Why is a characteristic function continuous at $0$?

Question

My lecture notes say:

$t \mapsto \exp(-t^2/2)$ is a characteristic function (of $\mathcal{N}(0,1)$), so it is clear that it is continuous at $0$.

So why does "being a characteristic function" imply "being continuous at $0$".

I think it should be something very obvious, but I do not get it.

Answer 1

The characteristic function of any random variable is continuous. Say $X$ is a random variable and $t_n\to t$. Then $$\Bbb E[e^{it_n X}]\to\Bbb E[e^{it X}]$$by the Dominated Convergence Theorem.