Converge Almost Surely v.s. Converge in r-th Mean

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Let $X$ and $\{X_n\}_{n\geq1}$ be random variables.  $X_n$ converge to $X$ almost surely dose NOT implies $X_n$ converge to $X$ in $r$-th mean, and vice versa.

(1) Counterexample for $X_n$ converge to $X$ almost surely dose NOT implies $X_n$ converge to $X$ in $r$-th mean.

$\bullet$ Counterexample.
Define for $r\geq2$ and $n\geq1$, $\mathscr{P}\{X_n=0\}=1-\frac{1}{n^r}$, $\mathscr{P}\{X_n=n\}=\frac{1}{n^r}$ and $X_i$'s are independent.  $$\mathscr{P}\left\{\bigcap_{n=m}^{n_0}X_n=0\right\}=\prod_{n=m}^{n_0}\left(1-\frac{1}{n^r}\right)\rightarrow0\mbox{ as }n_0\rightarrow\infty.$$  Thus for all $\varepsilon>0$, $$\underset{m\rightarrow\infty}{\lim}\mathscr{P}\left\{\bigcap_{n\geq m}|X_n|\leq\varepsilon\right\}=1\implies X_n\overset{a.s.}{\longrightarrow}0.$$

But, $\mathscr{E}|X_n|^r=n^r\frac{1}{n^r}=1$ for all $n$.  Hence, $X_n\not\overset{L^r}{\longrightarrow}0$.

$\Box$

(2) Counterexample for $X_n$ converge to $X$ in $r$-th mean dose NOT implies $X_n$ converge to $X$ almost surely.

$\bullet$ Counterexample.
Define for $r>0$ and $n\geq1$, $\mathscr{P}\{X_n=0\}=1-\frac{1}{n}$, $\mathscr{P}\{X_n=n^{\frac{1}{2r}}\}=\frac{1}{n}$ and $X_i$'s are independent.  $$\mathscr{E}|X_n|^r=\left(n^{\frac{1}{2r}}\right)^r\frac{1}{n}=\frac{1}{\sqrt{n}}\rightarrow0\mbox{ as }n\rightarrow\infty.$$ Thus, $X_n\overset{L^r}{\longrightarrow}0$.

But, consider for $m>>0$, $$\mathscr{P}\left\{\bigcap_{n=m}^{n_0}X_n=0\right\}=\prod_{n=m}^{n_0}\left(1-\frac{1}{n}\right)=\frac{m-1}{m}\frac{m}{m+1}\cdots\frac{n_0-1}{n_0}=\frac{m-1}{n_0}\rightarrow0$$ as $n_0\rightarrow\infty.$  Thus, $\mathscr{P}\left\{\bigcap_{n\geq m}|X_n|\leq\varepsilon\right\}\rightarrow0$ for all $\varepsilon>0$ and $m>0$.  That is, $X_n\not\overset{a.s.}{\longrightarrow}0$.

$\Box$