Uniform integrability of a backward submartingale

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.