Conditional expectation is $L_p$ norm reducing

Question

I am trying to follow the proof of \begin{align*} E \left(|E(X\mid \mathcal{B} )|^p \right) \le E(|X|^p) \end{align*} which goes as follows. Let $f(t)=|t|^p$ which is convex for $p\ge 1$ then \begin{align*} E \left(f(E(X\mid \mathcal{B} )) \right) &\le E \left(E(f(X) |\mathcal{B} ) \right) \text{ this is by Jensen's inequality}\\ &= E(f(X) ) \\ &=E(|X|^p) \end{align*} I don't undestand the second step where $E \left(E(f(X) \mid \mathcal{B} ) \right)=E(f(X) ) $. What happend to $\mathcal{B}$?

Answer 1

Using the definition of conditional expectation, we know that for any random variable $Y$ $$E(E(Y|\mathcal{B})) = \int_{\Omega} E(Y|\mathcal{B}) dP = \int_{\Omega} Y dP = E(Y)$$ This is called the tower property of conditional expectation and intuitively it "gets rid of the $\mathcal{B}." $ Now just apply this to $Y=f(X)$.