Conditional expectation is $L_p$ norm reducing 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 $\mathcal{B}$?

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 $Y=f(X)$.
