On the convergence in distribution of some random variables In this paper, they mention an equivalent result of the convergence in distribution of the random variable in page $11$.
I don't understand why the convergence in law of $Arg(R_n)$ to the uniform distribution of $[0,2\pi)$ is equivalent to $\lim_{n \to \infty} \mathbb{E}(f(Arg(R_n))=\int_{0}^{2\pi} f(x) dx$
for all continuous function with $f(0)=f(2\pi)$.
Why do we only consider functions with $f(0)=f(2\pi)$ ? I guess we should apply portmanteau lemma but I don't know how...
Thanks in advance.
 A: First of all, there is a mistake here, they must mean
$$\lim_{n \to \infty} \mathbb{E}(f(\operatorname{Arg}(R_n))=\frac{1}{2\pi}\int_{0}^{2\pi} f(x) dx$$
(plug in $f\equiv 1$ to see this). Secondly, what they mean to show here is much stronger than just weak convergence: indeed, from the above condition they wish to deduce that
$$\lim _{n \rightarrow \infty} \mathbb{P}\left(\operatorname{Arg}\left(R_{n}\right) \in[a, b]\right)=\frac{b-a}{2 \pi},$$
for $0 \leq a \leq b < 2\pi$. Let such $a,b$ be given and define
$$
f_k(x) := (1 - k\cdot d_{\mathbb{T}}(x,[a,b]))^{+},
$$
where $d_{\mathbb{T}}$ is the distance on the torus and $y^{+}$ denotes $\max(y,0)$. Note that $f_k$ is continuous and $2\pi$-periodic and therefore in particular satisfies $f_k(0) = f_k(2\pi)$. Furthermore, $f_k (x) = 1$ for $x \in [a,b]$ and $f_k \geq 0$. Hence $\mathbb{P}(\operatorname{Arg}(R_n) \in [a,b]) \leq \mathbb{E}[f_k(\operatorname{Arg}(R_n)]$ for every $k \in \mathbb{N}$. Furthermore, note that
$$\frac{1}{2\pi}\int_{0}^{2\pi}f_{k}(x)dx = \frac{b-a}{2\pi} + \frac{1}{2\pi k}.$$
Similarly, letting
$$
g_k(x) := (1 - k\cdot d_{\mathbb{T}}(x,[a+1/k,b-1/k]))^{+},
$$
we have $g_k \leq 1$ and $g_{k}(x) = 0$ for $x \in [0,2\pi)\setminus[a,b]$,
so $\mathbb{P}(\operatorname{Arg}(R_n) \in [a,b]) \geq \mathbb{E}[g_k(\operatorname{Arg}(R_n)]$ for every $k \in \mathbb{N}$. Furthermore,
$$\frac{1}{2\pi}\int_{0}^{2\pi}g_{k}(x)dx = \frac{b-a}{2\pi} - \frac{1}{2\pi k}.$$
It follows that
\begin{align*} \frac{b-a}{2\pi} - \frac{1}{2\pi k} &= \lim_{n\to\infty} \mathbb{E}[g_k(\operatorname{Arg}(R_n)] \leq \liminf_{n\to\infty} \mathbb{P}(\operatorname{Arg}(R_n) \in [a,b]) \leq \limsup_{n\to\infty} \mathbb{P}(\operatorname{Arg}(R_n) \in [a,b]) \\
&\leq \lim_{n\to\infty} \mathbb{E}[f_k(\operatorname{Arg}(R_n)]
= \frac{b-a}{2\pi} + \frac{1}{2\pi k}.
\end{align*}
Letting $k \to \infty$ proves the claim.
For the converse proof, note that $f$ is continuous on $[0,2\pi]$, and hence uniformly continuous. Let $\epsilon > 0$ be given and let $\delta$ be such that $|f(x)-f(y)| < \epsilon $ if only $|x-y| < \delta$. Let $x_k := k\delta$ for $k \in \mathbb{N}$ and define
\begin{align}
f_{\epsilon}(x) := \sum_{k=0}^{\lfloor 2\pi/\delta\rfloor} 1_{[x_k,x_{k+1})}(x) f(x_k).
\end{align}
Since $\sup_{x \in [0,2\pi)} |f(x) - f_{\epsilon}(x)| < \epsilon$ by construction, we have
$$
|\mathbb{E}(f(\operatorname{Arg}(R_n)) - \mathbb{E}(f_k(\operatorname{Arg}(R_n))| < \epsilon, \quad n \in \mathbb{N}.
$$
Furthermore,
\begin{align}
\lim_{n\to\infty}\mathbb{E}(f_{\epsilon}(\operatorname{Arg}(R_n)) &=\lim_{n\to\infty} \sum_{k=0}^{\lfloor 2\pi/\delta\rfloor}\mathbb{P}(\operatorname{Arg}(R_n) \in [x_{k},x_{k+1})) f(x_k)\\
 &= \frac{1}{2\pi}\sum_{k=0}^{\lfloor 2\pi/\delta\rfloor}f(x_k)(x_{k+1}-x_k),
\end{align}
Recognizing the right-hand side as a Riemann sum, we obtain
$$\lim_{\epsilon\to 0}\lim_{n\to\infty}\mathbb{E}(f_{\epsilon}(\operatorname{Arg}(R_n)) = \frac{1}{2\pi}\int_{0}^{2\pi} f(x)dx.$$
Since the convergence of $\mathbb{E}(f_{\epsilon}(\operatorname{Arg}(R_n))$
to $\mathbb{E}(f(\operatorname{Arg}(R_n))$ as $\epsilon \to 0$ is uniform in $n$, we may switch the limits and obtain
\begin{align}
\lim_{n\to\infty}\mathbb{E}(f(\operatorname{Arg}(R_n)) =\lim_{n\to\infty} \lim_{\epsilon\to 0}\mathbb{E}(f_{\epsilon}(\operatorname{Arg}(R_n)) = \lim_{\epsilon\to 0} \lim_{n\to\infty}\mathbb{E}(f_{\epsilon}(\operatorname{Arg}(R_n)) = \frac{1}{2\pi}\int_{0}^{2\pi} f(x)dx.
\end{align}
