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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

Can someone give me a hint why it is true?

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    $1_{T$\mathcal{F}_{t\wedge T}$-measurable. Don't we always know by time $t\wedge T$ whether $T$T$ is a stopping time.) – 2017-01-30
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    @ spaceisdarkgreen. Ah i see but we need the right continuity of the filtration for this to work. Right? – 2017-01-31
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    I don't think it needs right-continuity. Is $\{T< t\}$ not always in $\mathcal{F}_{t\wedge T}$ when $T$ is a stopping time? It's always in $\mathcal{F_t}$ and $t\wedge T\le t.$ Also with regard to your other question, they are probably using $H\in \mathcal{F_t}$ to mean $H$ is a $\mathcal{F}_{t}$-measurable RV. – 2017-01-31
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    @spaceisdarkgreen Thank you very much I got confused with the theorem that $\{ T – 2017-01-31
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    Ahh that makes sense. You'd need right continuity to prove the other direction of that iff. – 2017-01-31

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Since both $T,T\wedge t$ are stoping times, $1_{T