Say we have a probability space $(\Omega, A, \Pi)$ and a random field $f$ in $\mathbb{R}^d$ (or more simply a stochastic process on $\mathbb{R}$) where each sample path is square-integrable, that is for all $\omega \in \Omega$, we have $\int f(x,\omega)^2 d\mu(x) < \infty$. Here $\mu$ is the Lebesgue measure.
I would like to check whether the mean field $E[f]$ is also square-integrable or not. There are counterexamples where the expectation does not preserve continuity but I have the feeling that some kind of integrability might still hold. My current reasoning is the following:
$\begin{align*} \int E[f]^2 d\mu &= \int_{\mathbb{R}} \left(\int_{\Omega} f(x,\omega) d\Pi(\omega)\right)^2d\mu(x) \\ &\leq \int_{\mathbb{R}} \int_{\Omega} f(x,\omega)^2 d\Pi(\omega)d\mu(x)\quad \textrm{ since } E[f]^2 \leq E[f^2] \\ &\leq \int_{\Omega} \int_{\mathbb{R}} f(x,\omega)^2 d\mu(x) d\Pi(\omega)\quad \textrm{ using Fubini's theorem } \end{align*} $
Now, if we note $I(\omega) = \int_{\mathbb{R}} f(x,\omega)^2 d\mu(x)$, can we use the fact that the random integral $I$ is finite for any $\omega \in \Omega$ to justify that $ E[I] = \int_{\Omega}I(\omega)d\Pi(\omega) < \infty $ ?
If that does not work, could I be successful with just local square-integrability ?