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I have read some authors saying that the class of all Riemann integrable functions is "quite small". In what sense? How can we prove this statement?

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    $\begingroup$ In a general way, the class of all functions we know something about is "quite small". We know continuous functions, piecewise continuous functions, etc... But there are so much more functions that all we know is in fact a really small part of the mathematical reality :) $\endgroup$
    – krirkrirk
    Sep 17, 2015 at 22:05
  • $\begingroup$ See also math.stackexchange.com/questions/353452/…. The "rarely" might be a reference to the discontinuity only up to null-sets condition. $\endgroup$
    – parsiad
    Sep 17, 2015 at 22:09
  • $\begingroup$ You don't have a mathematical statement, so we can't prove it. But, in many senses, "most" functions are bad, "most" curves aren't smooth, etc., for precisely the same reasons that most real numbers aren't algebraic. Edit: Probably this isn't necessary, but it mostly comes down to cardinality arguments. When we restrict the members of some space in order to talk about them (i.e. Riemann integrable functions), sometimes we restrict our discussion to a countable subset of these things as a side effect. $\endgroup$
    – Eric Tressler
    Sep 17, 2015 at 22:10
  • $\begingroup$ @RobArthan I don't see how you can prove that "the cardinality of the set of Riemann integrable functions on the unit interval is c, the cardinality of the reals" . Is it really true ? $\endgroup$
    – krirkrirk
    Sep 17, 2015 at 22:26
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    $\begingroup$ I would assume this statement means that Riemann integrable functions form a meager subset of the space of Lebesgue integrable functions, though I don't know how to prove that. $\endgroup$
    – Eric Wofsey
    Sep 17, 2015 at 22:40

4 Answers 4

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There are a number of possible ways of viewing this statement. One might like to say that it is small in the sense of cardinality. This is not true: the indicator function of any subset of the standard Cantor set is Riemann integrable, and there are as many of these functions as there are real-valued functions of real variables. (An interesting corollary: there are Riemann integrable functions which are not Borel measurable.)

The simplest correct view that I can think of goes as follows. First consider the space $R_0[a,b]$, which consists of all $f : [a,b] \to \mathbb{R}$ such that the Riemann integrals $\int_a^b f(x) dx$ and $\int_a^b |f(x)| dx$ both exist as finite numbers. This is a pseudonormed vector space. To make it normed, consider the space $R[a,b]$, which is the quotient of $R_0[a,b]$ by the subspace of functions with $\int_a^b |f(x)| dx = 0$. Finally take the completion of $R[a,b]$. This completion is (isomorphic to) the Lebesgue space $L^1([a,b])$. Since there are Lebesgue integrable functions which are not Riemann integrable, it follows that $R[a,b]$ is incomplete. In fact in a sense it is "very" incomplete, in that there are a great deal of Lebesgue integrable functions which are not Riemann integrable, or even "essentially" Riemann integrable (i.e. Riemann integrable after a modification on a set of measure zero).

A concrete example of the latter point: $f_n(x)=x^{-1/2}$ for $x \in [1/n,1]$ and $0$ otherwise is Cauchy, because it converges in $L^1$ to a Lebesgue integrable function, but it does not converge to an element of $R[0,1]$, since the limit is not essentially bounded.

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  • $\begingroup$ Using the topology on $L^2[a, b]$, is it possible say that $R[a, b]$ is small in terms of toplogical term? $\endgroup$
    – user99914
    Sep 17, 2015 at 22:23
  • $\begingroup$ So is not possible to compare with some accuracy what some "changes" in these sets produces. Is not possible to state how bigger is a set in comparison with other. Only we can have is some ideas that these sets are bigger or smaller. That's it? I wanted to know, for example, how bigger is the set of all functions equivalent to Riemann integrable functions in comparison with the set of all Riemann-integrable functions. There exist some theory that can state affirmations like that? $\endgroup$
    – Filburt
    Sep 17, 2015 at 23:24
  • $\begingroup$ Further, can we talk about how many sets of measure non-zero exists? I know other examples that proves that this space is incomplete. But, to finish here, I want to know if we can talk about size in terms of numbers or no. And if no, if there exist a way that we can talk about size with precision. (apologize the bad english) $\endgroup$
    – Filburt
    Sep 17, 2015 at 23:25
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    $\begingroup$ @FernandoVieiraCostaJúnior There are as many sets of Lebesgue measure zero as there are subsets of $\mathbb{R}$. The classic proof of this proceeds in two steps: prove the Lebesgue measure is complete (all subsets of any measure zero set are measurable); then prove that the Cantor set is a set whose cardinality is the same as that of $\mathbb{R}$ and which has measure zero. As a result, the cardinality of the set of Riemann integrable functions is at least as big as the cardinality of any other subset of the real-valued functions of real variables. $\endgroup$
    – Ian
    Sep 17, 2015 at 23:27
  • $\begingroup$ @FernandoVieiraCostaJúnior So a cardinality argument will never work. You will have to resort to something else, like the topological description that others have given. I can imagine a description in which you choose some measure to put on some subset of the real-valued functions of real variables and show that the measure of the Riemann integrable functions is zero. I have no idea what this measure would be, but this has been done in other contexts (for example, the space of continuous functions which are differentiable at any point has Wiener measure zero). $\endgroup$
    – Ian
    Sep 17, 2015 at 23:29
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Here're two ways to state it.

Claim 1: Let $B$ be the Banach space of all bounded functions on the unit interval under pointwise addition, scalar multiplication and sup norm: $\|f\| = \sup \{|f(x)| : x \in [0, 1]\}$. Then, the class of Riemann integrable (or even Lebesgue integrable) functions in $B$ is of first category (in fact, it is even closed nowhere dense).

Claim 2: Let $L$ be the Banach space of all bounded Lebesgue integrable functions on the unit interval under pointwise addition, scalar multiplication and sup norm: $\|f\| = \sup \{|f(x)| : x \in [0, 1]\}$. Then, the class of Riemann integrable functions in $B$ is of first category (in fact, even closed nowhere dense).

The choice of the norm in Claim 2 is flexible. You can use an $L^1$-norm and the statement remains true.

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    $\begingroup$ Is there any difference between Claim 1 and Claim 2? $\endgroup$
    – uniquesolution
    Sep 17, 2015 at 22:57
  • $\begingroup$ But it also said (or even Lebesque integrable)... $\endgroup$
    – user99914
    Sep 17, 2015 at 22:58
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    $\begingroup$ The last comment about $L^1$-norm isn't correct since you can change any Riemann integrable map on a countable set to get a nowhere continuous function. $\endgroup$
    – hot_queen
    Sep 17, 2015 at 23:18
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    $\begingroup$ The Riemann integrable functions are definitely not nowhere dense under the $L^1$ norm. I could see them still being first category, however. $\endgroup$
    – Ian
    Sep 17, 2015 at 23:19
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    $\begingroup$ @Ian: For instance, every continuous function is Riemann integrable, and the continuous functions are well known to be dense in $L^1$. I'm quite certain it will be first category. There's a theorem that every linear subspace of a Banach space having the Baire property is first category, and the Riemann integrable functions are simple enough to define that I think they should be Borel or at least analytic. $\endgroup$
    – Nate Eldredge
    Sep 18, 2015 at 1:15
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For example the function $f:[0,1]\to\mathbb R$ defined by $f(x)=1, x\in [0,1]\setminus \mathbb Q $ and $f(x)=0, x\in \mathbb Q$. Then $f$ is not Riemann integrable, because every small Darboux sum is $0$ and every big Darboux sum is $1$. Now if you change the value of some irrational $x$ from $1$ to something else you get another function that is not Riemann integrable. In the same way you can get uncountably many functions that are not integrable. This might give you some idea, but definitely not in terms of cardinality. In other words, a function to be Riemann integrable, it has to have at most a set of discontinuities with Lebesgue measure $0$ (this is also sufficient).

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  • $\begingroup$ @krirkrirk What is the problem with the observation above ? $\endgroup$
    – Svetoslav
    Sep 17, 2015 at 22:28
  • $\begingroup$ I don't know ! I'm not the one who downvoted ^^ $\endgroup$
    – krirkrirk
    Sep 17, 2015 at 22:34
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A Riemann integrable function must have a set of discontinuities of measure 0. And the set of all continuous functions has the same size of ${\mathbb R}$, which is definitively smaller (in the sense of sizes of infinity) than the set of all functions from ${\mathbb R}$ to ${\mathbb R}$. So if you put that together, you get that the set of integrable functions is defined by a set of continuous functions (smaller infinite size), combined with measure zero sets of discontinuities (not smaller in the sense of size of infinities, but small in the sense of measure).

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  • $\begingroup$ The set of all continuous functions has the same size of $\mathbb R$ ? How would you prove that ? Giving that constant functions are continuous ones, and there are "exactly" as many constant functions that there are elements in $\mathbb R$ , I don't see how it's possible $\endgroup$
    – krirkrirk
    Sep 17, 2015 at 22:25
  • $\begingroup$ Continuous functions are uniquely defined by their values on the rationals. This is the set of all countable subsets of ${\mathbb R}$, which has the same size as ${\mathbb R}$. $\endgroup$
    – user2566092
    Sep 17, 2015 at 22:27
  • $\begingroup$ Actually the size of Riemann integrable function is strictly bigger than that of $\mathbb R$. (See Ian's answer) $\endgroup$
    – user99914
    Sep 17, 2015 at 22:28
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    $\begingroup$ @JohnMa I never said otherwise. I only said that the set of continuous functions was the same size as ${\mathbb R}$. The fact that you can have a measure zero set of discontinuities which is uncountable supports what both of us are saying. $\endgroup$
    – user2566092
    Sep 17, 2015 at 22:30
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    $\begingroup$ @JohnMa Perhaps you're right. The only definitively unambiguous thing about what I said is that the set of discontinuities has measure 0, which explains "rare". However, if we defined an arbitrary condition on arbitrary points that had measure zero, then this wouldn't in and of itself (IMO) qualify as rare. It's the fact that these functions really are continuous, up to a set of measure zero, that can arguably make them "rare". Perhaps other answers are better. $\endgroup$
    – user2566092
    Sep 17, 2015 at 22:40

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