This is exercise 4.3.8(c) from Abbott's Understanding Analysis 2nd ed. I'm attempting to prove this statement by contradiction. I found other solutions to this exercise which are brief, direct proofs (Functional Limits and Continuity). I first thought to try proof by contradiction and am checking if my solution is valid.
Notation From Understanding Analysis: $V_{\epsilon}(x)$ means $\epsilon$-neighborhood of $x$. $V_{\delta}(x)$ means $\delta$-neighborhood of $x$.
My Proof Attempt: Assume, for sake of contradiction, that $g(x)$ is not strictly positive for uncountably many points $x \in \mathbb{R}$. In other words, $g(x)$ is strictly positive for only countably (or finitely) many points.
Choose $\epsilon_0$ such that $0 < \epsilon_0 < g(x_0)$ (we can choose this $\epsilon_0$ due to the density of $\mathbb{Q}$ and $\mathbb{I}$ in $\mathbb{R}$). Since $g$ is continuous at $x_0$, there exists a $\delta > 0$ such that $x \in V_{\delta}(x_0)$ implies $g(x) \in V_{\epsilon_0}(g(x_0))$.
We know that $V_{\delta}(x_0)$ is uncountable because it is a nonempty open interval. Since, by assumption, there are only countably many points $x \in \mathbb{R}$ for which $g(x) > 0$, there must be uncountably many points $x \in V_{\delta}(x_0)$ for which $g(x) \leq 0$. Call the set of these uncountably many points $A$. But, for $x \in A \subseteq V_{\delta}(x_0)$,
$g(x) \notin V_{\epsilon_0}(g(x_0))$ due to our choice of $\epsilon_0$.
(For points $x$ such that $g(x) \in V_{\epsilon_0}(g(x_0))$, $0 < g(x_0) – \epsilon_0 < g(x) < g(x_0) + \epsilon$, but for $x \in A$, $g(x) \leq 0$ so $g(x)$ can't be in $V_{\epsilon_0}(g(x_0))$).
This contradicts the fact that $g$ is continuous at $x_0$. Therefore, our assumption is false and the original statement is true.
Best Answer
$g$ is continuous at $x_0$ and $g(x_0) >0$
Choose, $\epsilon =\frac{g(x_0)}{2}>0 $, then $\exists \delta>0$ such that $|x-x_0|<\delta$ implies $g(x) >\frac{g(x_0)}{2} >0$.
Hence, $g(x) >0 , \forall x\in (x-\delta, x+\delta) $
Since the interval $(x-\delta, x+\delta) $ is uncountable, $g(x) >0$ for uncountably many $ x$'s.