Why this sequence converges to $0$ almost surely

Question

I have this sequence of random variables: $$X_{1}(w) = \mathbb{1}_{(1/2,1]}(w), X_{2}(w) = \mathbb{1}_{(0,1/2]}(w), X_{3}(w) = \mathbb{1}_{(3/4,1]}(w), X_{4}(w) = \mathbb{1}_{(1/2,3/4]}(w) ...$$

I know and I understand that this sequence converges in Probability: $P(|X_{n}|>0) \longrightarrow 0$, $\textbf{but not almost sure}$, because since for any $\omega \in (0,1]$ and $N \in \mathbb{N}$ there exist $\textbf{n}$, $\textbf{m} \geq N$ such that $X_{n} = 1$ and $X_{m} = 0$.

But Why this sequence bellow converges to $0$ almost surely?

$$Y_{1}(w) = \mathbb{1}_{(0,1/2]}(w), Y_{2}(w) = \mathbb{1}_{(0,1/2]}(w), Y_{3}(w) = \mathbb{1}_{(0,1/4]}(w), Y_{4}(w) = \mathbb{1}_{(0,1/4]}(w) ...$$

Thanks guys!

Answer 1

So the probability space we're working on is $\Omega = (0, 1]$. Note for any $\omega \in \Omega$, there exists $K$ such that we have $\frac{1}{k} < \omega$ for all $k \geq K$. In particular if $Y (\omega)= 1_{(0, 1/k]}(\omega)$ then $Y(\omega) = 0$. Eventually all the $Y$'s take the form $1_{(0, 1/k]}$ for $k \geq K$, hence $\lim_n Y_n(\omega) = 0$ for all $\omega \in \Omega$. This gives almost sure convergence.

In this case we showed convergence for all $\omega \in \Omega$ but you actually only show it for all $\omega \in \Omega_0$ where $\Omega_0 \subset \Omega$ is of probability 1.