Convergence in probability implies convergence of characteristic functions

Question

If $X_n$ converge in probability to $X$, then they also converge weakly, and we can conclude by the continuity theorem that the characteristic functions $\phi_n$ of $X_n$ converge pointwise to the characteristic function $\phi$ of $X$. How can we reach the same conclusion without appealing to the continuity theorem?

Answer 1

If we had almost sure convergence, this would follow from the dominated convergence theorem ($|e^{i \lambda X_n}| \le 1).$

Assume the contrary. If $\phi_n(\lambda) \not\to \phi(\lambda)$, there is a subsequence $n_k$ so that $\phi_{n_k}(\lambda)$ stays distance $\epsilon$ away from $\phi(\lambda)$ for all $k$.

But $X_{n_k}$ still converges to $X$ in probability. By Borel-Cantelli it has a subsequence of its own, which converges to $X$ almost surely. Use dominated convergence to get a contradiction.

This is sometimes called the "double subsequence trick". We can also use it to prove a strengthening of the dominated convergence theorem: suppose $Y_n \to Y$ in probability, and there is an integrable random variable $Z$ with $|Y_n| \le Z$ a.s. for every $n$. Then $E Y_n \to EY$.

Answer 2

An alternative way of seeing this is as follows (solely for my own recording purposes). Let $Y_n(\theta)=\exp(e^{i \theta X_n})$. Note that, $|Y_n(\theta)|\leq 1$, and thus for every fixed $\theta$, the sequence $\{Y_n(\theta)\}_{n\geq 1}$ is uniformly integrable. This, together with $Y_n \to Y$ (using continuous mapping thm) yield the $Y_n\to Y$ in $L^1$ (using a result given in Williams' book).