# Weak convergence result in Levy's Continuity Theorem

I quote a part of Levy's Continuity Theorem and its proof.

Theorem
Let $$\left(\mu_n\right)_{n\geq1}$$ be a sequence of probability measures on $$\mathbb{R}^d$$, and let
$$\left(\hat{\mu}_n\right)_{n\geq1}$$ denote their Fourier transforms (aka characteristic functions). If $$\mu_n$$ converges weakly (that is, in distribution) to a probability measure $$\mu$$, then $$\hat{\mu}_n(u)\rightarrow\hat{\mu}(u)$$ for all $$u\in\mathbb{R}^d.$$

Proof
Suppose $$\mu_n$$ converges weakly to $$\mu$$. Since $$e^{iux}$$ is continuous and bounded in modulus, $$\hat{\mu}_n(u)={\displaystyle \int e^{iux}\mu_n(dx)}$$ converges to $$\hat{\mu}(u)={\displaystyle \int e^{iux}\mu(dx)}$$

My question is:
which is the result implicitly used so as to state that:

"Since $$f=e^{iux}$$ is continous and bounded in modulus $$\mu_n\xrightarrow{\mathcal{D}}\mu\Rightarrow\hat{\mu}_n(u)={\displaystyle \int e^{iux}\mu_n(dx)}\to\hat{\mu}(u)={\displaystyle \int e^{iux}\mu(dx)}\hspace{0.5cm}\text{"}\,?$$

The Portmanteau theorem says, among other things, that $$\mu_n \to \mu \mathrm{\ weakly}$$ iff $$\forall f \in C_b: \int fd \mu_n \to \int fd \mu$$
What you are asking for uses this with your particular choice for $$f$$.
• So, by Portmanteau NOT ONLY $\mu_n\xrightarrow{\mathcal{D}}\mu\Rightarrow\hat{\mu}_n(u)={\displaystyle \int e^{iux}\mu_n(dx)}\to\hat{\mu}(u)={\displaystyle \int e^{iux}\mu(dx)}$ BUT ALSO $\hat{\mu}_n(u)={\displaystyle \int e^{iux}\mu_n(dx)}\to\hat{\mu}(u)={\displaystyle \int e^{iux}\mu(dx)}\Rightarrow\mu_n\xrightarrow{\mathcal{D}}\mu$? Henceforth, are they equivalent? However, a doubt still remains, since here we start from convergence of probability measures and end up with convergence of a function of $x$, not of a function of the probability measures @YashaswiMohanty May 30 '20 at 10:39