Simplified Strong Law of Large Number by Using Truncating Function

Question

Consider $X_1,X_2,...$ be i.i.d. random variables with $E|X_i| <\infty$ and let $EX_i := \mu$ and $S_n := \sum_{i=1}^n X_i$. Now, consider the corresponding truncated random variables $Y_k := X_k 1_{|X_k| \le k}$ and put $T_n:=\sum_{i=1}^n Y_i$. Show that $T_n/n \to \mu$ almost surely.

Here is my try:

Observe that for $k \in \mathbb{N}$, $$ \begin{array}{l} \infty > E\left| {{X_k}} \right| = \int_0^\infty {P(\left| {{X_k}} \right| > y)} dy\\ \begin{array}{*{20}{c}} {}&{}&{}&{}&{}&{} \end{array} = \sum\limits_{k = 1}^\infty {\int_{k - 1}^k {P(\left| {{X_k}} \right| > y)} dy} \\ \begin{array}{*{20}{c}} {}&{}&{}&{}&{}&{} \end{array} \ge \sum\limits_{k = 1}^\infty {\int_{k - 1}^k {P(\left| {{X_k}} \right| > k)} dy} \\ \begin{array}{*{20}{c}} {}&{}&{}&{}&{}&{} \end{array} = \sum\limits_{k = 1}^\infty {P(\left| {{X_k}} \right| > k)\int_{k - 1}^k {dy} } = \sum\limits_{k = 1}^\infty {P(\left| {{X_k}} \right| > k)} \\ \Rightarrow \sum\limits_{k = 1}^\infty {P(\left| {{X_k}} \right| > k)} < \infty \end{array}$$ By Borel-Cantelli Lemma, we can conclude that $P(\left\{ {\left| {{X_k}} \right| > k} \right\}i.o.) = 0$; hence, it implies $$ \Rightarrow P(\left\{ {{X_k} \ne {Y_k}} \right\}i.o.) = 0$$

Now,I think I could infer that $ {\forall \omega ,\begin{array}{*{20}{c}} {} \end{array}\exists N\begin{array}{*{20}{c}} {} \end{array}s.t.\begin{array}{*{20}{c}} {} \end{array}k \ge N \Rightarrow {X_k(\omega)} = {Y_k(\omega)}} $. But then I failed to go further.

Answer 1

Here is a basic lemma from analysis:

Let $(x_n)$ and $(y_n)$ denote some sequences such that there exists some finite $N$ such that $y_n=x_n$ for every $n\geqslant N$. Then, $\frac1n\sum\limits_{k=1}^nx_k\to\mu$ if and only if $\frac1n\sum\limits_{k=1}^ny_k\to\mu$.

Can you see why this answers your question? To prove the lemma, use the triangular inequality to show that, for every $n$, $$\left|\frac1n\sum\limits_{k=1}^nx_k-\frac1n\sum\limits_{k=1}^ny_k\right|\leqslant\frac{c}n$$ with $$ c=\sum_{k=1}^N|x_k-y_k|$$