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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 $$\mu_n\xrightarrow{\mathcal{D}}\mu\Rightarrow\hat{\mu}_n(u)={\displaystyle \int e^{iux}\mu_n(dx)}\to\hat{\mu}(u)={\displaystyle \int e^{iux}\mu(dx)}\hspace{0.5cm}\text{"}\,?$$

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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 $f$.

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  • $\begingroup$ Thank you a lot. Could you please give me a reference as to the result you have mentioned in your answer? $\endgroup$ May 30 '20 at 11:23
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    $\begingroup$ Sure, check the book "Real analysis and probability" by Dudley, section 11.1 p 385 for a treatment of this result where the domain is a Polish space (separable completely metrizable space). $\endgroup$ May 30 '20 at 11:27
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The result is an application of the Portmanteau Theorem.

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  • $\begingroup$ 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 $\endgroup$ May 30 '20 at 10:39

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