Show a Function Obeys the Strong Law of Large Numbers

Question

Let $f_n$ be a sequence of random variables such that $\mu ( \{f_n = 0 \}) = 1 - \frac{1}{n^2}$ and $E(f_n) = 0.$ I have to prove that $f_n$ obeys the Strong Law of Large Numbers.

In order for a sequence to obey SLLN, we must prove $$\mu \left( \bigcap_{N=1}^{\infty} \bigcup_{n=N}^{\infty} \{ \omega: | \frac{1}{n} \sum_{j=1}^n f_j (\omega) | \ge \frac{1}{k} \} \right) = 0 \text{ } \forall k$$ or equivalently, $$\sum_{n=1}^{\infty} \mu \left( \{ \omega: | \frac{1}{n} \sum_{j=1}^n f_j (\omega) | \ge \frac{1}{k} \} \right) < \infty \text{ } \forall k$$

From the fact that $\mu ( {f_n = 0}) = 1 - \frac{1}{n^2}$, I can show via Borel-Cantelli lemma that $$\mu \left( \bigcup_{N=1}^{\infty} \bigcap_{N=n}^{\infty} \{ f_n = 0 \} \right) = 1$$ which means $$\mu \left( \bigcap_{N=1}^{\infty} \bigcup_{N=n}^{\infty} \{ |f_n| > 0 \} \right) = 0$$ but don't see how to use this info to actually prove SLLN.

Could somebody please provide any hints/ideas?

Answer 1

Use Borel-Cantelli: $$\sum_{k\in \mathbb{N}}\mu(\{f_k\neq 0\})=\sum_{k\in \mathbb{N}}\frac{1}{k^2}<\infty\implies \mu(\{f_n\neq 0 \textrm{ i.o.}\})=0\implies \\\implies \mu(\{f_n=0,\,\forall n \geq N\textrm{ for some }N\})=1$$ This implies that for a.a. $\omega$ $$\lim_{n \to \infty}\sum_{k\leq n}\frac{f_k(\omega)}{n}=\lim_{n \to \infty}\sum_{k\leq N(\omega)}\frac{f_k(\omega)}{n}=0$$ so that the average converges to $0$ $\mu$-a.s.

Answer 2

As indicated by the hint in the comment: Since $$\sum_{n\mathbb{N}}\mu(\{|f_n|>0)\leq\sum_{n\in\mathbb{N}}\frac1{n^2}<\infty$$ $\mu(\{|f_n|>0,\,\text{i.o}\})=0$. Then, for every $x\in X\setminus\{|f_n|>0,\,\text{i.o}\}$, there is $n_\omega\in\mathbb{N}$ such that $|f_n(\omega)|=0$ for all $n\geq n_\omega$. Therefore $$\lim_{n\rightarrow\infty}\frac{1}{n}\sum^n_{j=1}f_n(\omega)=\lim_{n\rightarrow\infty}\frac{1}{n}\sum^{n_\omega}_{j=1}f_j(\omega)=0$$