Given a probability space $(\Omega$, $\mathcal{F}$, $\mathbb{P})$, $(M_n)_{n\geq1}$ is a martingale. Let $\Psi \in \mathcal{F}_m$ and $n\geq m$. How can one prove that, by martingale property \begin{equation} \mathbb{E}\{M_n1_{\Psi}\}=E\{M_m1_{\Psi}\} \end{equation} ?
I was thinking about focusing on r.h.s. of the above equation. By martingale property, one would have \begin{equation} \mathbb{E}\{M_m1_{\Psi}\}=\mathbb{E}\{\mathbb{E}\{M_n|\mathcal{F_m}\}1_{\Psi}\} \end{equation} However, at this point, giving that the outer expectation is defined on the product $\mathbb{E}\{M_n|\mathcal{F}_m\}1_{\Psi}$ I am not sure I am allowed to apply the Law of Iterated Expectations on $\mathbb{E}\{\mathbb{E}\{M_n|\mathcal{F_m}\}\}$ so as to conclude that, given that $\mathbb{E}\{\mathbb{E}\{M_n|\mathcal{F_m}\}\}=M_n$, it holds true that: \begin{equation} \mathbb{E}\{M_m1_{\Psi}\}=\mathbb{E}\{\mathbb{E}\{M_n|\mathcal{F_m}\}1_{\Psi}\}=\mathbb{E}\{M_n 1_{\Psi}\} \end{equation}
Am I actually allowed to do so?