# True or False: convergence on L1 of a martingale Xn with E|Xn|=1

I have to prove whether the next statement is true or not:

'if {Xn} for n>=1 to infinitive it is such a martingale that for everything n>=1, Xn>=0 and E|Xn|=1, then the sequence {Xn} for n>=1 to infinitive converges on L1'.

I think I have to use Doob's martingale convergence theorem but I don't know how to use it to get convergence on L1, because as I understand with that theorem you get almost sure convergence.

A positive martingale $$(X_n)$$ converges almost surely toward a random variable $$X_{\infty}:= \lim_{n \rightarrow \infty}X_n$$. But in $$L^1$$, we do not have automatically a convergence, even if $$(X_n)$$ is bounded in $$L^1$$.
Let $$(Z_n)$$ be a non-symetric random walk on $$\mathbb{Z}$$ and $$(X_n)$$ a process defined by $$X_n = (\frac{q}{p})^{Z_n}$$ where $$p = 1-q$$, $$p\in]0,1[$$, $$p\not= 0.5$$, representing the probability of up and down of the random walk.
Then $$(X_n)$$ is a positive martingale converging almost surely toward $$X_{\infty}=0 \,a.s.$$ . If your statement was true, we would have $$E[X_0]=E[X_{\infty}]\Leftrightarrow 1=0$$ which is false.