René Schilling's proof for backwards submartingale $L^1-\lim_{n} w_{-n} = w_{-\infty}$ $\iff$ $\inf_{n \in \mathbb{N}_0} \int w_{-n}d\mu > -\infty$ I am reading René Schilling's Measures, Integrals, and Martingales. 
Let $(w_l, \mathscr{A}_l)_{l \in -\mathbb{N}_0})$ be a backwards submartingale and assume that $\mu|_{\mathscr{A}_{-\infty}}$ is $\sigma$-finite. Then we have the following.
(i) $\lim_{n \to \infty} w_{-n}=w_{-\infty} \in [-\infty, \infty)$ exists a.e.
(ii) $L^1-\lim_{n \to \infty} w_{-n} = w_{-\infty}$ if, and only if, $\inf_{n \in \mathbb{N}_0} \int w_{-n}d\mu > -\infty.$ If this is the case, then $(w_l,\mathscr{A}_l)_{l \in -\mathbb{N}_0 \cup -\infty}$ is a submartingale and $w_{-\infty}$ is a.e. real-valued. 
I cannot see why we get the equivalence condition in (ii). The proof gives the equivalences 
$$\sup_{n\in \mathbb{N}_0} \int |w_{-n}|\,d\mu < \infty \iff \inf_{n\in \mathbb{N}_0} \int w_{-n}\, d\mu > -\infty \iff \lim_{n \to \infty} \int w_{-n}\,d\mu \in \mathbb{R}.$$ From this I can see that we have the only if direction. But how do we get the necessity? 
All this shows is that $\lim_{n\to \infty} \int w_{-n}\,d\mu$. But how can we conclude that this limit must indeed be $\int w_{-\infty}\,d\mu$? 
P.S. I think I will be able to solve this problem, if I can show that if $u_n$ converges a.e. to $u$, and $||u_n||_p$ converges for $p\ge 1$, then $||u_n||_p \to ||u||_p$. Is this true?
 A: I got a solution to this problem from Rene Schilling. 
I here put his reply for anyone interested in the proof. 
We show that $\{f_n\}_{n \in -\mathbb{N}_0}$ is uniformly integrable. 
Fix $\epsilon>0$. Then since we assume that $\int f_n d\mu \downarrow I \in \mathbb{R}$, we have some $m = m_\epsilon$ such that for all $n \le m$: $\int f_n d\mu \le \int f_m d\mu \le \int f_n d\mu + \epsilon$. 
Now take some $w \in L^1(\mu)$ that is positive everywhere. Let $R>0$. Then we have 
\begin{align*}\int_{|f_n|>Rw}|f_n| &= \int_{f_n<-Rw} (-f_n)+\int_{f_n>Rw}f_n \\ &= \int_{f_n\ge -Rw}f_n-\int f_n + \int_{f_n>Rw}f_n \\ &\leq \int_{f_n \ge -Rw} f_m-\int f_m + \epsilon + \int_{f_n>Rw}f_m   \\ &\leq \int_{|f_n|>Rw}|f_m|+\epsilon \\ &= \int_{\{|f_n|>Rw\} \cap \{|f_m|>\frac{1}{2}Rw\}}|f_m|+\epsilon+ \int_{\{|f_n|>Rw\} \cap \{|f_m|\le\frac{1}{2}Rw\}}|f_m| 
\\ &\leq \int_{|f_m|>\frac{1}{2}Rw}|f_m|+\epsilon + \frac{1}{2}\int_{|f_n|>Rw}|f_n|.\end{align*}
Hence we have $$\int_{|f_n|>Rw} |f_n| \le 2\epsilon + 2 \int_{|f_m|>\frac{1}{2}Rw}|f_m|$$ for all $n < m=m_\epsilon$. Finally, using dominated convergence theorem, we can choose $R=R_\epsilon$ large enough that for all $n \ge m=m_\epsilon$, $\int_{|f_n|>\frac{1}{2}R_\epsilon} |f_n| < \epsilon$. Thus we have 
$$\sup_{n \in -\mathbb{N}_0} \int_{|f_n|>R_\epsilon w_\epsilon}|f_n| \le 4\epsilon.$$
