x \mapsto \int f(x, y) d \nu(y)$ is in $L^{p}(\mu)$, and $$ \left\|\int f(\cdot, y) d \nu(y)\right\|_{p} \leq \int\|f(\cdot, y)\|_{p} d \nu(y). $$

Could you verify it my proof of part (b) that $f(x, \cdot) \in L^{1}(\nu)$ for $\mu$-a.e. $x \in X$ and that the function $x \mapsto \int f(x, y) d \nu(y)$ is in $L^{p}(\mu)$ is correct? This would solidify my understanding of the integrability criterion of a function.

Assume $f$ takes values in a Banach space $(E, \|\cdot\|)$. I denote the $L_p$-norm by $\| \cdot \|_p$.

  • Because $\|f(\cdot, y)\|_{1} \le \|f(\cdot, y)\|_{p}$ for all $p\ge 1$ and $y \mapsto\|f(\cdot, y)\|_{p}$ is in $L_1$, so the map $y \mapsto\|f(\cdot, y)\|_1$ is also in $L_1$ [here I use the fact that if $f$ is integrable and $\|g\| \le \|f\|$ then $g$ is integrable]. This means $\int_Y \int_X \|f(x,y)\| \mathrm d \mu(x) \mathrm d \nu (y) < \infty$. Then by Fubini's theorem, $$\int_X \int_Y \|f(x,y)\| \mathrm d \nu (y) \mathrm d \mu(x) = \int_Y \int_X \|f(x,y)\| \mathrm d \mu(x) \mathrm d \nu (y)< \infty,$$ which in turn implies $\int_Y \|f(x,y)\| \mathrm d \nu (y) < \infty$ for $\mu$-a.e. $x \in X$. This means $f(x, \cdot) \in L_1(\nu)$ for $\mu$-a.e. $x \in X$.

  • We have $y \mapsto\|f(\cdot, y)\|_{p}$ is in $L_1(\nu)$, i.e., $$\int \left[\int \|f(x, y)\|^p d \mu(x)\right]^{1 / p} d \nu(y) <\infty.$$ Combining this with (a), we get $$\left[\int\left(\int \|f(x, y)\| d \nu(y)\right)^{p} d \mu(x)\right]^{1 / p} < \infty.$$ This means the map $x \mapsto \int \|f(x, y)\| d \nu(y)$ is in $L_p(\mu)$. On the other hand, $\|\int f(x, y) d \nu(y)\| \le \int \|f(x, y)\| d \nu(y)$. This in turn implies $x \mapsto \int f(x, y) d \nu(y)$ is in $L_p(\mu)$ [here I also use the fact that if $f$ is integrable and $\|g\| \le \|f\|$ then $g$ is integrable].

Update: I was wrong in saying $\|f(\cdot, y)\|_{1} \le \|f(\cdot, y)\|_{p}$ for all $p \ge 1$. I fix the proof as follows.

  • We have $y \mapsto\|f(\cdot, y)\|_{p}$ is in $L_1(\nu)$, i.e., $$\int \left[\int \|f(x, y)\|^p d \mu(x)\right]^{1 / p} d \nu(y) <\infty.$$ Combining this with (a), we get $$\left[\int\left(\int \|f(x, y)\| d \nu(y)\right)^{p} d \mu(x)\right]^{1 / p} < \infty.$$ This means $\int \|f(x, y)\| d \nu(y) < \infty$ for $\mu$-a.e. $x \in X$ or equivalently $f(x , \cdot) \in L_1(\nu)$ for $\mu$-a.e. $x \in X$.
Akira
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    Why is $\lVert f(., y) \rVert_{1} \leq \lVert f(., y) \rVert_{p}$? I don't think this is true in general (it is true in probability spaces though). – Mason Oct 31 '21 at 18:30
  • @Mason You are right!! A detailed discussion can be found here. I will try to fix it. – Akira Oct 31 '21 at 18:44
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    This remark that $|f|1$ cannot be bounded by $|f|_p$ was my point from your other question. It is true on probability spaces (and finite measure spaces with a constant, which would be sufficient for your argument), but it cannot be true in general. Also, a side note, it isn't clear to me if you mean $|\cdot|$ to mean absolute value of a function, or a certain $L^p$ norm. It would improve the clarity of your post if you wrote $L^p$ norms as $|\cdot|{L^p(\mu)}$ (with a subscript and the corresponding measure) and absolute value as $|\cdot|$ (simple vertical bars). – Alex Ortiz Oct 31 '21 at 19:13
  • Thank you @AlexOrtiz! I should be more precise. It seems that we only need the hypothesis $y \mapsto|f(\cdot, y)|{p}$ is in $L_1(\nu)$. In my (fixed) proof, I have not used the hypothesis $f(\cdot, y) \in L{1}(\mu)$ for $\nu$-a.e. $y \in Y$. In this version, I only assume in (b) that $f(x, \cdot)$ is $\nu$-integrable for $\mu$-a.e. $x \in X$. – Akira Oct 31 '21 at 19:35

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