View source for Non-measurable set

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A set that is not a [[Measurable set|measurable set]]. In more detail: A set  $  X $
belonging to a hereditary  $  \sigma $-
ring  $  H ( S) $
is non-measurable if

$$ 
\mu  ^ {*} ( X)  > \ 
\mu _ {*} ( X) ;
$$

here  $  S $
is the  $  \sigma $-
ring on which the measure  $  \mu $
is given, and  $  \mu  ^ {*} $
and  $  \mu _ {*} $
are the exterior and interior measures, respectively (see [[Measure|Measure]]).

For an intuitive grasp of the concept of a non-measurable set, the following  "effective constructions"  are useful.

Example 1. let

$$ 
K  =  \{ {( x , y ) } : {0 \leq  x \leq  1 , 0 \leq  y \leq  1 } \}
$$

be the unit square and define a measure  $  \mu $
on the sets

$$ 
\widehat{E}   =  \{ {( x , y ) } : {x \in E , 0 \leq  y \leq  1 } \}
,
$$

where  $  E $
runs through the Lebesgue-measurable sets of measure  $  m ( E) $, 
by setting  $  \mu ( \widehat{E}  ) = m ( E) $. 
Then the set

$$ 
X  =  \{ {( x , y ) } : {0 \leq  x \leq  1 , y = 1 / 2 } \}
$$

is non-measurable, since  $  \mu  ^ {*} ( X) = 1 $, 
$  \mu _ {*} ( X) = 0 $.

The oldest and simplest construction of a non-measurable set is due to G. Vitali (1905).

Example 2. Let  $  \mathbf Q $
be the set of all rational numbers. Then a set  $  X $(
called a Vitali set) having in accordance with the [[Axiom of choice|axiom of choice]] exactly one element in common with every set of the form  $  \mathbf Q + a $, 
where  $  a $
is any real number, is non-measurable. No Vitali set has the [[Baire property|Baire property]].

Example 3. Let  $  B $(
respectively,  $  C $) 
be the set of numbers of the form  $  n + m \xi $, 
where  $  \xi $
is an irrational number,  $  m $
and  $  n $
are integers with  $  n $
even (respectively,  $  n $
odd), and let  $  X _ {0} $
be a set obtained by means of the axiom of choice from the equivalence classes of the set of real numbers under the relation:

$$ 
x \sim y \  \textrm{ if } \  x - y \in A = B \cup C .
$$

Let  $  X = X _ {0} + B $. 
Then for every measurable set  $  E $:

$$ 
\mu _ {*} ( X \cap E )  =  0 ,\ \ 
\mu  ^ {*} ( X \cap E )  =  \mu ( E) .
$$

Yet another construction of a non-measurable set is based on the possibility of introducing a total order in a set having cardinality of the continuum.

Example 4. There exist a set  $  B \subset  \mathbf R $
such that  $  B $
and  $  \mathbf R \setminus  B $
intersect every uncountable closed set. Any such set (a Bernstein set) is non-measurable (and does not have the Baire property). In particular, any set of positive exterior measure contains a non-measurable set.

Apart from invariance under a shift (Example 2) and topological properties (Example 3) there are also reasons of a set-theoretical character why it is impossible to define a non-trivial measure for all subsets of a given set; such is, for example, Ulam's theorem (see [[#References|[2]]]) for sets of bounded cardinality.

No specific example is known of a Lebesgue non-measurable set that can be constructed without the use of the axiom of choice.

====References====
<table><TR><TD valign="top">[1]</TD> <TD valign="top">  P.R. Halmos,   "Measure theory" , v. Nostrand  (1950)  {{MR|0033869}} {{ZBL|0040.16802}} </TD></TR><TR><TD valign="top">[2]</TD> <TD valign="top">  J.C. Oxtoby,   "Measure and category" , Springer  (1971)  {{MR|0393403}} {{ZBL|0217.09201}} </TD></TR><TR><TD valign="top">[3]</TD> <TD valign="top">  B.R. Gelbaum,   J.M.H. Olmsted,   "Counterexamples in analysis" , Holden-Day  (1964)  {{MR|0169961}} {{ZBL|0121.28902}} </TD></TR></table>

====Comments====
See also [[Measure|Measure]]; [[Gödel constructive set|Gödel constructive set]]; [[Descriptive set theory|Descriptive set theory]].

In certain models of ZF (not ZFC) every set of real numbers is Lebesgue measurable (a result of Solovay), so the axiom of choice is needed to construct a Lebesgue non-measurable set.

For Ulam's theorem see [[Cardinal number|Cardinal number]].