I am getting confused with the difference between pointwise convergence and almost sure convergence (or if there should be any difference at all)...
Say that I have a random variable $X_n=n1_{[0,1/n]}$. In this case, $X_n$ converges to $0$ pointwise, and it also converges in probability, since:
$$ P(|X_n - 0| > \epsilon) = \frac{1}{n} \rightarrow 0 $$
However, even though $X_n$ converges pointwise, this is not almost sure convergence right? Since we have
$$ \sum P(|X_n - 0| > \epsilon) = \sum_{n=1}^{\infty} \frac{1}{n} = \infty $$
since the sum diverges, it means (using the Borel-Cantelli lemma) that it is not convergence almost surely.
So how should I relate the following?
- convergence pointwise
- convergence almost surely
- convergence almost everywhere
From the comments it is clear that the probability space is $\Omega=[0,1]$ endowed with the Borel-$\sigma$-algebra and the uniform measure, call it $\mathbb P$.
Let's clarify the definitions first. Almost everywhere convergence is the same as saying that there exists some $\Omega^*$ for which $\mathbb P(\Omega^*)=1$ and $X_n$ converges pointwise on $\Omega^*$. That is for all $\omega\in\Omega^*$ we have $X_n(\omega)\to X$ for some random variable $X$. In probability setting one usually uses "almost surely" instead of "almost everywhere". One also says converges surely if $\Omega^*=\Omega$, i.e. the whole space.
Using this definition we see that for all $\omega \in (0,1]:=\Omega^*$ we have $X_n(\omega)=0$ for $n$ large enough. Moreover $\mathbb P((0,1])=1$ which means that we have almost surely convergence to $0$.
Where does the application of Borel-Cantelli goes wrong in the question then? Well, for the use of Borel-Cantelli like that one needs to assume that the events are independent. However they are not. Just see that for $\varepsilon\in (0,1)$ the event $\{|X_n-0|\leq \varepsilon\}$ implies that $\{|X_m-0|\leq \varepsilon\}$ for all $m>n$ (You see this?).