Martingales and expectation on subset Given an $\mathcal{F}_n$ martingale $X_n$, it is well known that $E[X_n]=E[X_m]$ for all $n,m$. But why is it true that if $m\geq n$, for all $A \in \mathcal{F}_n$
$$E[X_m\mathbb{1}_A]=E[X_n\mathbb{1}_A]$$
 A: It is pretty straightforward from the hints in the comments, but here is the answer for the sake of completeness.
The tower rule states that, for a probability space $(\Omega,\mathcal{A},\mathbb{P})$ and a random variable $X$, such that $\min(\mathbb{E}[X
_+],\mathbb{E}[X
_-])<\infty$, if $\mathcal{H}\subset\mathcal{G},$ where $\mathcal{H}$ and $\mathcal{G}$ are sub $\sigma$-algebras of $\mathcal{A},$ then
$$ \mathbb {E} [\mathbb{E} [X\mid {\mathcal {G}}]\mid {\mathcal {H}}]=\mathbb{E} [X\mid {\mathcal {H}}].$$
Thus, setting $\mathcal{G}=\mathcal{F_n}$, $\mathcal{H}=\sigma(\Omega)$ and $X=X_m \mathbb{1}_A$ above, we have $$ \mathbb{E}\left[X_m\mathbb{1}_A\right]=\mathbb{E}\left[\mathbb{E}\left[X_m\mathbb{1}_A\mid \mathcal{F}_n\right]\right].$$ Moreover, since $\mathbb{1}_A$ is $\mathcal{F}_n$-measurable and $\{X_n\}$ is a martingale (hence $\mathbb{E}\left[X_m\mid \mathcal{F}_n\right]=X_n$, for $m\ge n)$,$$ \mathbb{E}\left[X_m\mathbb{1}_A\right]=\mathbb{E}\left[\mathbb{E}\left[X_m\mathbb{1}_A\mid \mathcal{F}_n\right]\right]=\mathbb{E}\left[\mathbb{1}_A\mathbb{E}\left[X_m\mid \mathcal{F}_n\right]\right]=\mathbb{E}\left[\mathbb{1}_AX_n\right],$$
for all $m\ge n$.
