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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.

dantopa
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user64742
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    What is your definition of "piecewise constant"? – Eric Wofsey Oct 01 '16 at 03:39
  • Why can't you be sure?! Have you tried to prove this? Where did you get stuck? – Mariano Suárez-Álvarez Oct 01 '16 at 03:43
  • I suggest you pick a definition of what a piecewise constant function is, two piecewise constant functions f and g, an arbitrary function h, and try to show that h(f(X),g(X)) is also satisfies the definition. – Mariano Suárez-Álvarez Oct 01 '16 at 04:17
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    @TheGreatDuck It's better to define the terms within the question to avoid confusion. In this case, what are the assumptions on the subsets which define the piecewise function. That helps preventing misunderstandings with other usages of piecewise like Limit of a Piecewise Function defined on Rationals and Irrationals. – dxiv Oct 01 '16 at 04:28
  • You'll have a harder time getting pertinent answers without a clear definition. In the most basic case where $f$ and $g$ are piecewise constant on a finite set of intervals each, then yes $f$ will be piecewise constant on the (finite) set defined by the intersections of the former. P.S. What exactly do you mean by a level step *unbounded* function, more to the point how can a constant function be unbounded? – dxiv Oct 01 '16 at 04:33
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    A good first step is to decide exactly what you mean by piecewise constant. Unless you do that, it is impossible for you to do anything more. – Mariano Suárez-Álvarez Oct 01 '16 at 04:44
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    @TheGreatDuck period where it is equal to infinity or negative infinity Sorry, no idea what that means. No real function proper is "equal to infinity", ever. It is doubly important to define your terms when you are using them in unconventional ways. – dxiv Oct 01 '16 at 04:49
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    Whatever it is that the "traditional view" on the matter is, if you cannot make precise what being piecewise constant means, then you are simply not being able to prove that anything is or is not piecewise constant. – Mariano Suárez-Álvarez Oct 01 '16 at 04:51
  • @TheGreatDuck Now that you accepted HenryW's (good) answer, do you understand what definition of piecewise constant he assumed you meant? – dxiv Oct 01 '16 at 04:54
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    You seem to believe that something is a piecewise constant function, but you don't know what you mean exactly by piecewise constant. That is your problem, not whether your hypothesis is correct it not! – Mariano Suárez-Álvarez Oct 01 '16 at 04:56

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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 $m, n$.

Henricus V.
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