Tag Archives: axiom-of-choice

A non-measurable set

We describe here a non-measurable subset of the segment \(I=[0,1] \subset \mathbb R\).

Let’s define on \(I\) an equivalence relation by \(x \sim y\) if and only if \(x-y \in \mathbb Q\). The equivalence relation \(\sim\) induces equivalence classes on \(I\). For \(x \in I\), it’s equivalence class \([x]\) is \([x] = \{y \in I \ : \ y-x \in \mathbb Q\}\). By the Axiom of Choice, we can form a set \(A\) by selecting a single point from each equivalence class.

We claim that the set \(A\) is not Lebesgue measurable.

For all \(q \in \mathbb Q\) we denote \(A_q = \{q+x \ : x \in A\}\). Let’s take \(p,q \in \mathbb Q\). If it exists \(z \in A_p \cap A_q\), it means that there exist \(u,v \in A\) such that
\[z= p+u=q+v\] hence \(u-v=q-p=0\) as \(u,v\) are supposed to be unique representatives of the classes of the equivalence relation \(\sim\). Finally if \(p,q\) are distincts, \(A_p \cap A_q = \emptyset\).

As Lebesgue measure \(\mu\) is translation invariant, we have for \(q \in \mathbb Q \cap [0,1]\) : \(\mu(A) = \mu(A_q)\) and also \(A_q \subset [0,2]\). Hence if we denote
\[B = \bigcup_{q \in \mathbb Q \cap [0,1]} A_q\] we have \(B \subset [0,2]\). If we suppose that \(A\) is measurable, we get
\[\mu(B) = \sum_{q \in \mathbb Q \cap [0,1]} \mu(A_q) = \sum_{q \in \mathbb Q \cap [0,1]} \mu(A) \le 2\] by countable additivity of Lebesgue measure (the set \(\mathbb Q \cap [0,1]\) being countable infinite). This implies \(\mu(A) = 0\).

Let’s prove now that
\[[0,1] \subset \bigcup_{q \in \mathbb Q \cap [-1,1]} A_q\] For \(z \in [0,1]\), there exists \(u \in A\) such that \(z \in [u]\). As \(A \subset [0,1]\), we have \(q = z-u \in \mathbb Q\) and \(-1 \le q \le 1\). And \(z=q+u\) means that \(z \in A_q\). This proves the inclusion. However the inclusion implies the contradiction
\[1 = \mu([0,1]) \le \sum_{q \in \mathbb Q \cap [-1,1]} \mu(A_q) = \sum_{q \in \mathbb Q \cap [-1,1]} \mu(A) =0\]

Finally \(A\) is not Lebesgue measurable.

A vector space not isomorphic to its double dual

In this page \(\mathbb{F}\) refers to a field. Given any vector space \(V\) over \(\mathbb{F}\), the dual space \(V^*\) is defined as the set of all linear functionals \(f: V \mapsto \mathbb{F}\). The dual space \(V^*\) itself becomes a vector space over \(\mathbb{F}\) when equipped with the following addition and scalar multiplication:
\begin{array}{lll}(\varphi + \psi)(x) & = & \varphi(x) + \psi(x) \\
(a \varphi)(x) & = & a (\varphi(x)) \end{array} \right. \] for all \(\phi, \psi \in V^*\), \(x \in V\), and \(a \in \mathbb{F}\).
There is a natural homomorphism \(\Phi\) from \(V\) into the double dual \(V^{**}\), defined by \((\Phi(v))(\phi) = \phi(v)\) for all \(v \in V\), \(\phi \in V^*\). This map \(\Phi\) is always injective. Continue reading A vector space not isomorphic to its double dual