I have to do this exercise:
Let $(X_n)$ be some sequence of random variables and let $X$ be some random variable such that $X_n \to X$ almost surely.
Show that, given $\epsilon > 0$, there is a set $A$ such that $P(A)\leq \epsilon$ and $X_n \to X$ uniformly on $\Omega \setminus A$.
I think that I can use these sets $E(n_k) = \bigcup_{m\geq n}\{\omega \in Ω: \mid X_m(\omega) − X(\omega)| \geq \frac1k\}$ and then obtain $A$ from there.
But I don't know how to start and furthermore what $X_n \to X$ uniformly means.
Best Answer
We say that $X_n\to X$ uniformly on a set $E$ if $\sup_{\omega\in E}\left\lvert X_n(\omega)-X(\omega)\right\rvert\to 0$.
Here are some steps:
For simplicity, assume that $X=0$ and $X_n\geqslant 0$ (consider $\left\lvert X_n-X\right\rvert$ instead of $X_n$).
Let $$E(n,k) = \bigcup_{m\geqslant n}\left\{\omega \in Ω: X_m(\omega) \geqslant \frac1k\right\}.$$ The assumption that $X_m\to 0$ implies that for all fixed $k$, $\Pr\left(\bigcap_{n\geqslant 1}E_{n,k}\right)=0$.
The sequence $\left(E_{n,k}\right)_{n\geqslant 1}$ is non-increasing for each $k$ hence there exists $n_k$ such that $\Pr\left( E_{n_k,k}\right)\lt \varepsilon 2^{-k}$
Define $A:=\bigcup_{k\geqslant 1}E_{n_k,k}$ and check that $A$ does the job.