I want to solve the following exercise in Probability and Measure, Billingsly [1994]
According to Example 25.8, if $X_n \Rightarrow X$, $a_n \rightarrow a$ and $b_n \rightarrow b$, then $a_nX_n+b_n \Rightarrow aX+b$. Prove this by means of characteristic functions.
I believe we should use the continuity theorem for charateristic functions. Using rules for these, we can WLOG assume $b_n=0$ and just worry about $a_nX_n$. Then we need to show that $$ \varphi_{a_nX_n}(t) \rightarrow \varphi_{aX}(t) $$ for all $t \in \mathbb{R}$. Taking such a $t$, we slide in a "mixed term" have $$ \vert \varphi_{a_nX_n}(t) - \varphi_{aX}(t) \vert \leq \vert \varphi_{a_nX_n}(t) - \varphi_{aX_n}(t) \vert + \vert \varphi_{aX_n}(t) - \varphi_{aX}(t) \vert = \vert \varphi_{X_n}(a_nt) - \varphi_{X_n}(at) \vert + \vert \varphi_{X_n}(at) - \varphi_{X}(at) \vert, $$ where the the second term vanishes by the continuity theorem. However the first term is a bit more tricky as we have a "double" dependence on $n$.
Can anyone help me see how and why this term vanishes?
You are on the right track by using the continuity theorem for characteristic functions. To show that the first term vanishes, you can use the fact that $a_n \rightarrow a$ and the uniform continuity of $\varphi_{X_n}$.
Since $X_n \Rightarrow X$, we have that $\varphi_{X_n}(t) \rightarrow \varphi_X(t)$ for all $t \in \mathbb{R}$. By Lévy’s continuity theorem, this implies that $\varphi_X$ is continuous. Since $\mathbb{R}$ is a metric space, this implies that $\varphi_X$ is uniformly continuous.
Now, let $\epsilon > 0$. Since $\varphi_X$ is uniformly continuous, there exists a $\delta > 0$ such that for all $s,t \in \mathbb{R}$ with $|s-t| < \delta$, we have $|\varphi_X(s) - \varphi_X(t)| < \epsilon/2$. Since $a_n \rightarrow a$, there exists an $N$ such that for all $n > N$, we have $|a_n - a| < \delta/|t|$ (assuming $t\neq 0$, otherwise both terms are equal to 1). Then for all $n > N$, we have $$|\varphi_{X_n}(a_nt) - \varphi_{X_n}(at)| = |\varphi_{X}(a_nt) - \varphi_{X}(at)| <\epsilon/2.$$ Thus, the first term vanishes as desired.