1 Vitali Set
The Vitali non-measurability theorem is a fundamental result in Measure Theory Notes constructed to prove that it is impossible to define a function on the power set of the real numbers \(\mathbb{R}\) that satisfies the intuitive properties of countable additivity, translation invariance and produces intuitive interval measure \(\mu([a,b])=b-a, \forall a\leq b\). The idea of the theorem can be extended to higher dimensions (see [[banarch-tarski-theorem]]) as well as similar results for other abstract sets although the form of the proof relies heavily on the nature of the set being considered.
We preemptively prove the following lemma proving that we obtain monotonicity from countable additivity of functions.
Given a set \(X\) and countably additive function \(\mu:2^X\to[0,\infty]\) then for all \(A,B\subseteq X\) \[ A \subseteq B\implies \mu(A)\leq \mu(B). \]
PROOF: Since \(A \subseteq B\) we can write \(B =A\cup(B\setminus A)\implies \mu(B)=\mu(A)+\mu(B\setminus A)\geq \mu(A)\) since measures are non-negative. \(\square\)
There exists no function \(\mu :2^\mathbb{R}\to[0,+\infty]\) that is countably additive, translation invariant and satisfies \(\mu([a,b])=b-a\) for all \(a\leq b\).
PROOF: Assume for the sake of contradiction that such a \(\mu\) exists. We define an equivalence relation \(\sim\) on \(\mathbb{R}\) \[ x\sim y\iff x-y \in \mathbb{Q}. \] The equivalence class of \(\sim\) to which \(x\) belongs is therefore given by \[ [x]_{\sim}:=\{ y \in \mathbb{R} : x\sim y \}. \] We make the claim that every such equivalence class contains an element in \([0,1]\). Assuming the claim is false then \(\exists x \in \mathbb{R}\) s.t. \([x]_{\sim}\cap[0,1]=\emptyset \implies \forall a \in [0,1],~x-a\not\in \mathbb{Q}\) i.e. \([x-1,x]\) contains no rational numbers which is impossible since the rationals are dense in the reals, giving the contradiction. Using the [[axiom-of-choice]], for each equivalence class we choose one element in \([0,1]\) belonging to that class and call the resulting set \(A\). Then we define \[ A_{q}=\{ A+q \}_{q \in \mathbb{Q}\cap [-1,1]}=\{ a+q:a\in A,q\in \mathbb{Q}\cap[-1,1] \}, \] hence each \(A_{q}\) contains the elements of \(A\) shifted by rational \(q \in [-1,1]\). We assume these are not disjoint i.e. \(\exists x \in A_{q}\cap A_{p}~s.t.~p \neq q\). This gives that \[ \begin{cases} x = a+p\quad\therefore\quad a=x-p \in A \\ x = a+q\quad \therefore\quad a = x-q \in A, \end{cases} \] \(\implies x-p\) and \(x-q\) are distinct elements of \(A\) from the same equivalence class giving the contradiction \(\rightarrow\leftarrow\). We therefore define the disjoint union \(B = \bigcup_{q \in\mathbb{Q}\cap [-1,1]}A_{q}\) and claim that \[ [0,1]\stackrel{(i)}{\subseteq} B \stackrel{(ii)}{\subseteq} [-1,2]. \] (i) Since \(\forall x \in \mathbb{R}\) we have \(x \in [a]\) for some \(a \in A\), in particular \(\forall x \in [0,1]\), \(x -a\in \mathbb{Q}\) for some \(a \in A\), i.e. \(x = a+q\) for some \(a \in A\) and \(q \in \mathbb{Q}\). Since \(x \in[0,1]\) and \(a \in A \subseteq [0,1]\), \(q = x-a\in[-1,1]\) and thus \(x \in B\). (ii) If \(b \in B\), then \(b = a+q\) for \(a \in A \subseteq [0,1]\), \(q \in [-1,1]\) and so \(b \in [-1,2]\). By monotonicity of countably additive functions \[ 1 = \mu([0,1])\leq \mu(B)\leq \mu([-1,2])=3. \] However, we also note that \[ \mu(B)=\sum_{q\in\mathbb{Q}\cap [-1,1]}\mu(A+q)=\sum_{q \in \mathbb{Q}\cap [-1,1]}\mu(A). \] Since \(\mu(B)\leq 3\) we must have that \(\mu(A)=0\), however this implies \(\mu(B)=0\) which contradicts our previous conclusion that \(\mu(B)\geq 1\). \(\rightarrow\leftarrow\) \(\square\)