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I "created" this problem approximately 10-14 hours ago.

I think that this problem could be used as some undergraduate exercise in the chapter on Riemann integrability.

The meaning of between is the following:

Suppose that $f$ and $g$ are two Riemann-integrable functions defined on the closed interval $I$ and that $f<g$.

$f<g$ means that for every $x \in I$ it is true that $f(x)<g(x)$.

The function $h$ is between $f$ and $g$ if and only if $f<h<g$.

I think that different approaches exist to solve this exercise, and, which one would be yours?

One intuitive idea is to choose for every $f$ and $g$ some $h$ which is very wildly discontinuous in such a way for Riemann integral of $h$ to not exist.

But, I didn´t turn this idea into a concrete proof yet.

How to prove this?

  • @DavidMitra But that one is Riemann-integrable. –  Nov 06 '19 at 13:50
  • A counterexample to this statement would be rather pathological. If $f$ and $g$ have distinct values within some interval, the statement is already true. – Peter Nov 06 '19 at 13:53
  • @DavidMitra That´s not allowed since $f<h<g$ everywhere in the interval. –  Nov 06 '19 at 13:55
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    Well, how about $h=(1/4)f+(3/4)g$ for the rationals and $h=(3/4)f+(1/4)g$ for the irrationals. The set of points where both $f$ and $g$ are continuous has positive measure; but at any such point, $h$ will be discontinuous, I think. – David Mitra Nov 06 '19 at 14:03
  • @DavidMitra That´s much much better, but I do not know how to prove that that construction always gives Riemann-nonintegrable $h$. –  Nov 06 '19 at 14:08
  • Have you seen this result? – David Mitra Nov 06 '19 at 14:11
  • @YamMir If $f$ and $g$ differ only in isolated points, the statement is false. – Peter Nov 06 '19 at 14:11
  • @DavidMitra Yes, a long time ago. So, your construction gives everywhere discontinuous $h$? –  Nov 06 '19 at 14:20
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    Perhaps not everywhere; but at least at any point $x$ where both $f$ and $g$ are continuous. So at every point in a set of positive measure, $h$ is discontinuous (so, not integrable). – David Mitra Nov 06 '19 at 14:23
  • @DavidMitra I didn´t study seriously measure theory but if $A$ is a set of positive measure, does that mean that $A$ is an almost-interval (in some precisely defined sense and meaning)? –  Nov 06 '19 at 14:31
  • OK, what am I missing? Why can't we just let $h$ be the pointwise-defined arithmetic mean of $f$ and $g,$ namely for each $x$ we define $h(x) = \frac{f(x) + g(x)}{2}?$ Usually in considerations like this (e.g. here or here), the in-between function $h$ is required to nicer than $f$ and $g$ are assumed to be. (MOMENTS LATER) Oh, I see. You want the in-between function to be non-Riemann integrable. Well, maybe my links could be helpful in suggesting search terms. – Dave L. Renfro Nov 06 '19 at 14:41
  • @DaveL.Renfro That $h$ is Riemann-integrable. –  Nov 06 '19 at 14:42
  • Take $I=[0,1]$ for convenience. Since $f<g,$ for every $x\in I,\ g(x)-f(x)$ contains an interval. So, if every $h$ between $f$ and $g$ were Riemann integrable, then the cardinality of them would be $\mathbb R^{\mathbb R}$, but it is well-known that the cardinality is $2^{\frak c}$ – Matematleta Nov 06 '19 at 16:20

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