Question taken from https://www.probabilitycourse.com/chapter7/7_2_7_almost_sure_convergence.php
Consider the sample space $S=[0,1]$ with a probability measure that is uniform on this space, i.e.,
$\begin{align}%\label{} P([a,b])=b-a, \qquad \textrm{ for all }0 \leq a \leq b \leq 1. \end{align}$
Define:
$\begin{equation} \nonumber X_n(s) = \left\{ \begin{array}{l l} 1 & \quad 0 \leq s < {\large \frac{n+1}{2n}} \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation}$
and
$\begin{equation} \nonumber X(s) = \left\{ \begin{array}{l l} 1 & \quad 0 \leq s < \frac{1}{2} \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation}$
Show $X_n \ \xrightarrow{a.s.}\ X$.
I understand the solution by finding the set $\begin{align}%\label{eq:union-bound} A= \left\{s \in S: \lim_{n\rightarrow \infty} X_n(s)=X(s)\right\} \end{align}$ and showing this has probability $1$. I tried to show almost sure convergence by using the rule:
If for all $\epsilon>0$, we have
$\begin{align}%\label{} \sum_{n=1}^{\infty} P\big(|X_n-X| > \epsilon \big) < \infty \end{align}$, then $X_n \ \xrightarrow{a.s.}\ X$.
$\begin{equation} \nonumber |X_n(s) - X(s)| = \left\{ \begin{array}{l l} |1 - 1| = 0 & \quad 0 \leq s < \frac{1}{2} \\ & \quad \\ |0 - 1| = 1 & \quad \frac{1}{2} \leq s < \frac{n+1}{2n} \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{equation}$
For $0 < \epsilon < 1:$
$P(|X_n(s) - X(s)| > \epsilon) = \frac{n+1}{2n} - \frac{1}{2} = \frac{1}{2n}$
Therefore $\begin{align}%\label{} \sum_{n=1}^{\infty} P\big(|X_n-X| > \epsilon \big) = \infty \end{align}$.
So does this show the implication does not go the other way?