Let $\Omega =[0,1],$ the sigma algebra: $F=B(\mathbb{[0,1]})$ and $\mathbb{P}=m$ (Lebesgue measure) $X_n$ is a random variable: $X_n= 2^n 1_{(0,\frac{1}{n})}, n \geq 1 $
Prove directly, that is, using the definition, that $X_n \overset{p}\to 0$
The definition of convergence in probability states that If $X, X_n$ are random variables $\forall \varepsilon \geq 0, \mathbb{P}(|X_n| > \varepsilon) \overset{n \to \infty}\to 0$
My lecture notes give this short proof, that I am struggling to understand:
$\forall \varepsilon \geq 0, \mathbb{P}(|X_n| > \varepsilon)=\mathbb{P}(X_n=2^n)=\mathbb{P}((0,\frac{1}{n}))=\frac{1}{n} \overset{n \to \infty}\to 0$
I don't understand how they go from $\mathbb{P}(|X_n| > \varepsilon)$ to $\mathbb{P}(X_n=2^n)$ to $ \mathbb{P}((0,\frac{1}{n}))$. I trying to separate it into cases according to the value of $\varepsilon$, but I have to consider that the indicator function eventually becomes zero for sufficiently large n and at a fixed $\omega \in \Omega$. Can someone explain it in detail?
Notice that for a fixed $n \in \mathbb{N}$, $X_n(\omega) > 0$ if and only if $\omega \in (0, \frac{1}{n})$. Thus for any $\epsilon > 0$, $P(|X_n| > \epsilon)= P((0,\frac{1}{n})) = \frac{1}{n}$.
Edit: changed $x$ to $\omega$ to make notation a bit clearer