Derivative of conditional expectation of integral of stochastic process

Question

Let $T>0$. Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space and let $\mathbb{F}=(\mathcal{F}_u)_{u\in[0;T]}$ be a filtration such that $\mathcal{F}_T=\mathcal{F}$. Let $(\alpha_u)_{u\geq 0}$ be a strictly positive, $\mathbb{F}$-adapted stochastic process.

I would like to show in a rigorous way that $$ \frac{d\mathbb{E}_{\mathbb{P}}\left[\int_t^\tau\alpha_u du\biggr|\mathcal{F}_t\right]}{d\tau}\Big|_{\tau=\tau'}=\mathbb{E}_{\mathbb{P}}\left[\alpha_{\tau'}\Big|\mathcal{F}_t\right], $$ where $t,\tau'\in(0;T)$ and $\tau'>t$.

I could only come up with a intuitive argument where $\alpha_u$ is constant over a small interval $[\tau;\tau+\Delta\tau]$, but I suspect there might be a more rigorous way to prove this statement. What are some theorems that one could use to construct such a rigorous argument? If so, which are those theorems? Is there maybe a general method for proving such statements?

EDIT: Imposing conditions on $(\alpha_u)_{u\geq 0}$ may be required to give a definitive answer to this question.

Answer 1

(I assume $(\alpha_u)_{u\ge 0}$ is progressively measurable and uniformly bounded.)

One can choose the conditional expectations $A_u:=\Bbb E_{\Bbb P}[\alpha_u\mid\mathcal F_t]$, $u\ge t$, so that $(A_u)_{u\ge t}$ is jointly $\mathcal F_t$-measurable in $\omega$ and Borel measurable in $u$. Then, by a version of Fubini's theorem $$ \Bbb E_{\Bbb P}\left[\int_t^\tau \alpha_u du\mid \mathcal F_t\right] =\int_t^\tau A_u du $$ in the sense that the integral on the right is a version of the conditional expectation on the left, for each $\tau>t$. The function $\tau\mapsto\int_t^\tau A_u du$ is absolutely continuous on $[t,\infty)$, and so has a pointwise derivative at a.e. $\tau'>t$. The same is therefore true of the left, and so $$ \lim_{\epsilon\downarrow 0}{\Bbb E_{\Bbb P}\left[\int_t^{\tau'+\epsilon} \alpha_u du\mid \mathcal F_t\right]-\Bbb E_{\Bbb P}\left[\int_t^{\tau'} \alpha_u du\mid \mathcal F_t\right]\over\epsilon} =\lim_{\epsilon\downarrow 0}{\Bbb E_{\Bbb P}\left[\int_{\tau'}^{\tau'+\epsilon}\alpha_u du\mid\mathcal F_t\right]\over \epsilon}=A_{\tau'} $$ for a.e. $\tau'>t$, almost surely.

Answer 2

Added note: If $H$ is bounded and $\mathcal F_t$-measurable, then $$ \eqalign{ \Bbb E_{\Bbb P}\left[\int_t^\tau\alpha_u du\cdot H\right] &=\int_t^\tau\Bbb E_{\Bbb P}\left[\alpha_u \cdot H\right]du\cr &=\int_t^\tau\Bbb E_{\Bbb P}\left[\Bbb E_{\Bbb P}[\alpha_u\mid\mathcal F_t]\cdot H\right]du\cr &=\int_t^\tau\Bbb E_{\Bbb P}\left[A_u\cdot H\right]du\cr &=\Bbb E_{\Bbb P}\left[\int_t^\tau A_u du\cdot H\right].\cr } $$ As $\int_t^\tau A_u du$ is $\mathcal F_t$-measurable, this shows that $\int_t^\tau A_u du$ is a version of the conditional expectation $\Bbb E_{\Bbb P}\left[\int_t^\tau \alpha_u du\mid \mathcal F_t\right]$.