Convergence in probability does not imply almost sure convergence?

Consider the Lebesgue measure $P$ on $[0,1]$ and define a sequence of random variables $(X_n)$ by letting $X_n$ be the characteristic function of the interval $$ \left[\frac{j}{2^k},\frac{j+1}{2^k}\right], $$ where $k:=\lfloor\log_2(n)\rfloor$ and $j\in\{0,\ldots,2^k-1\}$ satisfies $n=2^k+j$. This sequence converges in probability to zero but does not converge almost surely to zero.

We have $$ X_1=1_{[0,1]},\, X_2=1_{[0,1/2]},\, X_3=1_{[1/2,1]},\, X_4=1_{[0,1/4]},\, X_5=1_{[1/4,1/2]},\, X_6=1_{[1/2,3/4]},\ldots. $$ From this it is easy to see that, for every $\varepsilon>0$, $$ P(|X_n|>\varepsilon)\to 0. $$ Thus $(X_n)$ converges to zero in probability. However, given any $\omega\in[0,1]$, we have that $X_n(\omega)=1$ for infinitely many $n$. Therefore $(X_n)$ does not converge to zero almost surely.

This example is commonly called the typewriter sequence. Note that there are many subsequences of $(X_n)$ which converge almost surely to zero.

Convergence in probability : $X_n \xrightarrow{p} X$ if for all $\epsilon > 0$, $\Pr[|X_n - X| > \epsilon] \to 0$ as $n \to \infty$.

Convergence almost surely : $X_n \xrightarrow{a.s} X$ if $\Pr[X_n(w) \nrightarrow X(w)] = 0$.

Convergence almost surely implies convergence in probability. However, the other way is not true, the most popular counterexample being the following.

Let the space in consideration be $[0,1]$. We will define our sequence with double indices, $X_{nk}$, where $0 \leq k < n$ and $n \geq 1$. Reindexing can be done later.

Let $X_{nk} = 1_{[\frac kn, \frac{k+1}{n}]}$, and let $X \equiv 0$. I claim that $X_{nk} \xrightarrow{p} X$. This is because, if we choose any $\epsilon > 0$, then for each $n$ and $k$, $X_{nk}$ is non-zero only in an interval of $\frac 1n$. If we choose $n > \frac 1\epsilon$, then you can see that the probability that $X_{nk}$ is non-zero (or different from zero) is less than $\epsilon$, as desired.

However, for each $w \in [0,1]$ and $n \in \mathbb N$, there exists $k$ such that $w \in [\frac kn, \frac{k+1}{n}]$. Therefore, the sequence ${X_{nk}}$ consists of infinitely many zeros and infinitely many ones, and therefore does not converge for any $w \in [0,1]$. Since $[0,1]$ has probability non-zero (infact, probability one), we conclude that $X_{nk} \nrightarrow X$ almost surely.