Is it true that $\mathbb E[M_tX|\mathcal F_s]=M_s\mathbb E[X|\mathcal F_s]$?

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Suppose we have a martingale $\{M_t\}_{t\in[0,T]}$ on the natural filtered probability space $(\Omega,\mathcal F,\{\mathcal F_t\}_{t\in[0,T]},\mathbb P)$. $X$ is any measurable random variable with respect to $\mathcal F$. Is it true that for any $0\le s\le t\le T$,

$$\mathbb E[M_tX|\mathcal F_s]=M_s\mathbb E[X|\mathcal F_s]\ ?$$


I guess it should be correct. It is apparently true when $X$ is $F_s$-measurable. But for a general $\mathcal F$-measurable $X$, is it still true?

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This is not necessarly true. Take for instance $X=M_t$. Since $M_t^2$ is not necessarly a martingale, we cannot have $\mathbb{E}[M_t^2|F_s] = M_s\mathbb{E}[M_t|F_s] = M_s^2$.