Suppose $W$ is a Brownian motion, let $H_B$ be the hitting of $B \in \mathbb{R}$ and let $\tau$ be another stopping time that is taken to be minimal, i.e $(W_{t\wedge \tau})_{t \geq 0}$ is uniformly integrable.
Show that $\mathbb{E}[(W_{H_B} - W_\tau)\mathbb{1}_{H_B \leq \tau}]=0$.
The "solution" that I got and needs checking is the following:
We apply the optional stopping theorem which states that for a cadlag adapted integrable process $X$ and any stopping times $T$, $S$, we have that $\mathbb{E}[X_T|\mathcal{F}_S]=X_{T\wedge S}$ iff X is a uniformly integrable martingale. Since $W^\tau$ is UI a mart, we get that $\mathbb{E}[W_\tau|\mathcal{F}_{H_B}]=W_{\tau \wedge H_B}$. In particular $\mathbb{E}[W_\tau 1_A]=\mathbb{E}[W_{\tau \wedge H_B}1_A]$ for any $A$ in $\mathcal{F}_{H_B}$, thus it is sufficient to show that $(H_B\leq \tau )$ is in $\mathcal{F}_{H_B}$.
The definition that I have is $\mathcal{F}_{H_B}:=\{A \in \mathcal{F}_{\infty}:A \cap (H_B \leq t) \in \mathcal{F}_{t}\}$
Fix $t$, then $(H_B \leq \tau)=(H_B < \tau) \cup (H_B=\tau)$. Sufficient to show that both $(H_B < \tau)$ and $(H_B=\tau)$ are in $\mathcal{F}_{H_B}$.
Consider first $(H_B <\tau)=\cup_{q\in \mathbb{Q}}(H_B\leq q) \cap (q<\tau)$. Then $(H_B< \tau) \cap (H_B \leq t)=\big(\cup_{q\in \mathbb{Q}, q\leq t} (H_B \leq q\wedge t) \cap (q < \tau)\big) \cup (t < \tau)$. Clearly $(H_B \leq q \wedge t) \in \mathcal{F}_t$ and $(t < \tau) = (\tau \leq t)^C \in \mathcal{F}_t$. Therefore, $(H_B<\tau) \cap (H_B\leq t) \in \mathcal{F}_t$.
Now let's look at $(H_B=\tau)$. Note that $(H_B\leq t)\cap (\tau \leq t)=\big((H_B=\tau)\cap (H_B \leq t)\big) \cup \big( (H_B < \tau)\cap (\tau \leq t)\big) \cup \big((\tau < H_B) \cap (H_B \leq t)\big).$
Observe that all three sets in $\big(\big)$ are disjoint. It is obvious that $(H_B \leq t) \cap (\tau \leq t) \in \mathcal{F}_t$. From before $(H_B <\tau) \in \mathcal{F}_t$ so clearly $(H_B<\tau)\cap(\tau \leq t) \in \mathcal{F}_t$. Similarly, $(\tau < H_B) \cap (H_B \leq t) \in \mathcal{F}_t$. Hence we get that $(H_B = \tau) \cap (H_B \leq t) \in \mathcal{F}_t$. This concludes our case by case analysis.