Is any algebraic operation between two piecewise constant functions guaranteed to return a piecewise constant function?

Question

Let there be two piece-wise constant functions $f(x)$ and $g(x)$.

Is the function composition $h(f(x),g(x))$ always going to be a piece-wise constant function? I believe it is. Nothing would indicate otherwise, in my opinion, but I cannot be sure.

Answer 1

I will assume that your definition of a piecewise constant function is a function whose image is a finite or countable set. It can be shown that every such function can be represented in the form (called a simple function, set $m$ to $\infty$ for countable image.) $$ f = \sum_{j = 0}^{m-1} a_j \chi_{S_j} $$ where $a_j$ are coefficients, $S_j$ are disjoint sets whose union is the domain of $f$, and $\chi$ is the characteristic function.

Let $$f = \sum_{j = 0}^{m-1} a_j \chi_{S_j}, g = \sum_{k = 0}^{n-1} b_k \chi_{T_k}$$ be simple.

Then $$\begin{align*} h(f, g) &= h\left(\sum_{j = 0}^{m-1} a_j \chi_{S_j},\sum_{k = 0}^{n-1} b_k \chi_{T_k} \right) \\ &= \sum_{j = 0}^{m-1} h\left(a_j, \sum_{k=0}^{n-1} b_k \chi_{T_k} \right) \chi_{S_j} \\ &= \sum_{j=0}^{m-1} \sum_{k=0}^{n-1} h(a_j, b_k) \chi_{S_j \cap T_k} \end{align*}$$ is also simple.

This can be extended for infinite $m, n$.