Verifying convergence in probability and distribution. Suppose random variables $\{X_n\}$ are defined on $([0,1],\mathcal{B}([0,1]),\lambda)$ (where $\lambda$ is Lebesgue measure) as follows:
$X_1 = \mathbb{1}_{[0,1]}, X_2 = \mathbb{1}_{[0,1/2]}, X_3 = \mathbb{1}_{[1/2,1]}, X_4 = \mathbb{1}_{[0,1/3]}, X_5 = \mathbb{1}_{[1/3,2/3]}, X_6 = \mathbb{1}_{[2/3,1]}$, etc.
Exercise:
a) Verify that $X_n \stackrel{p}{\to} 0$
b) Verify that for any $w \in [0,1]$, $X_n \not\to 0.$
Questions:


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*a) How do I show that $\lim\limits_{n \to \infty}P(\left|X_n\right|>\epsilon)\to 0$? Intuitively it feels very clear that $X_n$ converges in probability: as $n$ goes to infinity the probability that the indicator functions is equal to $1$ goes to zero. I'm not sure however how to mathematically show this.

*b) What does $w$ do for our random variables? $X_n$ isn't dependent on $w$ right? So how can we write $X_n(w)$?  
 A: Notice that your random variables are given as:
$$X_{\frac{n(n-1)}2+1+j} = \mathbb{1}_{\left[\frac{j}{n}, \frac{j+1}{n}\right]}, \text{ for } j \in \{0, \ldots, n-1\}, n\in\mathbb{N}$$
So for any $\varepsilon \in \langle 0, 1\rangle$ we have
$$\mathbb{P}\left(\left|X_{\frac{n(n-1)}2+1+j}\right| > \varepsilon\right) = \mathbb{P}\left(\mathbb{1}_{\left[\frac{j}{n}, \frac{j+1}{n}\right]} > \varepsilon\right) = \mathbb{P}\left(\left[\frac{j}{n}, \frac{j+1}{n}\right]\right) = \frac1n \xrightarrow{n\to\infty} 0$$
For $\varepsilon \ge 1$ we have $\left\{\left|X_{\frac{n(n-1)}2+1+j}\right| > \varepsilon\right\} = \emptyset$, so the probabilities also converge to $0$. Therefore, $X_n \xrightarrow{\mathbb{P}} 0$.
Take $w \in [0,1]$. Notice that for any $n \in \mathbb{N}$ we have that $X_{\frac{n(n-1)}2+1+j}(w) = 1$ at exactly one or two $j \in \{0, \ldots, n-1\}$.
Namely, if $w \in \left\langle \frac{j_0}{n}, \frac{j_0+1}{n}\right\rangle$ for some $j_0$, then $X_{_{\frac{n(n-1)}2+1+j_0}}(w) = 1$, and $X_{\frac{n(n-1)}2+1+j}(w) = 0$ for every $j \ne j_0$. If $w = \frac{j_0}{n}$ for some $j_0$, then $X_{_{\frac{n(n-1)}2+1+(j_0-1)}}(w) = 1$, $X_{_{\frac{n(n-1)}2+1+j_0}}(w) = 1$ and $X_{_{\frac{n(n-1)}2+1+j}}(w) = 0$ for every $j \notin \{j_0-1, j_0\}$.
We conclude that $X_n(w)$ cannot converge to $0$ because $1$ occurrs infinitely many times in the sequence.
A: a) Note that 
$P(x_n=1)=\frac{1}{k}$ for all $\frac{(k-1)k}{2}<n\leq \frac{k(k+1)}{2}$.
Hence as $n\to\infty$ (because $k\to\infty$) 
$P(|X_n|>\epsilon)\to 0$.
b) Here note that 
for any $w$,
$X_n(w)=1$ for infinitely many $n\in\mathbb N$.
Hence $X_n(w)\not\to0$.
