Why does $X \in \mathcal{L}^p(\Omega,\mathcal{F},\mathbb{P}),p \in [1,\infty)$, implies that $\mathbb{E}[X\;|\mathcal{A}] \in \mathcal{L}^p(\Omega,\mathcal{A},\mathbb{P})$, where $\mathcal{A} \subset \mathcal{F}$ is a sub-$\sigma$-algebra?
The defintion of the conditional expectation says that $\mathbb{E}[X \mathbb{1}_A]=\mathbb{E}[\mathbb{E}[X|\mathcal{A}] \mathbb{1}_A]$ for all $A \in \mathcal{A}$ but why can we conclude from this that $\mathbb{E}[|X|^p \mathbb{1}_A]=\mathbb{E}[\mathbb{E}[|X|^p|\mathcal{A}] \mathbb{1}_A]$ for all $A \in \mathcal{A}$?
We will use: Jensen's inequality for conditional expectation: If $\phi$ is a convex function, then $$ \phi\big(\mathbb E[X\mid \mathcal A]\big) \le \mathbb E\big[\phi(X)\mid\mathcal A\big] $$
Now $p \in [1,\infty)$, so the map $t \mapsto |t|^p$ is convex.
Assume $X \in L^p(\Omega,\mathcal F,\mathbb P)$; that is $\mathbb E[|X|^p] < \infty$. By Jensen, $$ \big|\mathbb E[X\mid \mathcal A]\big|^p \le \mathbb E\big[|X|^p\mid\mathcal A\big] $$ so $$ \mathbb E\Big[\big|\mathbb E[X\mid \mathcal A]\big|^p\big] \le \mathbb E\Big[\mathbb E\big[|X|^p\mid\mathcal A\big]\Big] =\mathbb E\big[|X|^p\big] < \infty. $$ Thus $E[X\mid \mathcal A] \in L^p(\Omega, \mathcal A, \mathbb P)$.