Karatzas and Shreve solution to uniform integrability of backward martingale Let $\{\mathscr{F}_n\}_{n=1}^\infty$ be a decreasing sequence of sub-$\sigma$-fields of $\mathscr{F}$, i.e. $\mathscr{F}_{n+1} \subset \mathscr{F}_n \subset \mathscr{F}$, and let $\{X_n, \mathscr{F}_n\}$ be a backward submartingale, i.e. $E|X_n| < \infty$, $X_n$ is $\mathscr{F}_n$ measurable and $E(X_n | \mathscr{F}_{n+1}) \ge X_{n+1}$ a.s. $P$, for every $n \ge 1$. Then $l:= \lim_{n \to \infty} E(X_n) > -\infty$ implies that the sequence $\{X_n\}$ is uniformly integrable.
The solution to this problem is given below. However, I cannot understand how we get the final two inequalities. That is, how do we select $\lambda>0$ in such a way that 
$$\sup_{n>m} \int_{X_n < -\lambda} |X_m|dP < \epsilon/2?$$ And why do these choices of $m$ and $\lambda$ give 
$$\sup_{n>m} \int_{X_n^- > \lambda}X_n^- dP < \epsilon?$$
I would greatly appreciate any explanations.


 A: By assumption, the limit $\ell =\lim_{n \to \infty} \mathbb{E}(X_n)>-\infty$ exists, and so $(\mathbb{E}(X_n))_{n \in \mathbb{N}}$ is a Cauchy sequence, i.e. for any given $\epsilon>0$ there exists $m \in \mathbb{N}$ such that
$$|\mathbb{E}(X_m)-\mathbb{E}(X_n)| \leq \frac{\epsilon}{2}, \qquad n \geq m.\tag{1}$$
By the inequality in the second display (in the extract from the book), we have
$$\int_{\{X_n<-\lambda\}} X_n \, d\mathbb{P} \geq \mathbb{E}(X_n)-\mathbb{E}(X_m) + \int_{\{X_n<-\lambda\}} X_m \, d\mathbb{P} \tag{2}$$
and so
\begin{align*} \int_{\{X_n<-\lambda\}} X_n^- \, d\mathbb{P}&= \int_{\{X_n<-\lambda\}} (-X_n) \, d\mathbb{P}\\ &\stackrel{(2)}{\leq} \mathbb{E}(X_m)-\mathbb{E}(X_n) + \int_{\{X_n<-\lambda\}} (-X_m) \, d\mathbb{P} \\ &\stackrel{(1)}{\leq} \frac{\epsilon}{2} + \int_{\{|X_n|>\lambda\}}|X_m| \, d\mathbb{P} \tag{3} \end{align*}
for all $n \geq m$. The random variable $\{|X_m|\}$ is uniformly integrable. This implies that there exists some $\delta>0$ such that
$$\int_A |X_m| \, d\mathbb{P} \leq \frac{\epsilon}{2} \tag{4}$$
for any measurable set $A$ with $\mathbb{P}(A) \leq \delta$. On the other hand, we have $\sup_{n \in \mathbb{N}} \mathbb{P}(|X_n| >\lambda) \to 0$ as $\lambda \to \infty$ (see the first part of the proof in the book), and therefore we can choose $\lambda$ sufficiently large such that $\sup_{n \in \mathbb{N}} \mathbb{P}(|X_n|>\lambda) \leq \delta$. By $(4)$, this implies
$$\sup_{n \in \mathbb{N}} \int_{\{|X_n|>\lambda\}} |X_m| \, d\mathbb{P} \leq \frac{\epsilon}{2},$$
and so, by $(3)$,
$$ \sup_{n \geq m} \int_{\{X_n<-\lambda\}} X_n^- \, d\mathbb{P} \leq \epsilon,$$
which shows that $(X_n^-)_{n \geq 1}$ is uniformly integrable.
