Doob–Meyer decomposition Let $X_n = \sum_{m≤n} 1_{B_m} $and suppose $B_n \in F_n$. What is the Doob decomposition for $X_n$ ? I can write it down from the construction of the theorem, but is there any neat way showing the result?

It follows from the fact that $1_{B_k}$ is $\mathcal{F}_{n-1}$-measurable for $k \leq n-1$ that

$$\mathbb{E}(X_n \mid \mathcal{F}_{n-1}) = X_{n-1} + \mathbb{P}(B_n \mid \mathcal{F}_{n-1}).$$

Subtracting $A_n := \sum_{k=1}^n \mathbb{P}(B_k \mid \mathcal{F}_{k-1})$ on both sides yields

$$\mathbb{E}(X_n - A_n \mid \mathcal{F}_{n-1}) = X_{n-1} - A_{n-1}.$$

This shows that $M_n := X_n-A_n$ is a martingale. Consequently, the Doob decomposition is given by

$$X_n = \underbrace{\sum_{k=1}^n 1_{B_k}-\mathbb{P}(B_k \mid \mathcal{F}_{k-1})}_{\text{martingale}}+ \underbrace{\sum_{k=1}^n \mathbb{P}(B_k \mid \mathcal{F}_{k-1})}_{\text{predictable}}.$$