Convergence in distribution and sequence converging to $0$

Question

If $X_{n}\to 0$ in distribution . Then I am required to Find a sequence $x_{n}\to 0$. Such that $\mathbb{P}(|X_{n}|\geq x_{n})\to 0$.

My initial idea is using Skorohod's Representation theorem to find $Y_{n}$ which converges to $0$ almost surely Hence in probability also.

Then there exists $m\in\mathbb{N}$ such that $\mathbb{P}(|Y_{n}|\geq \frac{1}{n})<\frac{1}{n},$ $\forall n\geq m$. Then $\mathbb{P}(|X_{n}|\geq \frac{1}{n})=\mathbb{P}(|Y_{n}|\geq \frac{1}{n})$ . Then would this suffice?. I am having doubt regarding the previous step.

What is the correct way of approaching this?. Any help is appreciated.

Answer 1

Consider any sequence $\epsilon_n\searrow 0$. Since $X_n\xrightarrow{d}0$, for each $k\ge 1$, there exists $n_k\ge 1$ s.t. $\mathsf{P}(|X_n|\ge \epsilon_k)<\epsilon_k$ for all $n\ge n_k$. Then, you can construct $x_n$ recursively, i.e., set $x_1=\cdots=x_{n_1-1}=1$, $x_{n_1}=\cdots,x_{n_2-1}=\epsilon_1$, and so on.