Strong law of large numbers with finite fourth moment

Question

My textbook (Introduction to probability, bertsekas 2nd ed pg 294) showed that $\frac{(X_1+\dots+X_n)^4}{n^4}$ converges to zero with probability 1 and then said that it implies $\frac{(X_1+\dots+X_n)}{n}$ converges to zero with probability 1. I am not sure how this follows.

Answer 1

Let $\overline{X}_n:=\frac1n\sum_{i=1}^nX_i$.

Then for every $\omega\in\Omega$ we have: $$\lim_{n\to\infty}\overline{X}_n(\omega)=0\iff\lim_{n\to\infty}\overline{X}_n^4(\omega)=0$$So expressed in sets: $$\left\{\lim_{n\to\infty}\overline{X}_n=0\right\}=\left\{\lim_{n\to\infty}\overline{X}_n^4=0\right\}$$ so that:$$P(\lim_{n\to\infty}\overline{X}_n=0)=P(\lim_{n\to\infty}\overline{X}_n^4=0)$$