I have troubles getting a passage of Etemadi's SLLN demostration from page 244 of "Measure theory and probability theory" by Krishna and Soumendra.
Let $\{X_n\}$ be non-negative, pairwise independet, identically distributed with $\mathbb{E}[X_1]<\infty$.
The book defines $Y_i=X_i I(X_i \leq i)$ then, through an application of Borel-Cantelli lemma, it demonstrates that
\begin{gather*} \mathbb{P}(X_i=Y_i \ \ \text{for all $i$ but finitely many})=1 \end{gather*}
and it asserts that it is sufficent to have
\begin{gather*} \bar{Y}_n = \frac{1}n \sum_{i=1}^nY_i \rightarrow \mathbb{E}[X_1] \ \ \ \text{w.p.} \ \ 1 \end{gather*}
It sounds right to me, but I can't see a straightforward way to prove that. I'm looking for a proof, thanks.
For a proof that
$\mathbb P(X_i=Y_i $ for all $i$ but finitelly many $) = 1$ $\ \ \ $ (i)
or the proof of sufficiency of having
$\bar{Y_n} \to \mathbb E[X_1]$ $\ \ \ $ (ii).
If the first case, note that you have $$\sum_i \mathbb P(X_i \neq Y_i) = \sum_i \mathbb P(X_i > i) = \sum_i \mathbb P(X_1 > i) \le \mathbb E[X_1] < \infty$$ so by borel cantelli, you get $ \mathbb P( \limsup \{X_i = Y_i\}) = 0$, so that $\mathbb P(\liminf \{X_i \neq Y_i\}) = 1$, which is exactly $\mathbb P(X_i=Y_i $ for all $i$ but finitelly many $) = 1$.
If the latter, notice that $ \frac{1}{n}\sum_{k=1}^n X_k := \bar{X_n} = \bar{Y_n} + (\bar{X_n}-\bar{Y_n})$
And since (i) holds, then almost surely (so that for $\omega$ in set of measure $1$) you have
$$ |\bar{X_n}(\omega) - \bar{Y_n}(\omega)| \le \frac{1}{n} \sum_{k=1}^n |X_k(\omega)-Y_k(\omega)| \le \frac{1}{n}\sum_{k=1}^\infty |X_k(\omega)-Y_k(\omega)| = \frac{1}{n} \sum_{j=1}^m |X_{k_j}(\omega)-Y_{k_j}(\omega)| $$
Last equality due to having only finitelly many indices $k_j$ such that $X_{k_j}(\omega) \neq Y_{k_j}(\omega)$. But now, your sum is finite, independent of $n$, and before the sum you have term $\frac{1}{n}$ which goes to $0$, and that means $\bar{X_n}(\omega) - \bar{Y_n}(\omega) \to 0$.
Hence limit of $\bar{X_n}$ is very the same (in almost surely sense) so that of $\bar{Y_n}$, hence it sufficient to show $\bar{Y_n} \to \mathbb E[X_1]$ almost surely.