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I have a question regarding the proof of Doobs theorem (discrete case): It says that every adapted and integrable stochastic process $Y$ can be uniquely decomposed as the sum

$Y_t = M_t + A_t$ where

$A_{t+1} = \displaystyle{\sum_{n=0}^t \mathbb{E}[Y_{n+1}|\mathcal{F}_n] - Y_n}$ and

$M_{t+1} = \displaystyle{Y_0 + \sum_{n=0}^t Y_{n+1} - \mathbb{E}[Y_{n+1}|\mathcal{F}_n]}$

The proof claims that $A_{t+1}$ is $\mathcal{F}_t$-measureable (predictable) since $X_t$ is adapted, however I do not understand how this follows. In particular, I do not understand why $\mathbb{E}[Y_{n+1}|\mathcal{F}_n]$ is $\mathcal{F}_t$-measurable. I appreciate any help!

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1 Answer 1

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$\mathbb{E}[Y_{n+1}|\mathcal{F}_n]$ is $\mathcal{F}_n$-measurable and $\mathcal{F}_n \subseteq \mathcal{F}_t$ for $0 \leq n \leq t$. Hence, $\mathbb{E}[Y_{n+1}|\mathcal{F}_n]$ is $\mathcal{F}_t$-measurable

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