Convergence of random variables: $\sum_{n=1}^{\infty}\mathbb{E}(\lvert X_{n}-X\rvert^{r}) < \infty$ implies $X_n \to X$ almost surely

Question

Let there be a sequence of r.v.'s $\{X_{n}\}_{n\geq1}$. Show that if $\exists r \geq 1, \;\sum_{n=1}^{\infty}\mathbb{E}(\lvert X_{n}-X\rvert^{r}) < \infty$ then $X_{n} \xrightarrow[]{a.s.} X$

My ideas were using Fatou's lemma and divergence test ($\lim_{n\xrightarrow[]{} \infty} \mathbb{E}(\lvert X_{n}-X\rvert^{r})=0$) to squeeze the almost-surely condition somehow, but I'm stuck. I would appreciate help.

Answer 1

Let $(\Omega, \mathcal F, P)$ be the probability space. By monotone convergence, $$\int_\Omega \sum_{n=1}^\infty|X_n-X|^r \ dP =\mathbb E \left(\sum_{n=1}^\infty |X_n - X |^r \right) = \sum_{n=1}^\infty \mathbb E \left( |X_n - X|^r\right) < \infty$$ Therefore, the set $$ E := \left\{\omega \in \Omega : \sum_{n=1}^\infty |X_n(\omega) - X(\omega)|^r = \infty \right\}$$ has measure zero.

For any $\omega \in \Omega \setminus E$, the sum $\sum_{n=1}^\infty |X_n(\omega) - X(\omega)|^r $ is finite. Hence $$ \lim_{n \to \infty} |X_n(\omega) - X(\omega)|^r = 0,$$ and hence $$ \lim_{n \to \infty}X_n(\omega) \to X(\omega)$$ for any $\omega \in \Omega \setminus E$.