Let $I$ be an open interval and $ f:I \longrightarrow \mathbb{R}$ an increasing function, and define the functions $f^+$ (resp. $f^-$) by $f^+(x):=f(x^+)$ (resp. $f^-(x):=f(x^-)$), for all $x\in I$ (they are well-defined, by the properties of monotone functions). Then, putting $g=f^-$ and $h=f^+$ we have $$ x<y \Rightarrow g(x)\leq h(x)\leq g(y). $$
Now as the converse: is it true that if $g,h$ satisfy the above inequality, then there is an increasing function $f$ such that $g=f^-$ and $h=f^+$?
Thanks in advance
No. Counterexample: Let $g,h :[-1, 1] \to \mathbb R$ be defined by
$$ g(x) = \begin{cases} 0 & \text{ if } x < 0,\\ 1/3 & \text{ if } x = 0,\\ 1 & \text{ if } x > 0\end{cases}, \ \ \ h(x) = \begin{cases} 0 & \text{ if } x < 0,\\ 1/2 & \text{ if } x=0, \\ 1 & \text{ if } x> 0\end{cases}$$
Then $x<y$ implies $g(x) \le h(x) \le g(y)$. But there isn't such an $f$: if $f$ exists, then $g\le f\le h$ by construction, which implies
$$ f(x) = \begin{cases} 0 & \text{ if } x < 0,\\ 1 & \text{ if } x > 0.\end{cases}$$
This already implies $$ f^- (x) = \begin{cases} 0 & \text{ if } x \le 0,\\ 1 & \text{ if } x > 0\end{cases}, \ \ \ f^+ (x) = \begin{cases} 0 & \text{ if } x < 0,\\ 1 & \text{ if } x \ge 0\end{cases}$$ and thus $g\neq f^-$, $h\neq f^+$.