I am trying to show the following: Let $X_{n},X$ r.v so that $\sum_{n}E(|X_{n}-X|^{r})<\infty$ for some $r>0$. Show $X_{n}\rightarrow X$ a.s My Proof:
For hypothesis, exist some $r>0$ with $r\in \mathbb{N}$ so that
$\sum_{n}E(|X_{n}-X|^{r})<\infty$ then $\sum_{n}E(|X_{n}-X|^{r})$ converge. Now to convergence criterion we know that
$$E(|X_{n}-X|^{r})\rightarrow 0$$
i.e.
$$X_{n}\xrightarrow{L_{r}} X.$$
This follows readily from Chebyshev's inequality for any $\epsilon>0$ then
$$P[|X_{n}-X|\geq \epsilon]\leq \frac{E(|X_{n}-X|^{r})}{\epsilon^{r}}$$
Now we take series in both ways then
$$\sum_{n=1}^{\infty} P[|X_{n}-X|\geq \epsilon]\leq \sum_{n=1}^{\infty} \frac{E(|X_{n}-X|^{r})}{\epsilon^{r}} < \sum_{n=1}^{\infty} E(|X_{n}-X|^{r}) < \infty $$
therefore
$$\sum_{n=1}^{\infty} P[|X_{n}-X|\geq \epsilon]< \infty $$
by Borel-Cantelli lemma, we have
$$P(\limsup_{n\rightarrow \infty} |X_{n}-X|\geq \epsilon )=0$$
Let $w\in [\lim_{n\rightarrow \infty} X_{n}=X]$. Those $M\geq 1$ exist $N \geq 1$ so that $n\geq N$
$$|X_{n}(w)-X(w)|<\frac{1}{M}$$
That means that
$$w\in \bigcap_{M=1}^{\infty} \bigcup_{N=1}^{\infty} \bigcap_{n=N}^{\infty}\left( |X_{n}-X|\leq \frac{1}{M}\right)$$
therefore
$$P[\lim_{n\rightarrow \infty} X_{n}=X]=1-P\left[ \lim_{n\rightarrow \infty} X_{n}\neq X \right]=1-P\left[ \bigcup_{M=1}^{\infty} \bigcap_{N=1}^{\infty} \bigcup_{n=N}^{\infty} |X_{n}-X|\geq \frac{1}{M}\right]\geq 1-P \sum_{M=1}^{\infty} \left[ \bigcap_{N=1}^{\infty} \bigcup_{n=N}^{\infty} |X_{n}-X|\geq \frac{1}{M}\right] $$
Now
$$P[\lim_{n\rightarrow \infty} X_{n}=X]\geq 1-P \sum_{M=1}^{\infty} \left[ \bigcap_{N=1}^{\infty} \bigcup_{n=N}^{\infty} |X_{n}-X|\geq \frac{1}{M}\right]= 1-P \sum_{M=1}^{\infty} \left[\limsup_{n\rightarrow \infty}|X_{n}-X|\geq \frac{1}{M}\right]=1-0.$$
Then $$P[\lim_{n\rightarrow \infty} X_{n}=X ]=1$$
Therefore, $X_{n}\rightarrow X$ a.s .
Did I do anything wrong or is there something to improve?
Looks correct. A simpler but equivalent proof is to note that $E(\sum_{n}|X_n - X|^r) = \sum_n E(|X_n - X|^r) < \infty$ implies $\sum_{n}|X_n - X|^r < \infty$ a.s., so $|X_n - X|^r \to 0$ a.s.