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?

  • $\begingroup$ 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). $\endgroup$ – saz Mar 31 '20 at 6:23
  • $\begingroup$ @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... $\endgroup$ – nomadicmathematician Mar 31 '20 at 6:26
  • $\begingroup$ @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. $\endgroup$ – nomadicmathematician Mar 31 '20 at 6:31
  • $\begingroup$ I see, thanks for the additional information. $\endgroup$ – saz Mar 31 '20 at 9:35
  • $\begingroup$ @saz I have posted an answer I received from the author. $\endgroup$ – nomadicmathematician Mar 31 '20 at 17:41

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


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