Composing two discontinuous functions

A simple counterexample shows that $f\circ{g}$ is not necessarily discontinuous.

Let $f={\textbf{1}}_{\mathbb{Q}}=g$ , where ${\textbf{1}}_{\mathbb{Q}}$ is the indicator function for the rational numbers , defined by: ${\textbf{1}}_{\mathbb{Q}}(x)=\begin{cases} 1,&x\in\mathbb{Q} \\ 0,&x\not\in\mathbb{Q} \\ \end{cases}$ It is well-known that ${\textbf{1}}_{\mathbb{Q}}(x)$ is discontinuous at all $x$ , but $\left({\textbf{1}}_{\mathbb{Q}}\circ{\textbf{1}}_{\mathbb{Q}}\right)(x)$ is clearly the function that outputs $1$ for all $x$ , and thus the composition is continuous.