Suppose $W(t)$ is a stochastic process adapted to filtration $\mathcal{F}(t)$. That is, for each $t$, $W(t)$ is $\mathcal{F}(t)$-measurable.
I want to prove this claim: $E[W(t) \mid \mathcal{F}(t)] = W(t)$.
I'm having a hard time proving this. I know that through the definition of conditional expectations, the following two statements hold:
(i) $E[W(t) \mid \mathcal{F}(t)]$ is a $\mathcal{F}(t)$-measurable function
(ii) For all $A \in \mathcal{F}(t)$, $\int_A E[W(t) \mid \mathcal{F}(t)](\omega) d \mathbb{P}(\omega) = \int_A W(t)(\omega) d\mathbb{P}(\omega)$
But these two statements don't imply that $E[W(t) \mid \mathcal{F}(t)] = W(t)$. It only says that for all $A \in \mathcal{F}$, the integral in (ii) holds.
Can I please get a hint?
Update: I define $E[X \mid \mathcal{A}]$ as
(i) $E[X \mid \mathcal{A}]$ is a $\mathcal{A}$-measurable function
(ii) For all $A \in \mathcal{A}$, $\int_A E[X \mid \mathcal{A}](\omega) d \mathbb{P}(\omega) = \int_A X(\omega) d\mathbb{P}(\omega)$
Update 2: I kind of see it now, but I can only claim that $E[W(t) \mid \mathcal{F}(t)] = W(t)$ almost everywhere. $E[W(t) \mid \mathcal{F}(t)]$ and $W(t)$ can only differ on sets of measure zero? Am I right?
I can only prove that this claim holds almost everywhere.
Define $A = \{ E[W(t) \mid \mathcal{F}(t)] - W(t) > 0 \}$. Notice that $A \in \mathcal{F}(t)$. By (ii),
$$\int_A E[W(t) \mid \mathcal{F}(t)](\omega) - W(t)(\omega) d \mathbb{P}(\omega) = 0$$
Then $A$ has measure zero. Similarly, $ E[W(t) \mid \mathcal{F}(t)] < W(t)$ holds only on a set of measure zero. Then $ E[W(t) \mid \mathcal{F}(t)] = W(t)$ almost everywhere.