Convergence in probability if $\limsup_{n\to\infty}P(|X_n|>\epsilon)\leq c\epsilon$

Question

Let $(X_n)_n$ be a sequence of random variables. Assume that for every $\epsilon>0$, we have

$$\limsup_{n\to\infty}P(|X_n|>\epsilon)\leq c\epsilon,$$ where $c$ is a finite constant. Prove that $X_n$ converges to $0$ in probability.

I did like this.

$$0\leq \lim_{n\to\infty}P(|X_n|>\epsilon)\leq\limsup_{n\to\infty}P(|X_n|>\epsilon)\leq c\epsilon.$$

Let $\epsilon\to 0$. Then $\lim_{n\to\infty}P(|X_n|>\epsilon)\to 0$ as $n\to\infty$.

I feel that I did something wrong, but I don't know how to fix it. Any help would be appreciated.

Answer 1

Your proof is not correct because you have to fix an $\epsilon >0$ and show that $P(|X_n-X| >\epsilon) \to 0$ as $ n \to \infty$.

Let $\eta >0 $. Then $\lim \sup_n P(|X_n-X| >\epsilon) \leq P(|X_n-X| >\epsilon \wedge \eta)\leq c (\epsilon \wedge \eta)\leq c\eta$. This proves that $\lim \sup_n P(|X_n-X| >\epsilon)=0$ since we can let $\eta \to 0$.

[Notation: $x\wedge y=\min \{x,y\}$].

Answer 2

You may prove in this way:

Let $a>0$ be given. We go to prove that $\lim_{n}P\left(\left[|X_{n}|\geq a\right]\right)=0$. Let $\varepsilon\in(0,a)$ be arbitrary. Observe that $\left[|X_{n}|\geq a\right]\subseteq\left[|X_{n}|>\varepsilon\right]$, so $P\left(\left[|X_{n}|\geq a\right]\right)\leq P\left(\left[|X_{n}|>\varepsilon\right]\right)$. It follows that \begin{eqnarray*} \limsup_{n}P\left(\left[|X_{n}|\geq a\right]\right) & \leq & \limsup_{n}P\left(\left[|X_{n}|>\varepsilon\right]\right)\\ & \leq & c\varepsilon. \end{eqnarray*} Since $\varepsilon\in(0,a)$ is arbitrary, we have $\limsup_{n}P\left(\left[|X_{n}|\geq a\right]\right)=0$. Therefore, $\lim_{n}P\left(\left[|X_{n}|\geq a\right]\right)=0$.