martingale question 3 
Let $\{U(t)\}$ be a positive martingale, such that $\mathbb{E}(U(T))=1$. Then we define a new probability measure $Q$ such that $dQ=U(T)dP$. Prove that if $X\in\mathcal{F}_t$, $E_{Q}(X|\mathcal{F}_s)=E_{P}(\frac{U(T)}{U(S)}X|\mathcal{F}_t)$, for $s\leq t$.

I think we should use the tower property on $E_{Q}(X|\mathcal{F}_s)=E_{P}(\frac{U(T)}{U(S)}X|\mathcal{F}_t)$ and then take advantage of martingale properties. Can you please help me out?
 A: Let $(\Omega,\mathcal{F},\mathcal{F}_T,\mathbb{P})$ be a filtered Probability space and $(U_t)_{t\in[0,T]}$ a positive martingale w.r.t. $\mathbb{P}$, such that $\mathbb{E}_{\mathbb{P}}(U_T)=1$. Define a measure $\mathbb{Q}$ via the Radon Nikodym density 
$$
\frac{d\mathbb{Q}}{d\mathbb{P}}=U_T
$$
We start from an equivalent definition of conditional expectation, i.e. for all $A\in\mathcal{F}_s$
$$
\mathbb{E}_{\mathbb{Q}}[X \mathbf{1}_A]=\mathbb{E}_{\mathbb{Q}}[\mathbb{E}_{\mathbb{Q}}[X\vert\mathcal{F}_s]\mathbf{1}_A]
$$
So let $A\in\mathcal{F_s}$, then by definition of $\mathbb{Q}$ and measurability of $X$ and the indicator function of $A$ w.r.t. $\mathcal{F}_t$ we get 
\begin{eqnarray}
\mathbb{E}_{\mathbb{Q}}[X \mathbf{1}_A]&=&\mathbb{E}_{\mathbb{P}}[U_T X \mathbf{1}_A]
\\
&=&\mathbb{E}_{\mathbb{P}}[\mathbb{E}_{\mathbb{P}}[U_T X \mathbf{1}_A\vert\mathcal{F}_t]]
\\
&=&\mathbb{E}_{\mathbb{P}}[\mathbb{E}_{\mathbb{P}}[U_T\vert\mathcal{F}_t]X \mathbf{1}_A]
\end{eqnarray}
by the martingale property of $U$ and the towerlaw of conditional expectation this equals 
$$=\mathbb{E}_{\mathbb{P}}[U_t X\mathbf{1}_A]=\mathbb{E}_{\mathbb{P}}[\mathbb{E}_{\mathbb{P}}[U_t X\vert\mathcal{F}_s]\mathbf{1}_A]
$$
again by the definition of the measure $\mathbb{Q}$ we get
\begin{eqnarray}
&=&\mathbb{E}_{\mathbb{Q}}[(U_T)^{-1}\mathbb{E}_{\mathbb{P}}[U_t X\vert\mathcal{F}_s]\mathbf{1}_A]\\
&\overset{towerlaw}{=}&\mathbb{E}_{\mathbb{Q}}[\mathbb{E}_{\mathbb{Q}}[(U_T)^{-1}\mathbb{E}_{\mathbb{P}}[U_t X\vert\mathcal{F}_s]\mathbf{1}_A\vert\mathcal{F}_s]]\\
&\overset{measurability}{=}&\mathbb{E}_{\mathbb{Q}}[\mathbb{E}_{\mathbb{P}}[\mathbb{E}_{\mathbb{Q}}[(U_T)^{-1}\vert\mathcal{F}_s]U_t X\vert\mathcal{F}_s]\mathbf{1}_A]\\
\end{eqnarray}
Note that the inverse of $U$ is a martingale w.r.t. $\mathbb{Q}$. Hence we have shown
$$\mathbb{E}_{\mathbb{Q}}[X \mathbf{1}_A]=\mathbb{E}_{\mathbb{Q}}[\mathbb{E}_{\mathbb{P}}[(U_s)^{-1}U_t X\vert\mathcal{F}_s]\mathbf{1}_A]
$$
which by definition of the conditional expectation (above) implies 
$$\mathbb{E}_{\mathbb{Q}}[X \vert \mathcal {F}_s]=[\mathbb{E}_{\mathbb{P}}[(U_s)^{-1}U_t X\vert\mathcal{F}_s].
$$
