[proofplan] The predictable compensator is forced by the one-step conditional drifts of the submartingale. We define $A_n$ as the cumulative sum of $\mathbb E[X_k-X_{k-1}\mid\mathcal F_{k-1}]$, which is nonnegative by the submartingale property and predictable by construction. Subtracting this compensator from $X_n$ removes the conditional drift, producing a martingale. Uniqueness follows because the difference of two decompositions is simultaneously a predictable finite-variation process and a martingale, so each increment has conditional expectation equal to itself and equal to zero. [/proofplan] [step:Define the predictable compensator from one-step conditional drifts] For each $k\geq 1$, define the integrable $\mathcal F_{k-1}$-measurable random variable \begin{align*} D_k &= \mathbb E[X_k-X_{k-1}\mid\mathcal F_{k-1}]. \end{align*} This definition is valid because $X_k-X_{k-1}$ is integrable. Since $(X_n)_{n\geq 0}$ is a submartingale, \begin{align*} \mathbb E[X_k\mid\mathcal F_{k-1}] &\geq X_{k-1} \end{align*} almost surely, and therefore $D_k\geq 0$ almost surely. Define $(A_n)_{n\geq 0}$ by \begin{align*} A_0 &= 0, & A_n &= \sum_{k=1}^n D_k \quad \text{for } n\geq 1. \end{align*} For every $n\geq 1$, the increment $A_n-A_{n-1}=D_n$ is $\mathcal F_{n-1}$-measurable, so $A$ is predictable. Since $D_k\geq 0$ almost surely, $A$ is increasing. Each $A_n$ is integrable because it is a finite sum of integrable conditional expectations. [/step] [step:Subtract the compensator to obtain a martingale] Define a process $(M_n)_{n\geq 0}$ by \begin{align*} M_n &= X_n-A_n. \end{align*} For each $n$, $M_n$ is $\mathcal F_n$-measurable because $X_n$ is $\mathcal F_n$-measurable and $A_n$ is $\mathcal F_n$-measurable. It is integrable because both $X_n$ and $A_n$ are integrable. For $n\geq 1$, using $A_n=A_{n-1}+D_n$ and the fact that $A_{n-1}$ is $\mathcal F_{n-1}$-measurable, we compute \begin{align*} \mathbb E[M_n\mid\mathcal F_{n-1}] &= \mathbb E[X_n-A_n\mid\mathcal F_{n-1}] \\ &= \mathbb E[X_n\mid\mathcal F_{n-1}] - A_{n-1} - D_n \\ &= \mathbb E[X_n\mid\mathcal F_{n-1}] - A_{n-1} - \mathbb E[X_n-X_{n-1}\mid\mathcal F_{n-1}] \\ &= X_{n-1}-A_{n-1} \\ &= M_{n-1}. \end{align*} Thus $(M_n)_{n\geq 0}$ is a martingale, and by definition $X_n=M_n+A_n$ for every $n\geq 0$. [/step] [step:Prove uniqueness of the decomposition] Suppose \begin{align*} X_n &= M_n+A_n = N_n+B_n \end{align*} for every $n\geq 0$, where $M$ and $N$ are martingales, and $A$ and $B$ are predictable increasing integrable processes with $A_0=B_0=0$. Then \begin{align*} C_n &= A_n-B_n = N_n-M_n \end{align*} defines an integrable martingale $(C_n)_{n\geq 0}$, because it is the difference of two martingales. Its increment is \begin{align*} C_n-C_{n-1} &= (A_n-A_{n-1})-(B_n-B_{n-1}). \end{align*} For $n\geq 1$, this increment is $\mathcal F_{n-1}$-measurable because both $A$ and $B$ are predictable. Since $C$ is a martingale, \begin{align*} \mathbb E[C_n-C_{n-1}\mid\mathcal F_{n-1}] &= 0. \end{align*} But $C_n-C_{n-1}$ is already $\mathcal F_{n-1}$-measurable, so [Taking Out What is Known](/theorems/1151) gives \begin{align*} C_n-C_{n-1} &= \mathbb E[C_n-C_{n-1}\mid\mathcal F_{n-1}] = 0 \end{align*} almost surely. Hence $C_n=C_{n-1}$ almost surely for every $n$, and since $C_0=A_0-B_0=0$, induction gives $C_n=0$ almost surely for every $n$. Therefore $A_n=B_n$ and $M_n=N_n$ almost surely for every $n$. [/step]