Continuity of $f(x)=x \cdot g(x)$ on $\mathbb{R}_{\geq 0}$ where $g(x)$ has a positive upper bound

Question

I am trying to prove continuity of $f(x)=x \cdot g(x)$ where $g(x)$ has the positive upper bound $\bar c$. This means that the upper bound of $g(x)$ does not depend on $x$. Furthermore, it is known that $x \geq 0$ and $g(x) \geq 0$.

I have tried different approaches (Lipschitz definition, espilon-delta definition) but I am not sure if $f(x)$ is continuous on $\mathbb{R}_{\geq 0}$.

Is this an obvious case of continuity/discontinuity?

I am glad if someone can give me a hint

Answer 1

Take $g \equiv \chi_\mathbb{Q}$, that is

$$ g(x) = \cases{1 \text{ if $x$ is rational} \\ 0 \text{ otherwise}} $$

Now, $f(x) = xg(x)$ is not continuous. Can you see why?

A proof follows,

Take a sequence of rational numbers $(q_n)_{n \geq 1}$ that converge to $\sqrt{2}$. Now, we have that $f(q_n) \equiv 0 \not \rightarrow \sqrt{2} = f(\sqrt{2})$, and so $f$ is not continuous.