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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?

Martin Sleziak
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Ashok
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    $[0,1/3] \cup [2/3,1]$ 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. – Chris Eagle Nov 22 '11 at 13:03
  • Yes, you are right. So what I said shouldn't be the actual interpretation of this result. – Ashok Nov 22 '11 at 13:33

1 Answers1

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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.

http://www.digizeitschriften.de/dms/img/#navi

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.

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Dave L. Renfro
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  • 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. – Ashok Nov 23 '11 at 05:36
  • @Ashok: If by stronger you mean the set is nice in some way, this is true, but it's not much of a restriction since the simplest such constructions already give $F_{\sigma}$ examples. – Dave L. Renfro Nov 23 '11 at 16:47
  • The proof of results [1] and [2] in your post are exploiting the existence of a set such that both the set and its complement has a positive measure intersection with every subinterval of $[0,1]$. Whereas, the result I stated in my question is just talking about the existence of a set with the property that its intersection with every subinterval of $[0,1]$ has positive measure. Hope you understand. – Ashok Nov 24 '11 at 13:16
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    @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). – Dave L. Renfro Nov 25 '11 at 15:34
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    @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) – Dave L. Renfro Nov 25 '11 at 15:50
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    @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. – Dave L. Renfro Nov 25 '11 at 15:50
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    @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. – Dave L. Renfro Nov 28 '11 at 22:19