Constructing example where $X_n \to X$ almost surely but $\sum_{n=1}^\infty P(|X_n-X|>\varepsilon)=\infty$ for some $\varepsilon>0$

Question

It is well-known that if $(X_n)$ is a sequence of random variables and $X$ is another random variable, $$\sum_{n=1}^\infty P(|X_n-X|>\varepsilon)<\infty \quad\forall\,\varepsilon>0\implies X_n \stackrel{\text{a.s.}}\to X$$

I am wondering if it is possible to construct $(X_n)$ and $X$ such that $X_n \stackrel{\text{a.s.}}\to X$ but $\sum\limits_{n=1}^\infty P(|X_n-X|>\varepsilon)=\infty$ for some $\varepsilon>0$.

Suppose I choose $X=0$ with probability $1$.

Then $X_n \stackrel{\text{a.s.}}\to 0 \iff P(|X_n|>\varepsilon\,\text{ i.o.})=P\left(\lim\sup \{|X_n|>\varepsilon\}\right)=0$ for all $\varepsilon>0$.

But how can I choose the sequence $(X_n)$ so that $\sum\limits_{n=1}^\infty P(|X_n|>\varepsilon')$ diverges for some $\varepsilon'>0$?

I could not conclude anything from Borel-Cantelli lemma either.

Any suggestions would be great.

Answer 1

Take $\Omega=[0,1]$ and $P=m_{[0,1]}$ and $X_n=n1_{[0,\frac{1}{n}]}$

Then $X_n \to 0$ and take $\epsilon=1$