I was trying to understand the proof of Theorem 18 in Chapter 1 of the book on Stochastic Integration by Protter and there is some thing I don't understand
Why is $X_T \mathbb{1}_{T <t}$ is $\mathcal{F}_{t \wedge T}$ It is being used in the 4th line of the proof. I see that we can write $$X_T \text{ as } X_{t \wedge T} \text{ on the set } \mathbb{1}_{T<t} $$ and then it comes out of the conditional expectation since $X_{t \wedge T}$ is $\mathcal{F}_{t \wedge T}$ measurable since X is adapted and cadlag. But I somehow do not see why can I take the indicator function outside the conditional expectation. It seems impossible that one can do that.I am totally stuck . Moreover the statement in the proof $\text{ However }H \in \mathcal{F}_t \implies H \mathbb{1}_{T \geq t} \in \mathcal{F}_T$ doesnt make sense to me since its a product of a set and random variable(indicator).
Can someone give me a hint why it is true?
Since both $T,T\wedge t$ are stoping times, $1_{T<t}=1_{T\wedge t<t}\in\mathscr{F}_{T\wedge t}$ and $$X_T1_{T<t}=X_{T\wedge t}1_{T<t}=X_{T\wedge t }1_{T\wedge t<t}\in\mathscr{F}_{T\wedge t},$$ therefore $$\mathsf{E}[X_T1_{T<t}|\mathscr{F}_{T\wedge t}]=X_T1_{T<t}.$$