Uniform integrability of a backward submartingale Let $\{\mathcal{F}_n\}_n$ be a decreasing sequence of sub-$\sigma$-fields of $\mathcal{F}$($\mathcal{F}_{n+1}\subset\mathcal{F}_n$) and let $\{X_n\}_n$ be a backward submartingale($E[|X_n|]<\infty$, $X_n$ is $\mathcal{F}_n$ measurable and $E[X_n|\mathcal{F}_{n+1}]\geq X_{n+1}$).
If we know $l=\lim_{n\rightarrow\infty}E[X_n]>-\infty$, how to prove that $\{X_n\}_n$ is uniformly integrable? Many thanks!
I have an idea, it is enough to show: 


*

*$\sup_nE[|X_n|]<\infty$; 

*$\forall\epsilon>0$, $\exists\delta>0$ s.t. $\forall A\in\mathcal{F}$ such that $\mathbb{P}(A)<\delta$, we have $\sup_nE[1_{A}|X_n|]<\epsilon$. 
For $|X_n|=2X_n^+-X_n$, we have
$$E[|X_n|]=2E[X_n^+]-E[X_n]\leq 2E[X_0^+]-l\leq 2E[|X_0|]-l$$
which implies $\sup_nE[|X_n|]<\infty$.
Secondly, $\forall A\in\mathcal{F}$, we can suppose without loss of generality that $A\in\mathcal{F}_0$.
$$E[1_{A}|X_n|]=2E[[1_{A}X_n^+]-E[[1_{A}X_n]$$
Does someone have an idea for this term? Thanks a lot.....
 A: I think that I find a proof:
As I have showed, by Jensen's inequality we prove that $\{X_n^+\}_n$ is a backward submartingale. 
For any $K>0$, we have
$$KP(|X_n|\geq K)\leq E[|X_n|]=2E[X_n^+]-E[X_n]\leq 2E[X_1^+]-l<\infty$$
It follows that 
$$\sup_nP(|X_n|\geq K)\rightarrow 0,\ \ K\rightarrow\infty$$
Firstly,
$$E[1_{\{X_n^+\geq K\}}X_n^+]\leq E[1_{\{X_n^+\geq K\}}E[X_1^+|\mathcal{F}_n]]=E[1_{\{X_n^+\geq K\}}X_1^+]\leq E[1_{\{|X_n|\geq K\}}X_1^+]\rightarrow 0,\ \ K\rightarrow\infty$$
by the integrability of $X_1^+$, which implies the uniform intergrability of $\{X_n^+\}_n$.
Secondly,
$$0\geq E[1_{\{X_n^+\leq -K\}}X_n]=E[X_n]-E[1_{\{X_n^+>-K\}}X_n]\geq E[X_n]-E[1_{\{X_n^+>-K\}}X_N]=E[X_n]-E[X_N]+E[1_{\{X_n^+\leq -K\}}X_N],\ \ \forall n\geq N$$
So for any given $\epsilon>0$, we have take $N$ large enough that $-\epsilon/2\leq E[X_n]-E[X_N]\leq 0$ and for this fixed $N$ we can choose $K$ such that 
$$\sup_{n\geq N}E[1_{\{X_n^+\leq -K\}}|X_N|]\leq\epsilon/2$$
which implies the uniform intergrability of $\{X_n^-\}_n$, and we show $\{X_n\}_n$ is uniformly intergrable.
A: in the last step one can't choose $K$ large so that
$\sup_{n \geq N} \mathbb{E}[1_{\{X_n^+ \leq -K\}|X_N|}] \leq \frac{\epsilon}{2}$
and that is not just because of a typo (most likely you meant $\sup_{n \geq N} \mathbb{E}[1_{\{X_n \leq -K\}|X_N|}] \leq \frac{\epsilon}{2}$ ) 
It is because the negative part of the submartingale can't be bound that way
maybe you should take a look at
http://www.ma.utexas.edu/users/gordanz/notes/uniform_integrability.pdf
