Show that: $ \sum_{n\geq 1}{\frac{1}{n}(f_n(\omega)-g_n(\omega))}<\infty\qquad a.e $ Let $(\Omega,\mathcal{A},\mu)$ be a finite mesure space, and $\{f_n\}$  and $\{g_n\}$ are two $L^1$-bounded sequence, such that :
$$
\sum_{n\geq 1}{\frac{1}{n}(F_n(f_n)(\omega)-g_n(\omega))}<\infty\qquad a.e
$$
with: $F_n(f_n)=f_n1_{|f_n|\leq n}$
Show that:
$$
\sum_{n\geq 1}{\frac{1}{n}(f_n(\omega)-g_n(\omega))}<\infty\qquad a.e
$$
My effort:
according to $\sup_n\|f_n\|_1<\infty$, there exists $n_0\geq 1$, 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. 
Is what I wrote correct?
 A: Looking at your source, there are two different sequences of functions, $\{\xi_n\}$ and $\{\zeta_n\}$, that you both call $\{f_n\}$ in your question. It is the $\{\xi_n\}$ that are (merely) $L^1$-bounded, while the $\{\zeta_n\}$, a subsequence of the $\{\xi_n\}$, are the ones that you want to prove the following property about:

If the series $$\sum_{k=1}^\infty\frac{1}{k}(\zeta_k^{(k)} - \eta_k) \tag{A}\label{eqA}$$ converges a.s., then the series $$\sum_{k=1}^\infty\frac{1}{k}(\zeta_k - \eta_k) \tag{B}\label{eqB}$$ also converges a.s.

The $\{\zeta_n\}$ fulfill the following equation, shown in Lemma A.7.3:$$\sum_{k=1}^\infty P(|\zeta_k| \ge k) \lt \infty. \tag{1}\label{eq1}$$ From this it follows (as noted directly below the lemma) that$$\text{for almost all }\omega, |\zeta_k(\omega)| \gt k\text{ holds for only finitely many }k. \tag{2}\label{eq2}$$Therefore, for almost all $\omega$, the sequences $\{\zeta_n(\omega)\}$ and $\{\zeta_n^{(n)}(\omega)\}$ differ in only finitely many terms, and so do the two series \eqref{eqA} and \eqref{eqB}. But when two series differ in only finitely many terms, and one of them converges, so does the other.
Now why does \eqref{eq1} imply \eqref{eq2}? Assume \eqref{eq2} doesn't hold. Then there is a set $A \subseteq \Omega$ of positive measure such that for all $\omega \in A$, $|\zeta_k(\omega)| \gt k$ holds for infinitely many $k$. This implies$$A \subseteq \bigcup_{k=m+1}^\infty \{|\zeta_k| \gt k\}$$ for all $m \in \mathbb{N}$, and by continuity of measure there is an $N(m)\in\mathbb{N}$ such that$$\sum_{k=m+1}^{N(m)} P(|\zeta_k| \gt k) \gt \frac{1}{2}P(A).$$Now set $N_1 = 0$, and $N_{k+1} = N(N_k)$ for $k \ge 1$. Then$$\sum_{k=1}^\infty P(|\zeta_k| \ge k) = \sum_{i=1}^\infty \sum_{k=N_i+1}^{N_{i+1}} P(|\zeta_k| \ge k) \gt \sum_{i=1}^\infty \frac{1}{2}P(A) = \infty,$$which contradicts \eqref{eq1}.
