When reading a textbook, I couldn't understand how to conclude from the following condition that a stochastic process is predictable projection of the other. In the sequel I will try to formulate all relevant conditions and hopefully nothing essential is missed.
All processes mentioned below are cadlag. Say we have a Brownian filtration $\mathscr{F}_t$ and a bounded process $X$, which is $\mathscr{B}[0,T]\times\mathscr{F}_T$ measurable (here $T$ is a fixed real number). $X$ is not necessarily adapted. On the other hand, $\phi$ is a predictable process such that $\mathbb{E}[\int_0^T \phi_t^2 dt] < \infty$. Now, let's say we have that for any bounded $\mathscr{F}_t$-predictable process $u$, $$ \mathbb{E}\left[\int_0^T X_t u_t dt\right] = \mathbb{E}\left[\int_0^T \phi_t u_t dt\right] $$ How can we conclude from this equality that $\phi$ is the predictable projection of $X$? As far as I know, we need to establish that $$ \mathbb{E}\left[X_S\right] = \mathbb{E}\left[\phi_S\right],\mbox{ for any stopping time $S \leq T$} $$ But I can't see how this follows from the forgoing equality. Any hint will be greatly appreciated!