Show that $\lim_{n\rightarrow \infty}\frac{\mathsf {Var}[Y_1 + \dots + Y_n]}{n^2} = 0.$ Let $(X_n)_{n=1}^{\infty}$ be a sequence of pairwise independent, identically distributed random variables with finite mean. 
Let $Y_n = X_n\mathbf1\{{|X_n| \leq n\}}$, then 
$$ \sum_{n=1}^{\infty} P(Y_n \neq X_n) = \sum_{n=1}^{\infty} P(|X_n| >n)  = \sum_{n=1}^{\infty} P(|X_1|>n) \leq \mathbb{E}[|X_1|] < \infty. $$ 
Hence from Borel-Cantelli lemma $\lim_{n\rightarrow \infty}\frac{X_1 + \dots + X_n}{n} = \mathbb{E}[X_1]~a.s$  if and only if $\lim_{n\rightarrow \infty}\frac{Y_1 + \dots + Y_n]}{n} = \mathbb{E}[X_1]~a.s.$
Now I have to show that $\lim_{n\rightarrow \infty}\frac{\mathsf {Var}[Y_1 + \dots + Y_n]}{n^2} = 0$, unfortunately I don't know how. 
I would really appreciate any hints or tips.
 A: I come after the battle, but here it is : yet another answer,
\begin{align*} 
Var(Y_1+\ldots+Y_n) 
& =\sum_{i=1}^n Var(Y_i) \\
&=\sum_{i=1}^n Var(X_i 1_{|X_i|\le i}) \\
&=\sum_{i=1}^n Var(X_1 1_{|X_1|\le i}) \\
&\le \sum_{i=1}^n \mathbb E[X_1^2 1_{|X_1|\le i}] \\
& =  \mathbb E[X_1^2 \sum_{i=1}^n 1_{|X_1|\le i}] \\
& \le n  \mathbb E[X_1^2 1_{|X_1| \le n}]
\end{align*}
Therefore :
$$ \frac{Var(Y_1+\ldots+Y_n)}{n^2} \le \mathbb E\Big[\frac{X_1^2}{n} 1_{|X_1| \le n}\Big] $$
Now we have the bound
$$\frac{X_1^2}{n} 1_{|X_1| \le n} \le |X_1|$$
the left hand side converges to $0$ a.s., while the right hand side is an integrable bound (assuming $X_1$ has a finite expectation)
that allows to apply Dominated Convergence Theorem to the effect that :
$$\mathbb E\Bigg[ \frac{X_1^2}{n} 1_{|X_1| \le n}\Bigg] =o(1)$$
A: For this proof we need $E|X|<\infty$ (it seems that this is the case).
Notice that
$$E(Y_k^2)\leq \int_0^k 2tP(|X|>t)dt. $$
Hence, by Fubini's theorem
\begin{align}
\sum^\infty_{k=1}\frac{E(Y_k^2)}{k^2} &\leq \int_0^\infty \left(\sum^\infty_{k=1}\frac{1_{(t\leq k)}}{k^2}\right)2tP(|X|>t)dt\\
&  \int_0^\infty\left( \sum^\infty_{k=t}\frac{1}{k^2}\right)2tP(|X|>t)dt\\
& \leq \int_0^\infty \left( \sum^\infty_{k=t}\frac{2}{k(k+1)}\right)2tP(|X|>t)dt\\
& = \int_0^\infty \left( \sum^\infty_{k=t}\frac{2}{k} - \frac{2}{k+1}\right)2tP(|X|>t)dt\\
& \leq \int_0^\infty \frac{2}{t}2tP(|X|>t)dt = 4E|X|<\infty
\end{align}
So, we proved that
$$\sum^\infty_{k=1}\frac{E(Y_k^2)}{k^2} $$
converges. Then, by Kronecker's lemma,
$$\lim_{n\rightarrow \infty}\frac{1}{n^2}\sum^n_{k=1}E(Y_k^2) =0.$$
Additional thoughts: After reading the comments, I think this may help you. 
You can use Kolmogorov's maximal inequality (or a Martingale in $L_2$ convergence Theorem) to prove that if 
$$ \lim_{n\rightarrow \infty}\sum^n_{k=1}\frac{\text{Var}(Y_k^2)}{k^2}$$
converges, then 
$$\lim_{n\rightarrow \infty}\sum^n_{k=1}\frac{Y_k -EY_k}{k} $$
converges a.s..
Hence, by Kronecker's lemma, 
$$\lim_{n\rightarrow \infty}\frac{1}{n}\sum^n_{k=1}(Y_k -EY_k)=0. $$
Using Dominated Convergence Theorem, it's easy to show 
$$E Y_k \rightarrow EX. $$
Therefore, by Cesàro's Lemma, 
$$\frac{1}{n}\sum_{k=1}^n Y_k = EX,\ a.s. .$$
Now, you can use your Borel-Cantelli Argument to prove that the above expression implies
$$\lim_{n\rightarrow \infty}\frac{1}{n}\sum^n_{k=1}X_k=EX,\ a.s.. $$
This is basically the proof for the Strong Law of Large Numbers in the case i.i.d. with $E|X|<\infty.$
A: Since the $Y_i$ are independent, it is enough to show that $\frac{1}{n^2}\sum_{k=1}^n{Var(Y_k)} \rightarrow 0$. 
Now, $\mathbb{E}[Y_k^2]=\int_0^{k^2}{P(Y_k^2 > a)\,da}=\int_0^k{2uP(k > |X| > u)\,du} \leq \int_0^k{2uP(X > u)\,du}$. 
Denote as $F(u)=\int_u^{\infty}{P(|X|>a)\,da}=\mathbb{E}[|X|1(|X| > u)]$, then $uP(X>u)=F(u)-\frac{d}{du}(uF(u))$. 
As a consequence, $\mathbb{E}[Y_k^2] \leq 2\int_0^k{F(u)\,du}-kF(k) \leq 2\int_0^k{F(u)\,du}$. 
Note that $F$ is non-increasing and goes to $0$, so $\int_0^k{F(u)\,du}=o(k)$, hence $Var(Y_k) =o(k)$, which ends the proof. 
