Justifying a step in proving $M_{S\wedge T} = \mathbb{E}[M_T \mid \mathcal{F}_S ]$ $S,T$ are stopping times and $M$ is a (right) continuous martingale. My lecturer set this as an exercise and I am given a solution(essentially split $M_T = M_T \mathbf{1}_{S≤T} + M_T \mathbf{1}_{S>T}$ and use this property)  but there is a step I cannot justify, namely
$$ \mathbb{E}[ M_T \mathbf{1}_{T>S} \mid \mathcal{F}_T ] \overset{?}{=} \mathbb{E}[ M_T \mathbf{1}_{T>S} \mid \mathcal{F}_{T\color{red}{\wedge S}} ] $$
More generally, given an $L^1(Ω)$ random variable $X$, what conditions the on $\sigma$-algebras $\mathcal{F},\mathcal{G}$ guarantee the following equality?
$$ \mathbb{E}[X \mid \mathcal{F}] = \mathbb{E}[X \mid \mathcal{G}] $$
 A: Should the relationship in question be
$$
\mathbb{E}[ M_T \mathbf{1}_{T>S} \mid \color{red}{\mathcal{F}_S} ] \overset{?}{=} \mathbb{E}[ M_T \mathbf{1}_{T>S} \mid \mathcal{F}_{T\color{red}{\wedge S}} ]\,,
$$
in line with the primary problem of determining the general form of $\mathbb E[M_{T}\mid\mathcal F_{S}]$? If this is the case, notice that, on the event $[T\geqslant S]$, $M_T=M_{T\vee S}$ and $M_S=M_{T\wedge S}$. Also, $\mathcal F_{T\wedge S}\subseteq\mathcal F_T$ and $\mathcal F_S\subseteq \mathcal F_{T\vee S}$. Consequently,
$$\mathbb E[M_T{\bf 1}_{T\geqslant S}\mid\mathcal F_S]=\mathbb E[M_{T\vee S}{\bf 1}_{T\geqslant S}\mid\mathcal F_S]=\mathbb E[M_{T\vee S}\mid\mathcal F_S]{\bf 1}_{T\geqslant S}=M_S{\bf 1}_{T\geqslant S}=M_{T\wedge S}{\bf 1}_{T\geqslant S}=\mathbb E[M_{T}\mid\mathcal F_{T\wedge S}]{\bf 1}_{T\geqslant S}=\mathbb E[M_{T}{\bf 1}_{T\geqslant S}\mid\mathcal F_{T\wedge S}]$$

Remark: The relationships above hold because of the interplay between the martingale property and the random variables being measurable with respect to multiple sigma-algebras. 


*

*By the martingale property (for martingale $X$ and $t,r>s$),
$$
\mathbb{E}[X_t \mid \mathcal{F}_s]=X_s=\mathbb{E}[X_r \mid \mathcal{F}_s]
$$

*Also, for random variable $X$, measurable with respect to both $\mathcal G$ and $\mathcal F$, we have 
$$\mathbb{E}[X \mid \mathcal{F}] =X= \mathbb{E}[X \mid \mathcal{G}]$$ 
