Let $X$ and $Y$ be subsets of $\mathbb{R}$, and let $\mu$ be a measure on $X$ and $\nu$ a measure on $Y$. Let $f : X \times Y \rightarrow \mathbb{R}$ be $\mu$-summable and $\nu$-summable, i.e. $$\int_X \left| f(x,\cdot) \right| d\mu(x) < +\infty \hspace{1cm} \text{and} \hspace{1cm} \int_Y \left| f(\cdot,y) \right| d\nu(y) < +\infty.$$
- How can I prove that $f$ is $(\mu \times \nu)$-summable ?
I want to apply this result to the function $f(x,t) = \left| u(x+th) \right|^p$, where $\left| u \right| \in L^p(\mathbb{R})$, so that $$\int_{\mathbb{R}} \left( \int_0^1 \left| u(x+th) \right|^p dt \right) dx = \int_0^1 \left( \int_{\mathbb{R}} \left| u(x+th) \right|^p dx \right) dt,$$ using Fubini's theorem. But first, I must prove that $f$ is $(\mu \times \nu)$-measurable and $(\mu \times \nu)$-integrable. Unfortunately, I have no idea how to do that.
Thank you in advance for your help! :)