What is the cardinality of the set $\mathcal{R}[0,1]$ of all Riemann integrable real functions on [0,1]?

I expect it to be $2^\mathfrak{c}$. A function is Riemann integrable if and only if it is continuous almost everywhere and bounded. Since a set of measure 0 can be uncountable, I assume one can construct $2^\mathfrak{c}$ subsets of $[0,1]$ that have measure 0 yet are discontinuity sets of real functions. But then I also realize that there can be measure zero sets which are never discontinuity sets of a real function. So, I am stuck at this point and have no idea how to proceed.

  • 3
    $\begingroup$ This is a fairly common problem (it's in many books, on many U.S. Ph.D. Qualifying exams, etc.) but no one ever seems to mention who first proved this result. I eventually came across a paper where this result is proved --- around 1903 by Philip E. B. Jourdain. See here for the google-books digitization of the journal paper, and see here for an annotated version I prepared in 2007. $\endgroup$ May 23, 2018 at 20:09

1 Answer 1


Your guess is correct. The trick is to not worry about what the exact set of discontinuity is but instead just find a large family of sets which can only be discontinuous on some uncountable set of measure zero. For instance, if $C$ is the Cantor set and $A\subseteq C$ is any subset, the characteristic function of $A$ is Riemann integrable (since it is continuous at least on all of $[0,1]\setminus C$). There are $2^{\mathfrak{c}}$ such subsets, and so there are $2^{\mathfrak{c}}$ Riemann integrable functions.

  • 2
    $\begingroup$ This would show that there are at least this many Riemann integrable functions, no? You would also need to point out that there are at most $2^{\mathfrak{c}}$ functions (integrable or not). $\endgroup$
    – Teepeemm
    May 24, 2018 at 1:00
  • 1
    $\begingroup$ That's correct. I assumed from the context of the question that the asker was already aware of that. $\endgroup$ May 24, 2018 at 1:00
  • $\begingroup$ @Teepeemm The way I see is that the set of all real functions on $\mathbb{R}$ can be uniquely represented by a set S of ordered pairs $\{(a,b)\}$ where a is in domain and b is in range. This set S is clearly a subset of $\mathbb{R}^2$. The set of all real functions on $\mathbb{R}$ will thus be a subset of power set $\mathcal{P}(\mathbb{R}^2)$, which has cardinality $2^\mathfrak{c}$. So we get an upper bound. $\endgroup$
    – Prabhat
    May 24, 2018 at 4:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.