Editing Nowhere dense set (section)

== Nowhere dense sets with positive measure ==

A nowhere dense set is not necessarily negligible in every sense.  For example, if <math>X</math> is the [[unit interval]] <math>[0, 1],</math> not only is it possible to have a dense set of [[Lebesgue measure]] zero (such as the set of rationals), but it is also possible to have a nowhere dense set with positive measure. One such example is the [[Smith–Volterra–Cantor set]].

For another example (a variant of the [[Cantor set]]), remove from <math>[0, 1]</math> all [[dyadic fraction]]s, i.e. fractions of the form <math>a/2^n</math> in [[lowest terms]] for positive integers <math>a, n \in \N,</math> and the intervals around them: <math>\left(a/2^n - 1/2^{2n+1}, a/2^n + 1/2^{2n+1}\right).</math> 
Since for each <math>n</math> this removes intervals adding up to at most <math>1/2^{n+1},</math> the nowhere dense set remaining after all such intervals have been removed has measure of at least <math>1/2</math> (in fact just over <math>0.535\ldots</math> because of overlaps<ref>{{cite web| url = http://www.se16.info/hgb/nowhere.htm| title = Some nowhere dense sets with positive measure and a strictly monotonic continuous function with a dense set of points with zero derivative}}</ref>) and so in a sense represents the majority of the ambient space <math>[0, 1].</math> 
This set is nowhere dense, as it is closed and has an empty interior: any interval <math>(a, b)</math> is not contained in the set since the dyadic fractions in <math>(a, b)</math> have been removed.

Generalizing this method, one can construct in the unit interval nowhere dense sets of any measure less than <math>1,</math> although the measure cannot be exactly 1 (because otherwise the complement of its closure would be a nonempty open set with measure zero, which is impossible).<ref>{{Cite book|last=Folland|first=G. B.|url=http://hdl.handle.net/2027/mdp.49015000929258|title=Real analysis: modern techniques and their applications|publisher=John Wiley & Sons|year=1984|isbn=0-471-80958-6|location=New York|pages=41|hdl=2027/mdp.49015000929258}}</ref>

For another simpler example, if <math>U</math> is any dense open subset of <math>\R</math> having finite [[Lebesgue measure]] then <math>\R \setminus U</math> is necessarily a closed subset of <math>\R</math> having infinite Lebesgue measure that is also nowhere dense in <math>\R</math> (because its topological interior is empty). Such a dense open subset <math>U</math> of finite Lebesgue measure is commonly constructed when proving that the Lebesgue measure of the rational numbers <math>\Q</math> is <math>0.</math> This may be done by choosing any [[bijection]] <math>f : \N \to \Q</math> (it actually suffices for <math>f : \N \to \Q</math> to merely be a [[surjection]]) and for every <math>r > 0,</math> letting 
<math display="block">U_r ~:=~ \bigcup_{n \in \N} \left(f(n) - r/2^n, f(n) + r/2^n\right) ~=~ \bigcup_{n \in \N} f(n) + \left(- r/2^n, r/2^n\right)</math>
(here, the [[Minkowski sum]] notation <math>f(n) + \left(- r/2^n, r/2^n\right) := \left(f(n) - r/2^n, f(n) + r/2^n\right)</math> was used to simplify the description of the intervals). 
The open subset <math>U_r</math> is dense in <math>\R</math> because this is true of its subset <math>\Q</math> and its Lebesgue measure is no greater than <math>\sum_{n \in \N} 2 r / 2^n = 2 r.</math> 
Taking the union of closed, rather than open, intervals produces the [[Fσ set|F<sub>{{sigma}}</sub>-subset]]
<math display="block">S_r ~:=~ \bigcup_{n \in \N} f(n) + \left[- r/2^n, r/2^n\right]</math>
that satisfies <math>S_{r/2} \subseteq U_r \subseteq S_r \subseteq U_{2r}.</math> Because <math>\R \setminus S_r</math> is a subset of the nowhere dense set <math>\R \setminus U_r,</math> it is also nowhere dense in <math>\R.</math> 
Because <math>\R</math> is a [[Baire space]], the set
<math display="block">D := \bigcap_{m=1}^{\infty} U_{1/m} = \bigcap_{m=1}^{\infty} S_{1/m}</math>
is a dense subset of <math>\R</math> (which means that like its subset <math>\Q,</math> <math>D</math> cannot possibly be nowhere dense in <math>\R</math>) with <math>0</math> Lebesgue measure that is also a [[Nonmeager set|nonmeager subset]] of <math>\R</math> (that is, <math>D</math> is of the [[second category]] in <math>\R</math>), which makes <math>\R \setminus D</math> a [[Comeager set|comeager subset]] of <math>\R</math> whose interior in <math>\R</math> is also empty; however, <math>\R \setminus D</math> is nowhere dense in <math>\R</math> if and only if its {{em|closure}} in <math>\R</math> has empty interior. 
The subset <math>\Q</math> in this example can be replaced by any countable dense subset of <math>\R</math> and furthermore, even the set <math>\R</math> can be replaced by <math>\R^n</math> for any integer <math>n > 0.</math>