Let $f \colon \mathbb{R} \rightarrow \mathbb{R}$ be a simple function, that is, $f (x) = \sum_{i = 1}^{n} a_i \mathbb{1}_{A_i}(x)$ for all $x \in \mathbb{R}$, where each $a_i$ is a constant, each $A_i$ is a measurable set and $\mathbb{1}_{A_i}$ denotes the indicator function of the set $A_i$. If each $A_i$ is an interval, then $f$ is called a step function. While every step function is clearly a simple function, the reverse is not true. For instance, $f(x) = \mathbb{1}_{\mathbb{Q}}(x)$ is a simple function but not a step function ($\mathbb{Q}$ being the set of rational numbers).
Now, suppose that $f$ is a non-decreasing simple function, that is, $x \leq y$ implies $f(x) \leq f(y)$. Is $f$ necessarily a step function in this case? (I can't find a counterexample.)
Let $f$ be a simple function, so that its image is a finite set $\{a_1, \ldots, a_n\}$.
If $f$ is non-decreasing, then each set $I_j := f^{-1}(a_j)$ is connected, hence it is an interval. Namely, if $x_1, x_2 \in I_j$, with $x_1 < x_2$, then $$ a_j = f(x_1) \leq f(x) \leq f(x_2) = a_j, \qquad \forall x\in [x_1, x_2], $$ i.e. $[x_1, x_2]\subset I_j$.