For a discrete-time martingale $M_n$ ($\mathcal{F}_n$-measurable), fix $r>n$ and an event $F\in \mathcal{F}_n$, why do we have $$ E(M_n 1_{F})=E(M_r 1_{F}) $$ from the tower law and property of martingale?
My idea:
Since $r>n$, we have $\mathcal{F}_n\subset \mathcal{F}_r$. Then $M_n$ is $\mathcal{F}_r$-measurable?
Also, for event $F_{r-1}\in \mathcal{F}_{r}$, using martingale property $$ E(E(M_r 1_{F})\mid F_{r-1})=E(1_{F}E(M_r\mid F_{r-1}))=E(1_{F}M_{r-1})=\cdots=E(1_{F}M_{n})? $$
I don't believe you need the tower property, just the definition should do. We have that $\mathbb{E}[M_r|\mathcal{F}_n]$ is defined to be the unique random variable such that $\mathbb{E}[M_r|\mathcal{F}_n]$ is $\mathcal{F}_n-$measurable and $$\mathbb{E}[M_r 1_F] = \int_F M_r dP = \int_F \mathbb{E}[M_r|\mathcal{F}_n]dP $$ for all $F\in\mathcal{F}_n$. If $M_n$ is a discrete time martingale, we will have that $\mathbb{E}[M_r|\mathcal{F}_n] = M_n$. Thus the statement above becomes $$\mathbb{E}[M_r 1_F] = \int_F \mathbb{E}[M_r|\mathcal{F}_n]dP =\int_F M_n dP = \mathbb{E}[M_n1_F]$$