Editing Null set (section)

==Uses==

Null sets play a key role in the definition of the [[Lebesgue integration|Lebesgue integral]]: if functions <math>f</math> and <math>g</math> are equal except on a null set, then <math>f</math> is integrable if and only if <math>g</math> is, and their integrals are equal. This motivates the formal definition of [[Lp space|<math>L^p</math> spaces]] as sets of equivalence classes of functions which differ only on null sets.

A measure in which all subsets of null sets are measurable is ''[[complete measure|complete]]''. Any non-complete measure can be completed to form a complete measure by asserting that subsets of null sets have measure zero. Lebesgue measure is an example of a complete measure; in some constructions, it is defined as the completion of a non-complete [[Borel measure]].

===A subset of the Cantor set which is not Borel measurable===

The Borel measure is not complete. One simple construction is to start with the standard [[Cantor set]] <math>K,</math> which is closed hence Borel measurable, and which has measure zero, and to find a subset <math>F</math> of <math>K</math> which is not Borel measurable. (Since the Lebesgue measure is complete, this <math>F</math> is of course Lebesgue measurable.)

First, we have to know that every set of positive measure contains a nonmeasurable subset. Let <math>f</math> be the [[Cantor function]], a continuous function which is locally constant on <math>K^c,</math> and monotonically increasing on <math>[0, 1],</math> with <math>f(0) = 0</math> and <math>f(1) = 1.</math> Obviously, <math>f(K^c)</math> is countable, since it contains one point per component of <math>K^c.</math> Hence <math>f(K^c)</math> has measure zero, so <math>f(K)</math> has measure one. We need a strictly [[monotonic function]], so consider <math>g(x) = f(x) + x.</math> Since <math>g</math> is strictly monotonic and continuous, it is a [[homeomorphism]]. Furthermore, <math>g(K)</math> has measure one. Let <math>E \subseteq g(K)</math> be non-measurable, and let <math>F = g^{-1}(E).</math> Because <math>g</math> is injective, we have that <math>F \subseteq K,</math> and so <math>F</math> is a null set. However, if it were Borel measurable, then <math>f(F)</math> would also be Borel measurable (here we use the fact that the [[Image (mathematics)|preimage]] of a Borel set by a continuous function is measurable; <math>g(F) = (g^{-1})^{-1}(F)</math> is the preimage of <math>F</math> through the continuous function <math>h = g^{-1}</math>). Therefore <math>F</math> is a null, but non-Borel measurable set.