Weak convergence: equivalence of definitions Consider a sequence of random variables $(X_n)_{n\geq 0}$ and a random variable $X$. How to prove that the two following definitions of weak convergence are equivalent?
Def 1  $(X_n)_{n\geq 0} \overset{d}{\longrightarrow} X$ if for every continuous, bounded function $\varphi : \mathbb{R} \rightarrow \mathbb{R}$, $$\mathbb{E}[\varphi(X_n)] \overset{n\ \rightarrow \infty}{\longrightarrow} \mathbb{E}[\varphi(X)].$$
Def 2 $(X_n)_{n\geq 0} \overset{d}{\longrightarrow} X$ if for every bounded, Lipschitz-continuous function $\varphi : \mathbb{R} \rightarrow \mathbb{R}$, $$\mathbb{E}[\varphi(X_n)] \overset{n \rightarrow \infty}{\longrightarrow} \mathbb{E}[\varphi(X)].$$
 A: Assume that def 2. holds, then (exercise), for any $\epsilon>0$, their exist $n_\epsilon$ and $K_\epsilon>0$ such that $$\mathbb{P}(|X_n|> K_{\epsilon})\leq \epsilon,\ \forall n\geq n_\epsilon\text{ and }
\mathbb{P}(|X|> K_{\epsilon})\leq \epsilon.$$
Let $\varphi$ be a bounded continuous function on $\mathbb{R}$. For any $\epsilon>0$, their exists a Lipschitz function $\varphi_\epsilon$ such that
$$
\sup_{|x|\leq K_\epsilon} |\varphi(x)-\varphi_{\epsilon}(x)|\leq \epsilon.
$$
As a consequence, for all $n\geq n_\epsilon$,
\begin{align*}
|\mathbb{E}(\varphi(X_n))-\mathbb{E}(\varphi(X))|&\leq |\mathbb{E}(\varphi(X_n)\mathbf{1}_{|X_n|> K_\epsilon})-\mathbb{E}(\varphi(X)\mathbf{1}_{|X|> K_\epsilon})|\\
&\hspace{1cm} + |\mathbb{E}(\varphi(X_n)\mathbf{1}_{|X|\leq K_\epsilon})-\mathbb{E}(\varphi(X)\mathbf{1}_{|X_n|\leq K_\epsilon})|\\
&\leq 2\epsilon\|\varphi\|_{\infty}+2\epsilon + |\mathbb{E}(\varphi_\epsilon(X_n))-\mathbb{E}(\varphi_\epsilon(X)).|
\end{align*}
Then, if weak convergence as in def 2. holds, we easily deduce that the left hand side converges to $0$.
This being true for any continuous function $\varphi$, it implies weak convergence as in def 1.
