Below is a theorem from Protter's Stochastic Integration and Differential Equations. I don't get a line from the proof. Namely, how do we get $E[X_T 1_{T\ge t} | \mathscr{F}_{t \wedge T}] = E[X_T 1_{T\ge t} | \mathscr{F}_t]$? It seems like we need to use the fact that for $H \in \mathscr{F}_t$ we have $H1_{T\ge t} \in \mathscr{F}_T$. But I can't see how this can be applied here. I would greatly appreciate any help.
By definition, $\mathbb{E}(X_T 1_{\{T \geq t\}} \mid \mathcal{F}_t)$ is the $\mathcal{F}_t$-measurable random variable $Y \in L^1$ such that
$$\forall F \in \mathcal{F}_t\::\: \int_F Y \, d\mathbb{P} = \int_F X 1_{\{T \geq t\}} \, d\mathbb{P}.$$
We verify these properties for $Y:= \mathbb{E}(X_T 1_{\{T \geq t\}} \mid \mathcal{F}_{T \wedge t})$. Since $\mathcal{F}_{T \wedge t} \subseteq \mathcal{F}_t$, the random variable $Y$ is $\mathcal{F}_t$-measurable. If $F \in \mathcal{F}_t$, then
$$F \cap \underbrace{\{T \geq t\}}_{\in \mathcal{F}_t} \in \mathcal{F}_t$$
Moreover,
$$(F \cap \{T \geq t\}) \cap \{T \leq s\} = \begin{cases} \emptyset, & \text{if $s<t$}, \\ F \cap \{t \leq T \leq s\} \in \mathcal{F}_s, & \text{if $s \geq t$} \end{cases}$$
and so $F \cap \{T \geq t\} \in \mathcal{F}_T$. Hence, $$F \cap \{T \geq t\} \in \mathcal{F}_T \cap \mathcal{F}_t = \mathcal{F}_{T \wedge t}.\tag{1}$$ Since $\{T < t\} \in \mathcal{F}_{T \wedge t}$, we have $$1_{\{T < t\}} \mathbb{E}(X 1_{\{T \geq t\}} \mid \mathcal{F}_{T \wedge t}) = \mathbb{E}(X 1_{\{T \geq t\}} 1_{\{T<t\}} \mid \mathcal{F}_{T \wedge t}) = 0, \tag{2}$$
and so
\begin{align*} \int_F Y \, d\mathbb{P} &= \int_F \underbrace{1_{\{T<t\}} \mathbb{E}(X 1_{\{T \geq t\}} \mid \mathcal{F}_{T \wedge t})}_{\stackrel{(2)}{=}0} \, d\mathbb{P} + \underbrace{\int_{F \cap \{T\geq t\}} \mathbb{E}(X 1_{\{T \geq t\}} \mid \mathcal{F}_{T \wedge t}) \, d\mathbb{P}}_{\stackrel{(1)}{=} \int_{F \cap \{T \geq t\}} X 1_{\{T \geq t\}} \, d\mathbb{P}} \\ &= \int_F X 1_{\{T \geq t\}} \, d\mathbb{P} \end{align*}