# How to show martingale is bounded in $L^1$?

Fix a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_n)_{n\geq 0},\mathbb{P})$ and an $L^1$-bounded submartingale $X_n$.

We can show that, for $n\geq 0$, the sequence $(\mathbb E[X^+_p|\mathcal{F}_n],p\geq n)$ is increasing and converges to an a.s. limit $M_n$. We want to show that $(M_n)_{n\geq 0}$ is an $L^1$-bounded martingale.

$$\mathbb E[M_{n+1}|\mathcal{F}_n]=\mathbb E\lim_{p\to\infty}[\mathbb E[X_p^+|\mathcal{F}_{n+1}]|\mathcal{F}_n]$$

So it seems the right tool here is dominated convergence. By the conditional dominated convergence theorem, $M$ is a martingale if, for $n\geq0$, $\mathbb E|M_n|<\infty$.

How do we show that $M$ is $L^1$-bounded?

Thank you.

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The $L^1$-bounded submartingale $(X^+_p)$ has a Riesz decomposition $$X^+_p=Y_p+Z_p\hskip1.5cm (1)$$ where $(Y_p)$ is a martingale and $Z_p\to 0$ in $L^1$. Conditioning on ${\cal F}_n$ where $p\geq n$, we get $$\mathbb{E}(X^+_p\, |\,{\cal F}_n) =\mathbb{E}(Y_p\, |\,{\cal F}_n)+\mathbb{E}(Z_p\, |\,{\cal F}_n)=Y_n +\mathbb{E}(Z_p\, |\,{\cal F}_n).$$ Letting $p\to\infty$ shows that $\mathbb{E}(X^+_p\, |\,{\cal F}_n)$ converges to $Y_n$ in $L^1$. The martingale $(Y_n)$ is $L^1$ bounded by (1).

The Riesz decomposition follows because $(X^+_p)$ is a quasimartingale, see this reference.

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