 Solution

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.