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I'm reading the weak convergence section (Ch.$18$) in "Probability Essentials" by Protter and Jacod. Theorem $18.7$ states that

"$Xn \overset{D}{\to} X \iff \underset{n\to\infty}{\lim} E[g(X_n)] = E[g(X)]$ for all bounded Lipschitz continuous functions $g$."

But the corollary throws me off a little. It states:

"$Xn \overset{D}{\to} X \iff \underset{n\to\infty}{\lim} E[g(X_n)] = E[g(X)]$ for all bounded $\textbf{uniformly continuous}$ functions $g$."

I understand that Lipschitz continuity implies uniform continuity, but I thought the other way is not true. Why is it still true that $\underset{n\to\infty}{\lim} E[g(X_n)] = E[g(X)]$ for all bounded uniformly continuous functions $g$ $\implies Xn \overset{D}{\to} X$?

user207886
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In general, uniform continuity does not imply Lipschitz continuity, but it is possible to approximate uniformly continuous functions by Lipschitz-continuous functions (and in this business, it is often enough to have "good" approximations).

So let $g$ be bounded and uniformly continuous. For each $k \in \mathbb{N}$ we choose a cutoff-function $\chi_k \in C_c(\mathbb{R})$ such that $1_{B(0,k)} \leq \chi_k \leq 1_{B(0,2k)}$. Since $\chi_k \cdot g$ is a continuous function with compact support, there exists a Lipschitz-continuous function $\psi_k$ such that

$$\|\psi_k-\chi_k \cdot g\|_{\infty} \leq \frac{1}{k},$$

see e.g. this question or this question. Now $$\begin{align*} |\mathbb{E}g(X_n)-\mathbb{E}g(X)| &\leq |\mathbb{E}(g(X_n)-g \chi_k(X_n))| + |\mathbb{E}(g \chi_k(X_n)-\psi_k(X_n))| + |\mathbb{E}(\psi_k(X_n)-\psi_k(X))| \\ &+ |\mathbb{E}(\psi(X)- g \chi(X))| + |\mathbb{E}(\psi(X)-g\chi_k(X))| + |\mathbb{E}(g \chi_k(X)-g(X))| \\ &\leq \frac{2}{k} + |\mathbb{E}(\psi_k(X_n)-\psi_k(X))| + |\mathbb{E}(g(X_n)-g \chi_k(X_n)) + |\mathbb{E}(g \chi_k(X)-g(X))| \\ &\leq \frac{2}{k} + |\mathbb{E}(\psi_k(X_n)-\psi_k(X))| + \|g\|_{\infty} \left(\sup_{n \in \mathbb{N}} \mathbb{P}(|X_n| \geq k) + \mathbb{P}(|X| \geq k) \right). \end{align*}$$

Since $\psi_k$ is bounded and Lipschitz-continous, it follows from our assumption that

$$\limsup_{n \to \infty} |\mathbb{E}g(X_n)-\mathbb{E}g(X)| \leq \frac{2}{k} + \|g\|_{\infty} \left(\sup_{n \in \mathbb{N}} \mathbb{P}(|X_n| \geq k) + \mathbb{P}(|X| \geq k) \right).$$

Finally, we can let $k \to \infty$ and obtain

$$\limsup_{n \to \infty} |\mathbb{E}g(X_n)-\mathbb{E}g(X)| =0.$$

(In the last step, we have to use that the sequence $(X_n)_{n \in \mathbb{N}}$ is tight; this follows from the weak convergence $X_n \to X$.)

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