Pinko wrote: The non-decreasing functions $f,g: \mathbb{R}\rightarrow \mathbb{R}$ are such that $f(r)\leq g(r)$ for $\forall$ rational numbers $r$. Is it true that $f(x)\leq g(x)$ for $\forall$ real numbers $x$? No. Choose as counter-example : $f(x)=x\quad\forall x<\pi$ $f(\pi)=\pi+4$ $f(x)=x+5\quad\forall x>\pi$ $g(x)=x+1\quad\forall x<\pi$ $g(\pi)=\pi+3$ $g(x)=x+6\quad\forall x>\pi$

An even easier counterexample is $f(x) = 0$ for all $x<\sqrt{2}$, $f(x) = 1$ for all $x\geq \sqrt{2}$, with $g(x) = 0$ for all $x\leq \sqrt{2}$, $g(x) = 1$ for all $x>\sqrt{2}$. (The inequality $f(x) \leq g(x)$ is false only for $x=\sqrt{2}$).