Non-Measurable Sets
\[ \def\R{\mathbb R} \]
Following [1] and [2]. Working with Lebesgue measure \(\mu\) on the line.
Halmos’ Non-Measurable Set and Doob Example Set
Theorem 1 If \(E\) is a Lebesgue measurable set of positive, finite measure, and if \(\alpha\in [0,1)\), then there exists an open interval \(U\) such that \(\mu(E\cap U)\ge \alpha\mu(U)\)
Proof. \(\mu(E) = \inf\{ \mu(U) \mid E\subset U, U\text{\ open} \}\). So we can find open \(U_0\) only just bigger than \(E\) and hence so that \[ \alpha\mu(U) \le \mu(E). \] If \(U_n\) is a disjoint sequence of open intervals whose union is \(U_0\) (it is a union of intervals and the union must be countable because there is a rational between any two components), then \[ \alpha\sum_{n\ge 1} \mu(U_n) \le \sum_{n\ge 1} \mu(E\cap U_n) \] (\(E\subset U_0\) so \(E\cap U_0=E\)). Consequently \(\alpha \mu(U_n) \le \mu(E\cap U_n)\) for at least one \(n\), and \(U_n\) has the required properties.
Theorem 2 If \(E\) is a Lebesgue measurable set of positive measure, then there is an open interval \(U\) containing the origin and entirely contained in the difference set \(D(E)\).
Proof. Trivial if \(E\) is or contains an open interval. Generally, use Theorem 1 to find bounded open \(U\) so that \[ \mu(E\cap U) \ge \frac{3}{4}\mu(U). \] If \(x\in(-\frac{1}{2}\mu(U), \frac{1}{2}\mu(U))\), then \[ (E\cap U) \cup ((E\cap U) + x) \] is contained in \(U + (U+x)\) of length at most \(3\mu(U)/2\). If \((E\cap U)\) and \((E\cap U) + x\) are disjoint, then, since they have the same measure, we would have \[ \frac{3}{2}\mu(U) > \mu((E\cap U) \cup ((E\cap U) + x)) = 2\mu(E\cap U) \ge \frac{3}{2}\mu(U). \] Hence there is one point in their intersection, which gives \(x\in D(E)\), showing \((-\mu(U)/2, \mu(U)/2)\) is the required interval.
Theorem 3 If \(\xi\) is irrational, then \(A=\mathbb Z[\xi]\) is everywhere dense in the line. So are \(B=2\mathbb Z + \xi\mathbb Z\) and \(C=2\mathbb Z + 1 + \xi\mathbb Z\).
Proof. Write each element \(\xi\mathbb Z\) modulo \(1\) as \(x_i=n_i + i\xi\). If \(U\) is an open interval, then there is an integer \(k\) with \(\mu(U)>1/k\). Among the \(k+1\) numbers \(x_1,\dots,x_{k+1}\) in the interval there must be two with \(|x_i-x_j|<1/k\) by the pigeonhole principle. Hence some multiple of \(x_i-x_j\), which is an element of \(A\), must be in \(U\). For \(B\) reduce modulo \(2\). Finally, \(C=B+1\).
Aside: Weyl’s equidistribution theorem says \(\xi\mathbb Z\) modulo 1 is dense in \([0,1]\)
Theorem 4 There exists a non-Lebesgue measurable set \(E_0\).
Proof. Let \(E_0\) be one representative for each equivalence class of \(\mathbb R/A\), using the axiom of choice. Let \(F\subset E_0\) be Borel. Then \(D(F)\cap A\) cannot include any non-zero elements (otherwise two elements of \(E_0\) would be equivalent mod \(A\)) and so by Theorem 2 \(\mu(F)=0\) and hence \(\mu_*(E_0)=0\). Thus if \(E_0\) is Lebesgue measurable, it has measure zero.
If \(a_1,a_2\in A\) then \(E_0+a_1\) and \(E_0+a_2\) are disjoint (again, otherwise there are equivalent elements in \(E_0\)). Since \(E_0+A\) covers the line, and if \(E_0\) is measurable all translations would be measurable with measure \(0\), we derive a contradiction.
Theorem 5 There exists a subset \(M\) of the line such that for ever Lebesgue measurable set \(E\), \[ \mu_*(M\cap E)=0\quad\text{and}\quad \mu^*(M\cap E)=\mu(E). \]
Proof. \(M=E_0+B\): if \(F\) is Borel \(F\subset M\), then \(D(F)\) cannot contain any element of \(C\) (integer component of a difference is even) and therefore has inner measure zero. Then \[ M':=E_0+C = E_0 +(B+1) = M+1 \] has the same property. If \(E\) is Lebesgue measurable, then the monotone property of \(\mu_*\) implies \(\mu_*(M\cap E)=\mu_*(M'\cap E)=0\). Since \[ \mu_*(M\cap E) + \mu^*(M'\cap E)= \mu(E) \] if follows \(\mu^*(M\cap E)=\mu(E)\).
As an aside, [3] proves that differences of the Cantor set (which has measure \(0\)) fills the unit interval. The proof uses digit-by-digit subtraction in base 3.
Oxtoby and Bernstein Set
Background
A set \(A\) is dense in the interval \(I\) if \(A\) has non-empty intersection with every subinterval of \(I\) and is dense if it is dense in \(\R\). It is nowhere dense if it is dense in no interval; i.e., every interval has a subinterval contained in \(A'\). Nowhere dense sets are full of holes. A set \(A\) is nowhere dense iff its complement contains a dense open set iff \(\bar A\) 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 \(3^{-n}\)) 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 \(i\), put intervals around each point with a total length of \(2^{-(i+j)}\) to form \(G_i\) and then set \(B=\bigcap_i G_i\). \(B\) is clearly measure zero. Each \(G_i\) is a dense (contains all rationals) and open (union of open intervals) subset of \(\R\). Therefore \(G_i'\) is nowhere dense and \(B'=\bigcup_i G_i'\) is first category.)
A set \(B\) has the property of Baire if if can be represented in the form \(A=G\triangle P\), where \(G\) is open and \(P\) is of first category (or \(A=F\triangle Q\) where \(F\) is closed and \(Q\) is first category.)
A \(F_\sigma\) set is a countable union of closed sets and a \(G_\delta\) is a countable union of open sets. A set is measurable iff it can be represented as a \(F_\sigma\) plus a nullset or \(G_\delta\) minus a nullset (bracket between the two). Any closed set in \(\R\) is a \(G_\delta\) because it is the intersection of \(1/n\)-thickenings (relying on the metric on \(\R\)).
Results
Lemma 1 Any uncountable \(G_\delta\subset\R\) contains a nowhere dense closed set \(C\) of measure zero that can be mapped continuously onto \([0,1]\).
Proof.
Lemma 2 The class of uncountable closed subsets of \(\R\) has power \(c\).
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 \(c\) open sets, and closed sets (complements). There are at least \(c\) closed sets since there are that many closed intervals.
Theorem 6 (F. Bernstein) There exists a set \(B\subset\R\) such that both \(B\) and \(B'\) meet every uncountable closed subset of the line.
Proof. By Lemma 2, the class \(\mathcal F\) of uncountable closed subsets of the line can be indexed by the ordinal numbers less than \(\omega_c\), the first ordinal having \(c\) predecessors, \(\mathcal F = \{F_\alpha\mid \alpha<\omega_c\}\). Assume \(\R\) and each \(F_\alpha\) has been well ordered. Each \(F_\alpha\) has power \(c\) by Lemma 1 since any closed set is \(G_\delta\). Let \(p_1\), \(q_1\) be the first two members of \(F_1\). Let \(p_2\), \(q_2\) be the first two members of \(F_2\) different from both \(p_1\) and \(q_1\). If \(1<\alpha<\omega_c\) and if \(p_\beta\), \(q_\beta\) have been defined for all \(\beta<\alpha\), let \(p_\alpha\), \(q_\alpha\) be the first two elements of \(F_\alpha - \bigcup_{\beta<\alpha} \{p_\beta, q_\beta\}\). This set is non-empty (it has power \(c\)) for each \(\alpha\) and so \(p_\beta\), \(q_\beta\) are defined for all \(\alpha<\omega_c\). Put \(B=\{p_\alpha\}\). Since \(p_\alpha\in B\cap F_\alpha\) and \(q_\alpha\in B'\cap F_\alpha\) for each \(\alpha<\omega_c\), the set \(B\) has the property that both it and its complement meet every uncountable closed set (a Bernstein set).
Theorem 7 Any Bernstein set \(B\) is non-measurable and lacks the property of Baire. Every measurable subset of \(B\) or \(B'\) is a nullset, and any subset that has the property of Baire is of first category.
Proof. Let \(A\) be any measurable subset of \(B\). Any closed \(F\subset A\) must be countable (since all uncountable subsets meet \(B'\)) and hence \(m(F)=0\). Therefore \(m(A)=0\). If \(A\subset B\) has the property of Baire, then \(A=E\cup P\), \(E\) a \(G_\delta\) and \(P\) first category. Then \(E\) must be countable (since every uncountable \(G_\delta\) contains an uncountable closed set and therefore meets \(B'\)). Hence \(A\) is first category.
A set has power \(\aleph_1\) if it can be well ordered so that each element is preceded by only countably many elements; the elements of \(X\) can be put in one-to-one correspondence with the ordinal numbers less than the first uncountable ordinal.
Theorem 8 (Ulam) A finite measure \(\mu\) defined for all subsets of a set \(X\) of power \(\aleph_1\) 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 \(c\) vanishes identically if it is zero for points.
Proof.