Martingales bounded in $L^1$

Question

From the definition, for $X_n$ to be a martingale we require $E|X_n| < \infty$, for all $n\in\Bbb N$. Which implies that $X_n$ is bounded in $L^1$. However for theorems such as martingale convergence, it requires additionally that the martingale is in $L^1$. Surely this requirement would then be redundant?

Answer 1

By definition, a (discrete time) martingale $(X_n)$ must fulfill $\mathbb{E}[|X_n|]<\infty$ for all $n\in\mathbb{N}$. This does not imply $\sup_n \mathbb{E}[|X_n|]<\infty$, which is the definition of $L^1$-boundedness. Why?

Note that in general $\mathbb{E}[X]\neq\mathbb{E}[|X|]$ (just think of a strictly negative random variable $X$). Thus, in general $\mathbb{E}[|X_n|]=\mathbb{E}[|X_{n+1}|]$ must not hold even though $(X_n)$ is a martingale (which implies $\mathbb{E}[X_n]=\mathbb{E}[X_{n+1}]\quad\forall n$). For example, there exists a martingale $Y_n$ with $$\mathbb{E}[|Y_n|] = n\quad\forall n\in\mathbb{N}.$$ As a martingale $Y_n$ fulfills $\mathbb{E}[|Y_n|]<\infty$ for all $n$, but $\sup_n\mathbb{E}[|Y_n|]=\sup_n\mathbb{N}=\infty$. Thus $Y_n$ is a martingale that is not bounded in $L^1$.