Suppose we are working with some sets $\mathcal S_{1,2}$ of functions $[a,b]\to\mathbb R$. (Probably most often we would take $\mathcal S_1=\mathcal S_2=\mathcal S$.) Let us assume that we have some kind of integral such that all functions in $\mathcal S_{1,2}$ are integrable w.r.t this integral. (For example, we may assume that all these functions are Riemann integrable, Lebesgue integrable, Henstock-Kurzweil integrable, ...)

We can consider the following definition:

Definition. Let $f\colon[a,b]\to\mathbb R$ be a function. We say that $f$ is integrable if for every $\varepsilon>0$ there exist functions $g\in\mathcal S_1$ and $h\in\mathcal S_2$ such that $g\le f\le h$ and $$\int_a^b h(t) \; \mathrm{d} t - \int_a^b g(t) \; \mathrm{d} t \le \varepsilon.\tag{*}$$ (And we call the value $\sup\{ \int_a^b g(t) \; \mathrm{d} t; g\in\mathcal S_1, g\le f\} = \inf \{ \int_a^b h(t) \; \mathrm{d} t; h\in\mathcal S_2, f\le h\}$ the integral of $f$.)

Since the above definition depends on $\mathcal S_{1,2}$, it would be more precise to use the name $(\mathcal S_1,\mathcal S_2)$-integrable (or $\mathcal S$-integrable if the two sets are the same). But if they are clear from the context, if is probably better to go for brevity. (In fact, one could also say that it depends on the integral which we choose to evaluate in $(*)$. However, if we choose any reasonable integral such that all functions from $\mathcal S_{1,2}$ are integrable in this sense, then we should get the same value.)

If we choose $\mathcal S_1=\mathcal S_2$ to be the set of all step functions, we get exactly the Riemann integral. (The only difference is that the integrals in $(*)$ are in the definition of Riemann integral expressed as sums.) So we can consider Riemann integral as a basic motivating example for the above definition. Although actually I though about this question in connection with Perron integral - I will try to explain below in detail how they are related.

Question: What do we get using the above definition integral if we choose various nice classes of functions in place of $\mathcal S_{1,2}$? (If it is needed to characterize some integrals in this way, we might add assumption that $f$ is bounded.)

I am aware that phrased this way the question is rather vague and broad. But I suppose that at least for some integrals similar approaches are well-known. It seemed more natural to ask the question more generally than to ask series of questions about specific classes of functions.

Relation to Perron integral. This question came to me quite naturally after reading a bit about Perron integral in the book Van Rooij, Schikhof: A Second Course on Real Functions. (Maybe it would be more precise to say "Perron approach to Henstock–Kurzweil integral" - this name seems to be used more frequently for this integral nowadays.) The definition given there is that $f$ is Perron integrable if for any $\varepsilon>0$ there exist continuous functions $v$ and $u$ such that $D^+v\le f \le D^-u$ and $u(b)-u(a)\le v(b)-v(a)+\varepsilon$. Here $D^+$ and $D^-$ denotes the upper and lower derivative, respectively.

If $u$ and $v$ are nice functions (for example, if $D^+v$ and $D^-u$ are Riemann integrable) then this looks at least a bit similar to the above definition. (Here $D^+u$ and $D^-v$ are in the roles of $g$ and $h$.) Also there is an exercise in the same book which suggests that if we take continuous functions, we get Riemann integral. (Although this is not said in the exercise explicitly.)

Some interesting classes of functions. I might have missed something - and I did not try to check all details - but to me it seems quite plausible that (at least if I look only at bounded $f$):

  • If $\mathcal S$=continuous functions or $\mathcal S$=smooth functions, we get again Riemann integral. (See the exercise linked above.)
  • If we take some "large enough" integral, integrable functions add nothing new. Namely for $\mathcal S$=Lebesgue integrable functions, we get Lebesgue integral. Similarly for Perron integral. (Basically what we need here is monotone convergence theorem to hold for our integral.)
  • If we take upper semicontinuous functions for $\mathcal S_1$ and lower semicontinuous functions for $\mathcal S_2$ we should get Lebesgue integral. (Bounded Lebesgue integrable functions can be approximated, in some sense, by upper semicontinuous functions from below. See, for example, Lemma 20.1 in Van Rooij, Schikhof.)
  • $\begingroup$ It is sad that there is not even a single comment/answer to this question. $\endgroup$ – onurcanbektas Jul 8 '18 at 4:22

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