Yes. If a subsequence converges weakly to $\nu$, bounded continuity of $e^{iuy}$ gives
$$
\widehat\nu(u)=\lim_k\psi_{n(k)}(u)=\psi(u).
$$
The <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 $\nu$.

Solution

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

Two finite Borel measures on $\mathbb R$ with the same characteristic function are equal. One proof convolves both measures with centered Gaussians, applies Fourier inversion to the smooth densities, and then lets the Gaussian variance tend to zero.


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 Incoming links: Uniqueness theorem for characteristic functions