Tower Property for Expectations and Stopping Times Let $(\Omega,(\mathcal{F_t})_{t\geq0},P)$ be a filtered proability space with  $X\in L^1(P)$ and two stopping times $S$ and $T$.
Show that 
\begin{equation*}
\mathbb{E}(\mathbb{E}(X|\mathcal{F}_T)|\mathcal{F}_S)=\mathbb{E}(X|\mathcal{F}_{S\wedge T}),~P-a.s.
\end{equation*}
I am a bit stuck on this and tried spilting it into 2 cases, namely on $\{S\leq T\}$ and it's complement and usng the fact that $\mathcal{F}_s\cap\mathcal{F}_T=\mathcal{F}_{S\wedge T}$ but don't get far.
So any help is most welcomed and needs.  Thanks in advance
 A: Note that $\{T\leq S\}$, $\{S>T\}\in\mathcal{F}_T\cap\mathcal{F}_S$ and for any $A\in\mathcal{F}_T$ we have $A\cap\{T\leq S\}\in\mathcal{F}_S$.  Therefore  $A\cap\{T\leq S\}\in\mathcal{F}_T\cap\mathcal{F}_S$.  Thus by the definition on conditional expectation for any $A\in\mathcal{F}_T$,
$E(1_A.1_{\{T\leq S\}}E(X|\mathcal{F}_{T\wedge S})) =E(1_{A\cap\{T\leq S\}}X) 
 = E(1_{A\cap\{T \leq S\}}E(X|\mathcal{F}_T))=E(1_{A}.1_{\{T \leq S\}}E(X|\mathcal{F}_T))$.
Thus $1_{\{T\leq S\}}E(X|\mathcal{F}_{T})=1_{\{T\leq S\}}E(X|\mathcal{F}_{T\wedge S}))$  $-(*)$.
So then,
$1_{\{T\leq S\}}E(E(X|\mathcal{F}_{T}|\mathcal{F}_S))=E(1_{\{T\leq S\}}E(X|\mathcal{F}_T)|\mathcal{F}_S) =E(1_{T\leq S\}}E(X|\mathcal{F}_{T\wedge S})|\mathcal{F}_S)=1_{\{T\leq S\}}E(X|\mathcal{F}_{S\wedge T})$.
Switching the roles of $S$ and $T$ and replacing $X$ by $E(X|\mathcal{F}_T)$ in $(*)$ we similarly get, 
$1_{\{S<T\}}E(E(X|\mathcal{F}_T)|\mathcal{F}_S)=1_{\{S<T\}}E(E(X|\mathcal{F}_T)|\mathcal{F}_{T\wedge S})=1_{\{S< T\}}E(X|\mathcal{F}_{T \wedge S})$.
