weak convergence of dirac measure $\delta_{\frac 1 n}$ On wikipedia : https://en.wikipedia.org/wiki/Convergence_of_measures#Weak_convergence_of_measures it is written 

For example, the sequence where $P_n$ is the Dirac measure located at
  $\frac 1 n $ converges weakly to the Dirac measure located at $0$ (if
  we view these as measures on  $\mathbb {R}$ with the usual topology),
  but it does not converge strongly. This is intuitively clear: we only
  know that  $\frac 1 n $ is "close" to $0$ because of the topology of
  $\mathbb {R}$.

I don't understand that example. I am just starting that whole subject and I struggle to understand the nuances.
Can someone use that example to distinguish strong and weak convergence ?

My attempt leads to strong convergence of the dirac measure. Using the same definition as the ones on the wiki page :
$$\int \phi \ d \delta_{\frac 1 n} \rightarrow \int \phi \ d \delta_{0}$$
the same way that 
$$\lim \delta_{\frac 1 n} (A)= \delta_{0} (A)$$
as well as :
$$|| \delta_{\frac 1 n} - \delta_{0} ||_{TV} = 0$$
My reasoning is simply taking the limit and trivially 1/n converges to $0$... what am I doing wrong ? What am I thinking through wrongly ?
Please be gentle, I m not used to that field of maths, I really don't know what I should be careful about.
 A: The statements follow from the definitions of different convergence properties.
For weak convergence, we need to check 
$$\lim_{n\to \infty}\int_{\mathbb{R}} \phi \ d \delta_{\frac 1 n} =\int_{\mathbb{R}} \phi \ d \delta_{0},$$
for any given bounded continuous function $\phi$ on $\mathbb{R}$. This is true since by definition of Dirac measure and continuity of $\phi$, we have
$$\lim_{n\to \infty}\int_{\mathbb{R}} \phi \ d \delta_{\frac 1 n} =\lim_{n\to \infty} \phi({\frac 1 n})= \phi(0) =\int_{\mathbb{R}} \phi \ d \delta_{0}.$$
For strong convergence, we need to check $$\lim_{n\to\infty} \delta_{\frac 1 n} (A)= \delta_{0} (A),$$
for any given Borel measurable set $A\subset \mathbb{R}$. This is false since if we take $A=(0,1)$, then 
$ \delta_{\frac 1 n} (A)=1$ for all $n$ but $\delta_{0} (A)=0$.
Convergence in TV norm is even stronger than the strong convergence, since we need to check $$\lim_{n\to\infty}\sup_{A} |\delta_{\frac 1 n} (A)-\delta_{0} (A)|=0,$$
where the supremum is taken over all Borel measurable sets $A\subset \mathbb{R}$. The above counterexample shows that it is false.
