# 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?

• You need to show that $(w_l)_l$ is uniformly integrable. This is done using a reasoning similar to that in Thm. 24.6 (if you use the 2nd edition... it's the theorem preceding the theorem which you are trying to prove). – saz Mar 31 '20 at 6:23
• @saz I tried along the lines of (1) to (2) in 24.6. By using $w_0^+$ as the majorant, we can show with exactly the same steps that $w_{-l}^+$ is uniformly integrable. But then we still need $\int w_{-n} d\mu \to \int w_{-\infty} d\mu$ to conclude that $w_{-n}$ is uniformly integrable by Vitali's theorem. I cannot see how we can get this condition... – nomadicmathematician Mar 31 '20 at 6:26
• @saz I figured that Remark 22.8 can help, which states that if $u_n \to u$ a.e. then $\Vert u_n \Vert_1$ converging in $\mathbb{R}$ implies $|u_n|$ is Uniformly Integrable. If we take this, then we can use that $w_{-n}^-$ is a supermartingale, so $\Vert w_{-n}^-$ converges in $\mathbb{R}$ by the condition $\inf_n \int w_{-n} d\mu > -\infty$, and so $w_{-n}^-$ is uniformly integrable as well. However, I am not sure Remark 22.8 holds because the proof of (iii) $\doublearrow$ (ii) in 22.7 requires the $\lim_n \int |u_n|^p d\mu = \int |u|^p d\mu$. And this condition isn't available in 22.8. – nomadicmathematician Mar 31 '20 at 6:31
• I see, thanks for the additional information. – saz Mar 31 '20 at 9:35
• @saz I have posted an answer I received from the author. – nomadicmathematician Mar 31 '20 at 17:41

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 $$\sup_{n \in -\mathbb{N}_0} \int_{|f_n|>R_\epsilon w_\epsilon}|f_n| \le 4\epsilon.$$