People, I have no idea how to start proving weak converge. I know the definition that limit of distribution of $X_n$ should be equal to distribution of $X$. BUT no idea how to prove it. I think that there should be some hint with summation.
So we have: $P(X_n=i/n)=1/n$ for $i=1,\ldots,n$, and I need to show this: $X_n \rightsquigarrow X$.
I think it is obvious that if for example $$\begin{align*}m P(X_n \leq 5/10) &= P(X_n = 1/10) + P(X_n = 2/10) + P(X_n = 3/10) \\ &\quad + P(X_n = 4/10) + P(X_n = 5/10) \\ &= 5/10, \end{align*}$$ and $P(X\leq 5/10) = 5/10$ for $X$ uniformly distributed from $0$ to $1$.
So their distribution functions agree on $x=5/10$, and they agree on any value I choose. But in principle why this example says that we have weak convergence? Why strong convergence here does not work?
Picture for $X_{2n}(x)$ and $X_{2n+1}(x)$:
I can not see how both sequences converge to different values. But the value of distribution function is equal 1 if we take x=1. Am I right? This is what I conclude from the picture:
The distribution function $F_n := \mathbb{P}(X_n \leq x)$ equals
$$F_n(x) = \begin{cases} 0, & x < \frac{1}{n}, \\ \frac{1}{n}, & x \in \big[\frac{1}{n},\frac{2}{n} \big) \\ \frac{2}{n}, & x \in \big[ \frac{2}{n}, \frac{3}{n} \big) \\ \vdots & \\ 1, & x \geq 1. \end{cases}$$
Letting $n \to \infty$ shows $$F_n(x) \to \begin{cases} 0, & x < 0, \\ x, & x \in [0,1] \\ 1 & x>1 \end{cases}$$ Since the right-hand side is the distribution function of a $U[0,1]$-distributed random variables, this finishes the proof.
To see that pointwise convergence does in general not hold consider the following sequence of random variables on $[0,1]$ (endowed with the Lebesgue measure)
$$X_{2n}(x) := \begin{cases} \frac{1}{n}, & x \in \big[0,\frac{1}{n} \big) \\ \frac{2}{n}, & x \in \big[ \frac{1}{n}, \frac{2}{n} \big) \\ \vdots & \\ 1, & x \in \big[ \frac{n-1}{n}, 1 \big]. \end{cases}$$
and
$$X_{2n+1}(x) := \begin{cases} 1, & x \in \big[0,\frac{1}{n} \big) \\ 1- \frac{1}{n}, & x \in \big[ \frac{1}{n}, \frac{2}{n} \big) \\ \vdots & \\ \frac{1}{n}, & x \in \big[ \frac{n-1}{n}, 1 \big). \end{cases} $$
Then $X_n \sim F_n$, but $(X_n)_{n \in \mathbb{N}}$ does not converge pointwise.