Editing Meagre set (section)

===Meagre subsets and Lebesgue measure===

There exist nowhere dense subsets (which are thus meagre subsets) that have positive [[Lebesgue measure]].{{sfn|Narici|Beckenstein|2011|pp=371-423}} 

A meagre set in <math>\R</math> need not have [[Lebesgue measure]] zero, and can even have full measure.  For example, in the interval <math>[0,1]</math> [[fat Cantor set]]s, like the [[Smith–Volterra–Cantor set]], are closed nowhere dense and they can be constructed with a measure arbitrarily close to <math>1.</math>  The union of a countable number of such sets with measure approaching <math>1</math> gives a meagre subset of <math>[0,1]</math> with measure <math>1.</math><ref>{{cite web |title=Is there a measure zero set which isn't meagre? |url=https://mathoverflow.net/questions/43478 |website=MathOverflow}}</ref>

Dually, there can be nonmeagre sets with measure zero.  The complement of any meagre set of measure <math>1</math> in <math>[0,1]</math> (for example the one in the previous paragraph) has measure <math>0</math> and is comeagre in <math>[0,1],</math> and hence nonmeagre in <math>[0,1]</math> since <math>[0,1]</math> is a Baire space. 

Here is another example of a nonmeagre set in <math>\Reals</math> with measure <math>0</math>:
<math display=block>\bigcap_{m=1}^{\infty}\bigcup_{n=1}^{\infty} \left(r_{n}-\left(\tfrac{1}{2}\right)^{n+m}, r_{n}+\left(\tfrac{1}{2}\right)^{n+m}\right)</math>
where <math>r_1, r_2, \ldots</math> is a sequence that enumerates the rational numbers.