Suppose, we define $E_F[X]= \int X dF$ and $E_P[X] =\int X dP$.
We know that there exits the follow $L_p, \ p \ge 1$ inequality \begin{align*} E_P |E_P(X| \mathcal{B})|^p \le E_P[|X|^p] \end{align*} I was woundering if the following inequality holds: \begin{align*} E_P (|E_F(X| \mathcal{B})|^p) \le E_{???}(|X|^p) \text{ note the mismatch} \end{align*} where $???$ is either $P$ or $F$?
We can still mimic a lot from the original proof. Let $f(t)=|t|^p$ which is convex for $p \ge 1$ therefore \begin{align*} E_P (f(E_F(X| \mathcal{B}))) \le E_P (E_F(f(X)| \mathcal{B})) \end{align*}
But what to do next? I think maybe law of total expectation?