Study the a.s convergence of $X_n=X^n$, where $X\sim U(0,1)$
My attemp:
My definition of a.s convergence is : A sequence of r.v $X_n$converges a.s to $X$ $\iff$ $\mathbb{P}(X_n\not\to X)=0$.
$\mathbb{P}(X_n\not\to X)=\mathbb{P}(X^n\not\to X)=0$
If $X \in [0,1), X^n\to 0$
If $X =1, X^n\to 1$
If $X \in [1,+\infty), X^n\to \infty$
If $X \in [-\infty,0), X^n\to +\infty,$ if $n$ is even and to $-\infty $if $n$ is odd$
I am on the right track here? should I say that $(X^n\not\to X) = \mathbb{R}\setminus (0,1)$ ? or is it just $(X^n\not\to X) = \{1\}$ because a uniform variable takes values only on $[0,1]$ and I should not be analyzing the values outside of it? I am also not sure if there is convergence to 1 for X=1, an to 0 for X in [0,1)
You are making it too complicated. $1=P(0<X<1)\leq P(X^{n} \to 0)$ because $t^{n}\to 0$ whenever $0<t<1$. This finishes the proof.