Let's consider the following result.

There exists a Borel set $A\subset [0,1]$ such that $0<m(A\cap I)<m(I)$ for every subinterval $I$ of $[0,1]$, where $m$ is Lebesgue measure.

Can one interpret the above result that there are sets with positive measure which need not equal to any interval up to a set of measure zero? Does this result also have some interesting consequences?

  • 2
    $\begingroup$ $[0,1/3] \cup [2/3,1]$ is a much simpler example of a set of positive measure which isn't equal to any interval up to a set of measure zero. $\endgroup$ – Chris Eagle Nov 22 '11 at 13:03
  • $\begingroup$ Yes, you are right. So what I said shouldn't be the actual interpretation of this result. $\endgroup$ – Ashok Nov 22 '11 at 13:33

Here are two applications of "nice sets" (measurable, Borel, or $F_{\sigma}$) such that both the set and its complement have a positive measure intersection with every interval.

[1] What follows is taken from my answer at Is every Lebesgue measurable function on $\mathbb{R}$ the pointwise limit of continuous functions?

There exist measurable functions $f:{\mathbb R} \rightarrow {\mathbb R}$ that are not almost everywhere equal to any Baire $1$ function $g:{\mathbb R} \rightarrow {\mathbb R}.$ [Consider the characteristic function of a set such that both the set and its complement has a positive measure intersection with every interval. Oxtoby's book Measure and Category, 2nd edition, p. 37 gives a very nice construction of such a set that also happens to be $F_{\sigma}.$ Rudin gives the same construction in Well-distributed measurable sets, American Mathematical Monthly 90 (1983), 41-42.]

[2] Section I.1.b (pp. 11-13) of Bruckner's paper (see below) gives a brief discussion of nowhere monotone differentiable functions and outlines a proof of the existence of a nowhere monotone function with a bounded derivative through the use of a homeomorphic change of scale applied to the nowhere monotone absolutely continuous function $f-g,$ where $f$ and $g$ are the indefinite integrals of the characteristic functions of a set $E$ and its complement, and where $E$ has the property that both $E$ and its complement intersect every subinterval of $[0,1]$ in a set of positive measure. Exercise 18.31 (p. 296) in Hewitt/Stromberg's 1965 text and Problem 4.29(f) (p. 158) of Benedetto's 1976 text ask the reader to prove that $f-g$ formed in this way is nowhere monotone and absolutely continuous.

Andrew M. Bruckner, Current trends in differentiation theory, Real Analysis Exchange 5 (1979-80), 9-60.

John J. Benedetto, Real Variable and Integration, Mathematische Leitfaden. Stuttgart: B. G. Teubne, 1976, 278 pages.

Edwin Hewitt and Karl Robert Stromberg, Real and Abstract Analysis, Graduate Texts in Mathematics #25, Springer-Verlag, 1965/1975, x + 476 pages.

(ADDED THE NEXT DAY) I looked through some of my stuff at home early this morning and found the following additional references (full bibliographic information is further below). In a rough sort of way, it seems to me that each of the additional applications below still roughly belongs to one of the two applications I've already given.

additions to [1]

Foran's book, Section 6.1, pp. 261-262.

Hahn/Rosenthal's book, the remarks just before Article 11.3.22 on p. 147.

Stromberg's book, Exercise 13(c) on p. 309.

additions to [2]

Coffman's abstract.

Wise/Hall's book, Example 2.26 on p. 63.

[3] Goffman's paper (just below the middle of p. 544) uses such sets to show that a certain type of generalized Riemann integral (based on the upper and lower Burkill integral formulations when sets are neglected from a $\sigma$-ideal that contains at least one set of positive outer Lebesgue measure) can fail to be integrable in this sense but still be integrable in the Lebesgue sense.

[4] Settari gives the following example (see MR 36 #5892). Let $(X,\tau)$ be a metrizable topological space and let $D(X)$ be the collection of all metrics on $X$ that generate the topology $\tau.$ Then $D(X)$ can be partially ordered by defining $d_{1} \leq d_{2}$ if and only if there exists $\alpha > 0$ such that $d_{1} \; \leq \; \alpha \cdot d_{2}$ on $X \times X.$ Settari shows that there exist $d_1, \; d_2 \in D\left([0,1]\right)$ such that $d_1$ and $d_2$ have no common lower bound (i.e. there does not exist $d_3 \in D\left [0,1] \right)$ such that $d_{3} \leq d_{1}$ and $d_{3} \leq d_{2}$). This is accomplished by using two sets of the type under consideration whose union is $[0,1].$

[5] Such sets are constructed in ${\mathbb R}^n$ in Gardiner/Pau (Lemma 3 on p. 1130) for use in the proof of their Corollary 4.

Additional References

Charles Vernon Coffman, The existence of absolutely continuous nowhere monotone functions (conference abstract), American Mathematical Monthly 72 #8 (October 1965), p. 941 (abstract #5).

Stephen J. Gardiner and Jordi Pau, Approximation on the boundary and sets of determination for harmonic functions, Illinois Journal of Mathematics 47 (2003), 1115-1136.

Casper Goffman, A generalization of the Riemann integral, Proceedings of the American Mathematical Society 3 (1952), 543-547.

James Foran, Fundamentals of Real Analysis, Monographs and Textbooks in Pure and Applied Mathematics #144, Marcel Dekker, 1991, xii + 473 pages.

Hans Hahn and Arthur Rosenthal, Set Functions, The University of New Mexico Press, 1948, ix + 324 pages.

A. Settari, Directedness of the set of quasimetrics [Slovak], Acta Facultatis Rerum Naturalium Universitatis Comenianae, Mathematica 15 (1967), 53-60.


Karl R. Stromberg, Introduction to Classical Real Analysis, Wadsworth International, 1981, ix + 575 pages.

Gary L. Wise and Eric B. Hall, Counterexamples in Probability and Real Analysis, Oxford University Press, 1993, xii + 211 pages.

  • $\begingroup$ Thanks for the answer. It will take sometime for me to look into all that you have mentioned. But the two things which you have mentioned seem to be using a much stronger result that the existence of a set such that both the set and its complement has a positive measure intersection with every interval. $\endgroup$ – Ashok Nov 23 '11 at 5:36
  • 1
    $\begingroup$ @Ashok: Sorry, I didn't notice the absence of a complement requirement. Your property is called "metrically dense" or "measure dense". I like "measure dense", but "metrically dense" is much more common in the literature. Google-book and google-scholar searches for these terms will bring up a lot of applications. Also, in the same way that one can prove the Cantor-Bendixson theorem (Lindelof's method of proof), one can prove that any subset of the reals can be written as the union of a measure zero set and a set that is measure dense at each of its points (obvious meaning for the local notion). $\endgroup$ – Dave L. Renfro Nov 25 '11 at 15:34
  • 1
    $\begingroup$ @Ashok: Here's a proof of what I just stated. Given $E \subseteq {\mathbb R}^n$, let $E^{*}$ be the set of points belonging to $E$ at which $E$ is measure dense and put $E_{*} = E - E^{*}.$ Using the fact that $A \Delta B$ has measure zero implies “$A$ is measure dense at $x$ if and only if $B$ is measure dense at $x$", it suffices to prove that $E_{*}$ has measure zero. (proof continues) $\endgroup$ – Dave L. Renfro Nov 25 '11 at 15:50
  • 1
    $\begingroup$ @Ashok: (continuation of proof) Each $x \in E_{*}$ has a neighborhood $N_x$ such that $N_{x} \cap E$ has measure zero. Hence, each of the (possibly) smaller sets $N_{x} \cap E_{*}$ has measure zero. The collection of these neighborhoods (as $x$ varies over $E_{*}$) is an open covering of $E_{*}.$ Now take a countable subcover and use the fact that a countable union of measure zero sets has measure zero to show $E_{*}$ has measure zero. $\endgroup$ – Dave L. Renfro Nov 25 '11 at 15:50
  • 1
    $\begingroup$ @Ashok: I happened to look at your question again (I've been away a few days), and unless I'm really missing someting obvious, your question is in fact about the existence of a set such that both the set and its complement has a positive measure intersection with every subinterval. The complement part follows from the requirement that $m(A \cap I) < m(I)$ (i.e. the strict inequality aspect). I was in a hurry a few days ago when I left the previous few comments, including the one beginning with "Sorry", so I didn't notice this. $\endgroup$ – Dave L. Renfro Nov 28 '11 at 22:19

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.