On the Wikipedia page for 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 $\frac{1}{n}$ converges to $0$. What am I doing wrong ? What am I thinking through wrongly?
Please be gentle, I am not used to that field of maths, I really don't know what I should be careful about.
Best Answer
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.