Increasing Function and Convex Set Question

Question

Consider a function $0 \le f(x) \le 1$ which is increasing in $x \in [a,b]$, I was wondering can I say that $f(x) \le \epsilon$ for $0< \epsilon <1$ defines a convex set?

I think the answer should be yes but not sure.

Answer 1

Let $x_0 = \sup \{x \in [a, b] : f(x) \leq \epsilon\}$, assuming the latter set is nonempty. Since $f$ is increasing, $f(x) \leq \epsilon$ for every $x \in [a, x_0)$. It follows that $\{f(x) \leq \epsilon\}$ is either $[a, x_0)$ or $[a, x_0]$, both of which are convex. If the above set is empty, and so there is no such $x_0$, then $\{f(x) \leq \epsilon\} = \emptyset$, and I can't recall whether we call this set convex or not.