If $M_t$ is a martingale, prove $\Bbb E \left[ M_T\int_t^T h_s ds |F_t\right] = \Bbb E \left[ \int_t^T M_s h_s ds |F_t\right]$ If $M_t$ is a martingale,  for $0<t<T$, prove  $\Bbb E \left[ M_T\int_t^T h_s ds |F_t\right] = \Bbb E \left[ \int_t^T M_s h_s ds |F_t\right]$.
I can think of
$LHS=M_T \Bbb E \left[ \int_t^T h_s ds |F_t\right] $
RHS look similar to Ito isometry but not quite.
Anyone could give me a hint?
 A: Hints: If $\ \Bbb E[\int_{t}^{T}|h_s|\,\text ds],\,\Bbb E[\int_{t}^{T}|M_sh_s|\,\text ds]<\infty\ $ then, in succession, use the Fubini-Tonelli theorem, the tower property of conditional expectation, the martingale property, and Fubini-Tonelli one more time, as follows:
$$ \Bbb E \left[ M_T\int_t^T h_s \text ds \, \vert\, \mathcal F_t\right] = \int_t^T \Bbb E \left[\, \ldots\, |\,\mathcal F_t\right] \text ds  = \int_t^T \Bbb E \left[\, \Bbb E [\,\ldots\, \vert\, \ldots]\,|\,\mathcal F_t\right] \text ds \\= \int_t^T \Bbb E \left[\, \ldots\Bbb  E [\ldots\, \vert\, \mathcal \ldots]\,|\,\mathcal F_t\right] \text ds = \int_t^T \Bbb E \left[\, \ldots\,|\,\mathcal F_t\right] \text ds = \Bbb E \left[\,\int_t^T  M_sh_s \text ds\,|\,\mathcal F_t\right] $$

:) No peeking now :


$${\bf \text{The solution:}}\\\ \Bbb E \left[ M_T\int_t^T h_s \text ds \, \vert\, \mathcal F_t\right] = \int_t^T \Bbb E \left[\, M_T h_s\, |\,\mathcal F_t\right] \text ds  = \int_t^T \Bbb E \left[\, \Bbb E [\,M_T h_s\, \vert\, \mathcal F_s]\,|\,\mathcal F_t\right] \text ds \\= \int_t^T \Bbb E \left[\, h_s\Bbb  E [M_T\, \vert\, \mathcal F_s]\,|\,\mathcal F_t\right] \text ds = \int_t^T \Bbb E \left[\, M_sh_s\,|\,\mathcal F_t\right] \text ds = \Bbb E \left[\,\int_t^T  M_sh_s \text ds\,|\,\mathcal F_t\right] $$

