Example of weak convergence I'd like to understand better the concept of weak convergence.
I know that a sequence of probability measures $\mu_n$ converges weakly to $\mu$ if $\int{f d\mu_n}$ converges to $\int{f d\mu}$ for each $f$ which is continuous and bounded. 
Could you please give me an example of a sequence of probability measures $\mu_n$ that converges weakly to $\mu$ and find a function $f$ such that $\int{f d\mu_n}$ does not converge to $\int{f d\mu}$?
 A: Let $P(X_n=1/n)=1$, let $P(X=0)=1$.  The probability distribution of $X_n$ is $\mu_n=\delta_{1/n}$,  the point-mass measure concentrated at $1/n$, that of $X$ is the point mass at $0$, namely $\mu=\delta_0$.
We can check that $\mu_n$ converges to $\mu$ weakly: since $\int f d\mu_n = f(1/n)$ and $\int f d\mu=f(0)$, for continuous and bounded $f$ the desired limit holds: $\lim_{n\to\infty} f(1/n)=f(0)$.
But for $f=\chi_{\{0\}}$, say, the indicator function of the singleton set $\{0\}$, we have $\int f d\mu_n = 0$ and $\int f d\mu = 1$, so the desired limit does not hold.  This $f$ is bounded but not continuous.
A: Take on $[0,1]$ the measures $\mu_n=\frac{1}{n}\sum_{k=1}^n\delta_{k/n}$.
They converge weakly to the Lebesgue measure $\mu=\lambda^1$ restricted to $[0,1]\,$, because the Riemann sum $\mu_n(f)=\frac{1}{n}\sum_{k=1}^n f(\frac{k}{n})$ converges to its integral $\mu(f)=\int_{[0,1]}f(x)\,dx$ for every bounded continuous function $f$.
Let's break the continuous condition, but be sure not to take a Lipschitz function (see Portmanteau theorem https://en.wikipedia.org/wiki/Convergence_of_measures)
Let $f$ be the Dirichlet function $f=\chi_{\mathbb{Q}\cap[0,1]}$.
You have $1=\mu_n(f)\neq\mu(f)=0\quad\forall n\in\mathbb{N}$
