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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.

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    $\begingroup$ 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. $\endgroup$
    – Dave L. Renfro
    May 23, 2018 at 20:09

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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.

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    $\begingroup$ 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
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    $\begingroup$ That's correct. I assumed from the context of the question that the asker was already aware of that. $\endgroup$
    – Eric Wofsey
    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

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