I have a sequence of random variables $X_n$ defined on $\Omega =[0,1]$ by $X_n(\omega)=\omega n$. Let $\mathbb{P}$ be the Lebesgue measure on $[0,1]$.
Does $X_n \rightarrow 0$ in probability?
I know that if $X_n$ converges in probability to a constant then $ \lim_{n \to \infty}\mathbb{P}(|X_n-c|\gt\epsilon) = 0$
I am confused on the next step for proving or disproving this.
Thank you
Best Answer
Hint : if you want to show that $X_n$ doesn't converge in probability to $0$, all you have to do is make sure that $\mathbf{P}(|X_n|>\epsilon)$ does not goes to zero. But $$\mathbf{P}(|X_n|> \epsilon) = \mathbf{P}(\omega : |\omega n| > \epsilon ) = \mathbf{P}(\omega : \omega > \epsilon / n)$$ which can be computed : if $\mathbf{P}$ is the Lebesgue measure, then $$\mathbf{P}(\omega : \omega > \epsilon / n) = 1- \frac{\epsilon}{n} $$