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The following theorem is well known:

Theorem: A function $f: [a,b] \to \mathbb R$ is Riemann integrable if and only if its set of discontinuities has Lebesgue measure zero.

Now if we change Riemann integrable to Lebesgue integrable, this theorem is false since $\chi_{\mathbb Q}$ is Lebesgue integrable and has full measure discontinuities. We also know that if we change Lebesgue measure to Jordan measure, then having Jordan zero-measure discontinuities is sufficient for integrability. My question:

Does every Riemann integrable function necessarily have zero Jordan measure discontinuities?

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    $\begingroup$ The point here is that Jordan measurable implies Lebesgue measurable, and the measures coincide, but there are Lebesgue measurable sets that are not Jordan measurable, for example $\Bbb Q\cap [0,1]$. Not also that the Jordan measure is defined for bounded sets only. $\endgroup$ – Pedro Tamaroff Jul 17 '14 at 16:19
  • $\begingroup$ @PedroTamaroff Thank you for your response, but the utility of your comment is dubious. Is your intent to say that the discontinuities of a function need not be Jordan measurable, so in a sense the question is not well-defined? $\endgroup$ – Tyler Holden Jul 17 '14 at 17:21
  • $\begingroup$ Indeed. Rene's answer shows this is the case, since the discontinuities of his function are the rationals of $[0,1]$. (Remember we integrate over bounded closed intervals following Riemann!) $\endgroup$ – Pedro Tamaroff Jul 17 '14 at 17:24
  • $\begingroup$ Yes, I wasn't worried about bounded intervals. I more or less implied that I fixed one $[a,b]$ in my theorem statement :) Anyway, thank you. I see what you were now trying to say. $\endgroup$ – Tyler Holden Jul 17 '14 at 17:28
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Let ${\mathcal D}(f)$ be the set of points at which $f$ is not continuous. The correct result is:

Theorem: Let $f:[a,b] \rightarrow {\mathbb R}$ be bounded. Then the following are equivalent:

(a) $\;f$ is Riemann integrable on $[a,b].$

(b) $\;{\mathcal D}(f)$ is a countable union of closed Jordan measure zero sets.

(c) $\;{\mathcal D}(f)$ is an ${\mathcal F}_{\sigma}$ Lebesgue measure zero set.

Nothing special is needed to prove this other than an examination of the usual proof characterizing the Riemann integrability of a function in terms of the zero measure of the set of points at which the oscillation of the function is greater than or equal to any specified positive real number. (Since these sets are closed, it doesn't matter whether we say Jordan measure or Lebesgue measure.) The countable union aspect shows up when we take the union of all such sets corresponding to the positive real numbers $\frac{1}{n}$ (as $n$ varies over the positive integers) in order to form the set of all points at which the function is not continuous.

A surprisingly little known consequence of this theorem is that the discontinuity set for any Reimann integrable function has both Lebesgue measure zero and is of the first (Baire) category (i.e. is meager). This is because any closed measure zero set (again, since "closed" is used, both Jordan and Lebesgue measures agree) is nowhere dense (a nearly trivial observation), and hence a countable union of closed measure zero sets is a first category set.

Incidentally, there exist sets that are both Lebesgue measure zero and first category (i.e. the sets are small in both the Lebesgue measure sense and the Baire category sense) which are too big to be covered by any discontinuity set of a Riemann integrable function, or even by the union of the discontinuity sets of countably many Riemann integrable functions. (I say "covered" because the issue is size, not Borel type. I'm disallowing someone from simply picking a non-${\mathcal F}_{\sigma}$ set, and letting this be the example by saying that the discontinuity set always has to be an ${\mathcal F}_{\sigma}$ set.) For more about this issue, specifically the fact that the $\sigma$-ideal of sets generated by the discontinuity sets of Riemann integrable functions is properly contained in the $\sigma$-ideal of sets that are simultaneously Lebesgue measure zero and first category, see my 30 April 2000 sci.math essay HISTORICAL ESSAY ON F_SIGMA LEBESGUE NULL SETS.

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I think

$$f(x)=\begin{cases} 0&x \ \text{is irrational}\\ \frac{1}{q}& x=\frac{p}{q}\\ \end{cases}$$

is an counterexample, its discontinuities are the rationals, all you have to show is that they dont have Jordan measure zero. A finite number of open interval covering all rationals will leave only finitely many points uncovered.

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  • $\begingroup$ Exactly: math.ualberta.ca/~xinweiyu/217.1.13f/217-20131107.pdf $\endgroup$ – Siminore Jul 17 '14 at 16:13
  • $\begingroup$ Between this and my remark to Pedro's comment above, I see that any possible counter example must discontinuities which are not Jordan measurable. Thank you. $\endgroup$ – Tyler Holden Jul 17 '14 at 17:23
  • $\begingroup$ Why is $\mathbb{Q}$ not jordan measurable ? $\endgroup$ – Rene Schipperus Jul 17 '14 at 17:23
  • $\begingroup$ $\mathbb Q$ can't be Jordan measurable as a whole because it is unbounded and so certainly can't be covered by finitely many sets. Also, its boundary on any interval has full measure, and so is not measurable. $\endgroup$ – Tyler Holden Jul 17 '14 at 17:24
  • $\begingroup$ also its inner measure is $0$. $\endgroup$ – Rene Schipperus Jul 17 '14 at 17:26

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