Let $Y_n:=n^2{\bf 1}_{[0,1/n]}(U)$ where $U\sim \mathcal{U}([0,1]).$
I don't see why we have the equality $$P(\{\omega\in\Omega\;;\lim_{n\to\infty}Y_n=0\})=P(U\in(0,1])\;?$$
I have another question I have some issues to understand convergence almost surely of random variables.
For exemple let $(X_n)_n$ be a sequence of random variables such that the distribution is uniform on $[-\frac{1}{n},\frac{1}{n}]$, how can I look at the convergence almost surely of $X_n$ ?
Best Answer
If $U(w) \in (0,1]$, then there exists $N$ such that $U(\omega) > 1/N$. Then for all $n\geq N$ we have $Y_{n}(\omega)=0$, and therefore $\lim_{n\to\infty}Y_{n}(\omega)=0$. Conversely, if $U(w)=0$ then $Y_{n}(\omega)=n^{2}$ for all $n$, and so $Y_{n}(\omega)\to \infty$.
Thus $Y_{n}(\omega)\to 0$ iff $U(\omega)\in(0,1]$.
For the second question, you really need to be more specific about the underlying probability space and probability measure, but I'll just assume we're working on $[-1,1]$ with normalized Lebesgue measure (i.e. divide by 2, which is the length of the interval $[-1,1]$).
I claim that $X_{n}\to 0$ a.s. Indeed, if $\omega\neq 0$, then there exists $N$ such that $|\omega|>1/N$ for all $n\geq N$. Therefore $\lim_{n\to\infty}X_{n}(\omega)=0$ if $\omega\neq 0$. Since $\mathbb{P}([-1,1]\setminus\{0\})=1$, we have $X_{n}\to 0$ a.s.