Convergence in Distribution implies Convergence in Expectation uniformly for an equicontinuous family If $X_n \xrightarrow{\text{d}} X$, then for any equicontinuous family $\{f_\theta:\theta\in \Theta\}$, satisfying sup$\{|f_\theta(x)|:x\in \mathbf{R},\theta\in \Theta\}\}<\infty$, prove that $E[f_\theta(X_n)]\rightarrow E[f_\theta(X)]$ uniformly in $\theta\in \Theta$.
Now I know if $X_n \xrightarrow{\text{d}} X$, $E[f_\theta(X_n)]\rightarrow E[f_\theta(X)]\hspace{1ex}\forall \theta\in\Theta$. And I sense I have to use Arzela-Ascoli Theorem, but I don't have compact set as domain. Any help?
 A: Arzelà-Ascoli is a good idea; in order to apply it we need to truncate the functions $f_{\theta}$ in a suitable way.
Step 1: Find suitable level of truncation Fix $\epsilon>0$. By the monotone convergence theorem, there exists $R>0$ such that
$$\mathbb{P}(|X| \geq R) \leq \epsilon; \tag{1}$$
without loss of generality, we may assume that $\mathbb{P}(|X|=R)=0$. Since $X_n \to X$ weakly we have
$$\mathbb{P}(|X_n| \geq R) \to \mathbb{P}(|X| \geq R) \leq \epsilon,$$
and therefore we can choose $N \in \mathbb{N}$ such that
$$\mathbb{P}(|X_n| \geq R) \leq 2 \epsilon \quad \text{for all $n \geq N$.}$$
We can enlarge $R>0$ in such a way that
$$\mathbb{P}(|X_j| \geq R) \leq 2 \epsilon \quad \text{or all $j=1,\ldots,N-1$}$$
(note that we only have finitely many $j$'s here!), and therefore we get
$$\mathbb{P}(|X_n| \geq R) \leq 2 \epsilon \quad \text{for all $n \geq 1$.} \tag{2}$$
Step 2: Apply Arzelà-Ascoli For fixed $\epsilon>0$ let $R>0$ be as in Step 1 (i.e. such that $(1)$ and $(2)$ hold). Choose a function $\chi \in C_b(\mathbb{R})$, $0 \leq \chi \leq 1$ such that $\chi(x)=1$ for $|x| \leq R$ and $\chi(x)=$ for $|x| \geq R+1$. The family $$\{\chi \cdot f_{\theta}; \theta \in \Theta\}$$ is uniformly bounded and equicontinuous. Moreover, the support of each function $g_{\theta} := \chi \cdot f_{\theta}$ is contained in the closed ball of radius $R+1$. By the Arzelà-Ascoli theorem, this implies that the above family is compact in $C_b(\mathbb{R})$. In particular, we can choose $\theta_1,\ldots,\theta_m \in \Theta$ such that $$\{g_{\theta}; \theta \in \Theta\} \subseteq \bigcup_{j=1}^m B(g_{\theta_j},\epsilon). \tag{3}$$ For each $j=1,\ldots,m$ there exists $N_j$ such that $$|\mathbb{E}g_{\theta_j}(X_n)-\mathbb{E}g_{\theta_j}(X)| \leq \epsilon \quad \text{for all $n \geq N_j$};$$ set $N := \max\{N_1,\ldots,N_m\}$.
Step 3: Prove uniform convergence Fix $\theta \in \Theta$. Clearly, we have
$$|\mathbb{E}f_{\theta}(X_n)-\mathbb{E}f_{\theta}(X)| \leq I_1+I_2$$
where
$$\begin{align*} I_1 &:= |\mathbb{E}(f_{\theta}(X_n) \chi(X_n))- \mathbb{E}(f_{\theta}(X) \chi(X))| \\ I_2 &:=|\mathbb{E}(f_{\theta}(X_n) (1-\chi(X_n)))- \mathbb{E}(f_{\theta}(X) (1-\chi(X)))| \end{align*}$$
We estimate the terms separately.
Estimate of $I_1$: According to $(3)$ there exists $j \in \{1,\ldots,m\}$ such that $\|g_{\theta}-g_{\theta_j}\|_{\infty} \leq \epsilon$. Hence, $$I_1 = |\mathbb{E}g_{\theta}(X_n)-\mathbb{E}g_{\theta}(X)| \leq 2 \|g_{\theta}-g_{\theta_j}\|_{\infty} + |\mathbb{E}g_{\theta_j}(X_n)-\mathbb{E}g_{\theta_j}(X)| \leq 3 \epsilon \tag{4}$$ for all $n \geq N$.
Estimate of $I_2$: Since $\chi(x)=0$ for all $|x| \leq R$ and $0 \leq \chi \leq 1$, we clearly have
$$I_2 \leq \|f_{\theta}\|_{\infty} \left( \mathbb{P}(|X_n| \geq R) + \mathbb{P}(|X| \geq R) \right).$$
It follows from $(1)$ and $(2)$ that
$$I_2 \leq3  \|f_{\theta}\|_{\infty} \epsilon.$$
If we set $M:= \sup_{\theta \in \Theta} \|f_{\theta}\|_{\infty}$ (which is finite by assumption) then $$I_2 \leq 3 M \epsilon.$$
Combining our estimates, we find that
$$|\mathbb{E}f_{\theta}(X_n)-\mathbb{E}f_{\theta}(X)| \leq 3 (M+1) \epsilon$$
for all $n \geq N$. Since $\epsilon>0$ is arbitrary and $N$ does not depend on $\theta$, this proves the uniform convergence with respect to $\theta \in \Theta$: $$\sup_{\theta \in \Theta} |\mathbb{E}f_{\theta}(X_n)-\mathbb{E}f_{\theta}(X)| \xrightarrow[]{n \to \infty} 0.$$
Remark: Note that the result implies in particular the following statement.

Theorem  If $X_n \to X$ in distribution then $$\sup_{\xi \in K} |\mathbb{E}\exp(i \xi X_n)-\mathbb{E}\exp(i \xi X)| \xrightarrow[]{n \to \infty} 0$$ for any compact set $K$, i.e. the characteristic functions converge uniformly on compact sets.

