# How to prove $X_n$ converges to $0$ in probability measure

I'm new to measure theory. I'm pretty struggle with the following question. Really need help to formally prove the question.

$$U\sim \mathrm{Uniform}(0,1),$$ $X_n=\sqrt{n}I_{(0,1/n)}(U)$, how to prove $X_n \rightarrow 0$ in probability measure?

• Let $p_n := P(X_n >0)$, which presumably you can compute. Convergence to zero in measure simply means that $p_n \to 0$ as $n \uparrow \infty$. (More generally, you'd replace $X_n >0$ with $|X_n - X| > \epsilon$ to get a sequence $p_n(\epsilon)$. Convergence in measure (of $\{X_n\}$ to $X$) then requires that for every (fixed) $\epsilon >0$, $p_n(\epsilon) \to 0.$ Note that in principle you have to do the same in the above question as well, but since $P(|X_n|> \epsilon) \le P(X_n >0)$ in the case of your question, the latter going to zero is sufficient.) Nov 15, 2017 at 4:32
• In your case, $X_n \ge 0$ surely so if its absolute value is bigger than some number bigger than $0$, then $X_n$ itself must be bigger than $0$. Thus $\{|X_n| > \epsilon\} \subseteq \{X_n > 0\}.$ The claim then follows from monotonicity of measure. Nov 16, 2017 at 1:16

You can also prove that $(X_n)_{n \in \mathbb{N}}$ converges in $L^1(\mathbb{P})$ (in-mean convergence : $\displaystyle\lim_{n \to \infty} \mathbb{E}(|X_n - 0|) \to 0$). Then, you use the fact that in-mean convergence implies probability convergence.