Doob's decomposition and submartingale with bounded increments

Question

Let $(X_n)_{n \geq 0}$ be a submartingale defined on some filtered probability space $(\Omega, \mathcal{F}, ({\mathcal{F}}_n)_{n \geq 0}, \mathbb{P})$. It is a standard fact that $X_n = X_0 + M_n + A_n$, where $(M_n)_{n \geq 0}$ is a martingale null at $0$ and $(A_n)_{n \geq 0}$ is an increasing previsible process, also null at $0$.

Suppose that there exists a positive constant c such that $|X_{n+1} - X_n| \leq c$ for all $n \geq 0$. Then we need to prove the following: \begin{equation} \bigg\{ \sup_{n \geq 0 } X_n < + \infty \bigg\} \subseteq \bigg( \big\{ (X_n) \text{ converges in } \mathbb{R} \big\} \cap \big\{A_\infty < +\infty \big\} \bigg). \end{equation}

Answer 1

Define a sequence of stopping times $(\tau_k)_k$ by

$$\tau_k := \inf\{n \geq 0; X_n \geq k\}.$$

From

$$M_{n \wedge \tau_k} = X_{n \wedge \tau_k}-X_0 - \underbrace{A_{n \wedge \tau_k}}_{\geq 0} \leq 2k$$

it follows that $(M_{n \wedge \tau_k})_{n \in \mathbb{N}}$ is a martingale which is bounded above. Consequently, by a standard convergence theorem, the limit

$$\lim_{n \to \infty} M_{n \wedge \tau_k}$$

exists almost surely. For $\omega \in \{\sup_n X_n < \infty\}$ we have $\omega \in \{\tau_k = \infty\}$ for $k$ sufficiently large; hence, $\lim_{n \to \infty} M_n(\omega)$ exists. From

$$A_n(\omega) = X_n(\omega)-X_0(\omega)-M_n(\omega) \leq 2 \sup_n X_n(\omega) - \inf_n M_n(\omega)<\infty$$

we see that $(A_n(\omega))_n$ is bounded above. Since $(A_n(\omega))_n$ is increasing, we find that $A_{\infty}(\omega) = \lim_{n \to \infty} A_n(\omega)$ exists. Obviously, this implies that

$$\lim_{n \to \infty} X_n(\omega) = X_0(\omega) + \lim_{n \to \infty} A_n(\omega)+ \lim_{n \to \infty} M_n(\omega).$$

Answer 2

Remember that when you prove the Doob's decomposition theorem you define:

$$A_n = \sum_{k=1}^n \mathbb{E}[X_k | \mathcal{F}_{k-1}]-X_{k-1}$$ and $$M_n =X_0+ \sum_{k=1}^n X_{k}-\mathbb{E}[X_k | \mathcal{F}_{k-1}]$$

Being $X_n$ a submartingale, you have $A_n \ge 0$ this implies $M_n \le X_n$ which implies $\sup_nM_n < + \infty$ if we are in the event $\{\sup_nX_n < \infty \}$.

Now $M_n$ is a bounded martingale then it converges. And we have $A_n \le 2\sup X_n -\inf M_n < \infty$ being $A_n$ increasing you have the result.