I have an article as follows
Why are they enough to prove that $ \sum_{n=1}^\infty \dfrac{X_n \textbf{1}_{\{|b_n|< |X_n|\}}}{b_n} $ converges almost surely? I want to know why must prove $ \sum_{n=1}^\infty P(|X_n|\geq |b_n|)<\infty$ to have $ \sum_{n=1}^\infty \dfrac{X_n \textbf{1}_{\{|b_n|\geq |X_n|\}}}{b_n} $ converges almost surely. Please help me.
The author of that proof skips a lot of steps!
It seems that the author is secretly making use of the Borel-Cantelli lemma. Equation (*) shows that $\sum_{n=1}^{\infty} Pr[|b_n|\leq |X_n|] < \infty$, which means (by Borel-Cantelli) that, with prob 1, the event $|b_n|\leq |X_n|$ only happens for a finite number of integers $n$. This also means that $|b_n|<|X_n|$ only happens for a finite number of integers $n$.
Then: I think this proof is trying to prove part (i), which means they want to conclude that $\sum_{n=1}^{N}X_n/b_n$ converges almost surely to some (possibly random) real number as $N\rightarrow\infty$. So you can write: $$ \sum_{n=1}^{N}\frac{X_n}{b_n} = \sum_{n=1}^{N} \frac{X_n}{b_n} 1\{|b_n|<|X_n|\} + \sum_{n=1}^{N}\frac{X_n}{b_n}1\{|b_n|\geq|X_n|\} \: \: (Eq. **)$$ Since, with prob 1, the event $|b_n|<|X_n|$ only happens for a finite number of integers $n$, we can say there is some random $M$ such that $M$ is the last integer for which it holds. So, the first summation on the right-hand-side of (Eq. **) stays at a fixed value and never changes once $N>M$: $$ \lim_{N\rightarrow\infty} \sum_{n=1}^N\frac{X_n}{b_n}1\{|b_n|<|X_n|\} = \sum_{n=1}^M \frac{X_n}{b_n}1\{|b_n|<|X_n|\} $$ So the limit of the first term on the right-hand-side of (Eq. **) indeed converges almost surely as $N\rightarrow\infty$. That is why the author is saying it is "enough" to prove that the second term on the right-hand-side also converges almost surely as $N\rightarrow\infty$.