Let $f: \mathbb{R} \to \{0, 1\}$ be a non-decreasing, right-continuous function. Then $\sup\{x: f(x) = 0\} = c$, where for all $x \geq c$, $f(x) = 1$.

Question

Let $f: \mathbb{R} \to \{0, 1\}$ be a non-decreasing, right-continuous function with $\lim_{x \to -\infty}f(x) = 0$ and $\lim_{x \to +\infty}f(x) = 1$. Then $\sup\{x: f(x) = 0\}$ is equal to the value $c$ such that for all $x \geq c$, $f(x) = 1$.

We observe that $S = \{x: f(x) = 0\}$ is bounded from above by a value $c \in \mathbb{R}$ such that for all $x \geq c$, $f(x) = 1$ ($c$ exists due to the limit at $+\infty$ and right continuity of $f$). Thus $s:= \sup S$ exists. Of course, $s \leq c$ by definition.

Could someone give me a hint on how to show $s \geq c$?

Answer 1

Just define $c$ to the supremum of $S$. If $x>c$ then $x \notin S$ so $f(x)=1$. By right continuity it follows that $f(x)=1$ for $x=c$ also.