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Following [1] and [2]. Working with Lebesgue measure on the line.
Halmos’ Non-Measurable Set and Doob Example Set
Theorem 1 If is a Lebesgue measurable set of positive, finite measure, and if , then there exists an open interval such that
Proof. . So we can find open only just bigger than and hence so that If is a disjoint sequence of open intervals whose union is (it is a union of intervals and the union must be countable because there is a rational between any two components), then ( so ). Consequently for at least one , and has the required properties.
Theorem 2 If is a Lebesgue measurable set of positive measure, then there is an open interval containing the origin and entirely contained in the difference set .
Proof. Trivial if is or contains an open interval. Generally, use Theorem 1 to find bounded open so that If , then is contained in of length at most . If and are disjoint, then, since they have the same measure, we would have Hence there is one point in their intersection, which gives , showing is the required interval.
Theorem 3 If is irrational, then is everywhere dense in the line. So are and .
Proof. Write each element modulo as . If is an open interval, then there is an integer with . Among the numbers in the interval there must be two with by the pigeonhole principle. Hence some multiple of , which is an element of , must be in . For reduce modulo . Finally, .
Aside: Weyl’s equidistribution theorem says modulo 1 is dense in
Theorem 4 There exists a non-Lebesgue measurable set .
Proof. Let be one representative for each equivalence class of , using the axiom of choice. Let be Borel. Then cannot include any non-zero elements (otherwise two elements of would be equivalent mod ) and so by Theorem 2 and hence . Thus if is Lebesgue measurable, it has measure zero.
If then and are disjoint (again, otherwise there are equivalent elements in ). Since covers the line, and if is measurable all translations would be measurable with measure , we derive a contradiction.
Theorem 5 There exists a subset of the line such that for ever Lebesgue measurable set ,
Proof. : if is Borel , then cannot contain any element of (integer component of a difference is even) and therefore has inner measure zero. Then has the same property. If is Lebesgue measurable, then the monotone property of implies . Since if follows .
As an aside, [3] proves that differences of the Cantor set (which has measure ) fills the unit interval. The proof uses digit-by-digit subtraction in base 3.
Oxtoby and Bernstein Set
Background
A set is dense in the interval if has non-empty intersection with every subinterval of and is dense if it is dense in . It is nowhere dense if it is dense in no interval; i.e., every interval has a subinterval contained in . Nowhere dense sets are full of holes. A set is nowhere dense iff its complement contains a dense open set iff has no interior points.
A set if first category if it can be represented as a countable union of nowhere dense sets, otherwise it is second category. Baire’s theorem says the complement of a set of first category is dense, no interval is first category, and the intersection of any sequence of dense open sets is dense. The union of a countable family of first category sets is first category.
Any countable set is first category and measure zero. The Cantor set is first category (intersection of sets of length ) and measure zero, but uncountable.
The line can be decomposed into a set of first category and a set of measure zero. (Enumerate the rationals, for each , put intervals around each point with a total length of to form and then set . is clearly measure zero. Each is a dense (contains all rationals) and open (union of open intervals) subset of . Therefore is nowhere dense and is first category.)
A set has the property of Baire if if can be represented in the form , where is open and is of first category (or where is closed and is first category.)
A set is a countable union of closed sets and a is a countable union of open sets. A set is measurable iff it can be represented as a plus a nullset or minus a nullset (bracket between the two). Any closed set in is a because it is the intersection of -thickenings (relying on the metric on ).
Results
Lemma 1 Any uncountable contains a nowhere dense closed set of measure zero that can be mapped continuously onto .
Proof.
Lemma 2 The class of uncountable closed subsets of has power .
Proof. The class of open intervals with rational endpoints is countable, and every open set is the union of some subclass (there is a rational between the subintervals). Hence there are at most open sets, and closed sets (complements). There are at least closed sets since there are that many closed intervals.
Theorem 6 (F. Bernstein) There exists a set such that both and meet every uncountable closed subset of the line.
Proof. By Lemma 2, the class of uncountable closed subsets of the line can be indexed by the ordinal numbers less than , the first ordinal having predecessors, . Assume and each has been well ordered. Each has power by Lemma 1 since any closed set is . Let , be the first two members of . Let , be the first two members of different from both and . If and if , have been defined for all , let , be the first two elements of . This set is non-empty (it has power ) for each and so , are defined for all . Put . Since and for each , the set has the property that both it and its complement meet every uncountable closed set (a Bernstein set).
Theorem 7 Any Bernstein set is non-measurable and lacks the property of Baire. Every measurable subset of or is a nullset, and any subset that has the property of Baire is of first category.
Proof. Let be any measurable subset of . Any closed must be countable (since all uncountable subsets meet ) and hence . Therefore . If has the property of Baire, then , a and first category. Then must be countable (since every uncountable contains an uncountable closed set and therefore meets ). Hence is first category.
A set has power if it can be well ordered so that each element is preceded by only countably many elements; the elements of can be put in one-to-one correspondence with the ordinal numbers less than the first uncountable ordinal.
Theorem 8 (Ulam) A finite measure defined for all subsets of a set of power vanishes identically if it is equal to zero for every one-element subset.
Proof.
A finite measure defined for all subsets of a set of power vanishes identically if it is zero for points.
Proof.
References
1.
Halmos, P.R.: Measure theory. Springer (1974)
2.
Oxtoby, J.C.: Measure and category: A survey of the analogies between topological and measure spaces. Springer Science & Business Media (2013)