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Unfortunately, my book (Royden-Fitzpatrick's Real Analysis) offers no definition of a step function but merely compares its role in Riemann integration to the role played by linear combinations of characteristic functions in Lebesgue integration. From what I gather, these are simple function in which he defines as follows:

A real-valued function $\varphi$ is defined on a measurable set $E$ is simple provided it is measurable and takes only a finite number of values. This means that $\varphi = \sum_{i=1}^n c_k \chi_{E_k}$ where $E_k = \{x \in E \mid \varphi(x) = c_k\}$

However, here and here the step function is defined in the very same way, the only difference being that the $E_k$ are taken to be intervals. But I have read elsewhere that the set of step functions and simple functions do not coincide. What am I misunderstanding? The reason I ask is because the following problem I am working could be rendered trivial depending on how one interprets these terms (at least it seems it could be):

Let $I$ be a closed, bounded interval and $E$ a measurable subset of $I$. Let $\epsilon > 0$. Show that there is a step function $h$ on $I$ and a measurable subset $F$ of $I$ for which $h = \chi_E$ on $F$ and $m(I-F) < \epsilon$.

I suppose my question comes down to: how am I to understand how the author is using the term "step function" so that the problem makes sense but isn't triviaL?

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    $\begingroup$ I don't think you are misunderstanding anything. A step function is a special case of a simple function, in which the sets $E_k$ are intervals. For a simple function, the $E_k$ can be any measurable sets. So, every step function is a simple function, but not vice versa. $\endgroup$ – Bungo Oct 20 '17 at 15:32
  • $\begingroup$ @Bungo Oh. Well that's reassuring. It is customary to take the sums as finite? $\endgroup$ – user193319 Oct 20 '17 at 15:54
  • $\begingroup$ Yes, the sums are finite for both step functions and simple functions. These types of functions are often used to approximate other functions via limiting arguments. It would be a pain if we had to incorporate an additional limiting argument because the step/simple functions themselves were defined with an infinite sum. $\endgroup$ – Bungo Oct 20 '17 at 16:00
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I think everything is fine , Step function is just a special case of simple function with the sets in simple function taken to be just intervals, You can also conclude that every Step function is a Simple function but Converse is not true.

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