TITLE: Weakly convergent distributions on $\mathbb{R}$ with densities relating to Lebesgue-measure that do not converge
QUESTION [5 upvotes]: Show that there exist weakly convergent distributions on $\mathbb{R}$ that have a density relating to the Lebesgue measure $\lambda$ but the densities do not converge. Hint: $f_n(x):=1+\cos(2\pi n x)$


To be honest I do not understand this task completely.
Nevertheless I tried to use the given hint. Maybe
$$
\nu_n(x):=f_n(x)\lambda(x)
$$
are those distributions on $\mathbb{R}$? At least $\nu_n\ll\lambda$ and both $\lambda$ and $\nu_n$ are $\sigma$-finite so that from Radon-Nikodým it follows that there is $\lambda$-a.s. a density of $\nu_n$ relating to $\lambda$. But I do not know if this is meant and especially if the $\nu_n$ do weakly converge...
Could you please help me?
Sincerely yours
math12

REPLY [2 votes]: You are on the right track.
Your measures $\nu_n$ define distributions by the rule
$$\nu_n(g)=\int_\mathbb{R} g(x) d\nu_n(x)=\int_\mathbb{R} f_n(x)g(x) d\lambda(x)$$
for $g\in C_0^\infty(\mathbb{R})$.
Now your job is to notice that the $\nu_n$ converge distributionally, i.e. there is a distribution $\nu$ s.t.
$$\nu_n(g)\rightarrow \nu(g)$$
for all test functions $g$, but on the other hand, the sequence of functions $(f_n)_n$ does not converge (say, pointwise).
Hint: Use the Riemann-Lebesgue lemma.