Non-measurable set

A set that is not a 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).

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 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.

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 [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

[1]	P.R. Halmos, "Measure theory" , v. Nostrand (1950) MR0033869 Zbl 0040.16802
[2]	J.C. Oxtoby, "Measure and category" , Springer (1971) MR0393403 Zbl 0217.09201
[3]	B.R. Gelbaum, J.M.H. Olmsted, "Counterexamples in analysis" , Holden-Day (1964) MR0169961 Zbl 0121.28902

Comments

See also Measure; Gödel constructive set; 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.