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Let $(\Omega,\mathcal{A},\mu)$ be a finite mesure space, and $\{f_n\}$ and $\{g_n\}$ two $L^1$-bounded sequences, such that : $$ \sum_{n\geq 1}{\frac{1}{n}(F_n(f_n)(\omega)-g_n(\omega))}<\infty\qquad a.e $$ with: $F_n(f_n)=f_n1_{|f_n|\leq n}$

Can we say that: $$ \sum_{n\geq 1}{\frac{1}{n}(f_n(\omega)-g_n(\omega))}<\infty\qquad a.e $$

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  • $\begingroup$ We have $\sup_n\|f_n\|_1<\infty$. Then, there exists $n_0\in\mathbb{N}$, such that: for all $n\geq 1$ we have $$|f_n|\leq n_0~~ a.e.~$$Then, for all $n\geq n_0$: $$F_n(f_n)=f_n$ hence, we have the desired result. $\endgroup$
    – Parc John
    Commented May 19, 2020 at 4:01
  • $\begingroup$ The argument you suggest seems to assume a bound on $\|f_n\|_1$ rather than $\|f_n\|_\infty$. $\endgroup$
    – Yuval Peres
    Commented May 19, 2020 at 4:22

1 Answer 1

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Let $f_n$ be independent random variables where $f_n$ takes values $0, 2n$ with $\mu(f_n=2n)=1/n$ for each $n$. Take $g_n=0$ for all $n$. Then $F_n(f_n)=0$ for all $n$ so the hypothesis holds, but the conclusion fails: $\sum_n (f_n/n) = 2\sum_n 1_{\{f_n=2n\}}$ is infinite a.e. by the Borel-Cantelli Lemma.

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  • $\begingroup$ But in book Two-Scale Stochastic Systems Asymptotic Analysis and Control by Kabanov, Yu.M., Pergamenshchikov, S, on page 254, the author says that: the series $$\sum_{k=1}^{\infty}{\zeta_{k}^{(k)}-\eta_k}$$ converges a.s with $\zeta^{(c)}=\zeta 1_{\{|\zeta|\leq c\}}$. Since only a finite number of $\zeta_{k}^{(k)}(\omega)$ is different from $\zeta_{k}(\omega)$, the series $$\sum_{k=1}^{\infty}{\zeta_{k}-\eta_k}$$ also converges a.s. $\endgroup$
    – Parc John
    Commented May 19, 2020 at 15:39
  • $\begingroup$ Maybe you can attach a scan of the relevant pages. There could be additional assumptions there that would imply the desired conclusion. $\endgroup$
    – Yuval Peres
    Commented May 23, 2020 at 20:51

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