Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 201 / 3

Prokhorov's theorem says that a sequence of Borel probability measures on $\mathbb{R}$ is relatively compact for weak convergence of probability measures exactly when it is tight: for every $ϵ>0$, some compact $K$ satisfies ${\sup }_n{\mu }_n(\mathbb{R}\setminus K)<ϵ$.

Put $z=y/\lambda$. Direct integration givesFor $|z|\ge 1$, the right side has a positive infimum $c_0$ because $\sin z/z<1$ there and it tends to one at infinity. Thus the claimed inequality holds with $C=c_0^{-1}$.

The masses ${\mu }_n(\mathbb{R})={\psi }_n(0)$ converge and are therefore bounded. Part b and Tonelli theorem giveThe integrands are uniformly bounded. By pointwise convergence and the continuity of $\psi$ at zero, the right side can be made uniformly small for all sufficiently large $n$ by taking $\lambda$ large; finitely many remaining measures are individually tight. Thus $({\mu }_n)$ is tight. Applying Prokhorov's theorem after normalizing the masses, or adjoining missing mass at one fixed point, gives a weakly convergent subsequence.

Yes. If a subsequence converges weakly to $\nu$, bounded continuity of $e^{iuy}$ givesThe uniqueness theorem for characteristic functions makes $\nu$ unique. Tightness implies that every subsequence has a further weakly convergent subsequence, and every such limit is $\nu$. This subsequence criterion proves that the entire sequence converges weakly to $\nu$.