Motivations for Measure Theory

Proof: Let \(E \subseteq S\) & let \(\mu \in S\).

claim: The subset E and the set \(\mu E = \{ \mu z: z\in E \}\) are congruent.

proof of claim: Consider a function \(f:E \rightarrow \mu E \ni f(z) = \mu z~ \forall~ z \in E\). Proving that f is bijective is enough to prove the congruence.

Injectivity: - suppose not, let \(y \in \mu E\) and let \[ x_1,x_2 \in E \ni f(x_1) = f(x_2) = y \] \[\rightarrow \mu x_1 = \mu x_2 = y\] \[\rightarrow x_1 = x_2 \] Surjectivity: \(\forall~ x \in \mu E, x \in \{ \mu z : z \in E \} \rightarrow x = \mu z\) for some \(z \in E, \forall~ x \in \mu E\). \[ \therefore E \cong \mu E \]

hence by our 3rd property, \(P_r(E) = P_r(\mu E)~ \forall~ \mu \in S, E \subseteq S\)

let \(\mu = e^{i\theta}\), we see that \(\mu E\) is the set E, rotated by an angle \(\theta\).

Now consider T ⊂ S, \[ T = \{ e^{2\pi i t} : t \in \mathbb{Q} \} \] and observe that the set T is countable.

now in order to proceed, we must learn a new mathematical object, equivalence classes.

Equivalence classes: Suppose you have an equivalence relation '\(\sim\)' between sets A and B then \(\forall~ a \in A\), the equivalence class of a i.e. \([a]\) is given as: \[ [a] = \{ b \in B : a \sim b \} \] equivalence classes are mainly used to denote some sort of equality between 2 mathematical objects.

let S\T = {equivalence classes in S where \(z \sim w \leftrightarrow z = \mu w\) for some \(\mu \in T\)} for example: Suppose \(z = e^{2\pi i \frac{1}{5}}\) then \[ \Rightarrow [z] = \{ \mu z : \mu \in T \} \]

\[ \Rightarrow [z] = \{ e^{2\pi i q} e^{2\pi i \frac15} : q \in \mathbb Q \} \]

\[ \Rightarrow [z] = \{ e^{2\pi i (q+\frac15)} : q \in \mathbb Q \} \] \(\forall~ z \in S\)

Choose* exactly 1 representative element \(\psi\) from each equivalence class & let \(\Phi = \{ \psi \} \subset S\) be the collection of all representatives.

note * = This is exactly where we make use of the axiom of choice. The axiom of choice says that given a family of non-empty sets, \(E_i\) and an index set \(I\), there exists a function \(f:I \rightarrow \bigcup \limits_{i \in I}E_i \ni f(i) \in E_i~ \forall~ i\).
Note that this is not so obvious because the number of sets in the family that we have considered may not be countable at all! and as a matter of fact, the indexing set is also not required to be countable (as someone who never thought that the index set could be something other than the set of natural numbers, this came as a shock) A simple way of proving the existence of such uncountable index sets is to just let \(I = E\) where E is uncountable and let f be the identity function.

claim: \(S = \bigcup \limits_{\mu \in T} \mu \Phi\)

proof of claim: Let \(x \in \Phi \ni x = e^{2\pi i r}, r \in \mathbb{R}\) then x lies in some equivalence class containing elements of the form \(\{ e^{2\pi i (q + r)} : q \in Q \}\) thus \(\forall\) such x, \(\mu x\) gives us all the points achievable by rotating by \(\mu\) i.e. \(\mu \Phi\). \(\bigcup \limits_{\mu \in T} \mu \Phi\) just gives all the points achievable by rotating by any value of \(\mu \in T\) i.e. S.

claim: if \(\mu_1, \mu_2 \in T, \mu_1 \Phi \cap \mu_2 \Phi = \phi\)

proof of claim: Suppose \(q_1 \neq q_2, q_1,q_2 \in \Phi \ni \mu_1 q_1 = \mu_2 q_2\) \[ \Rightarrow q_1 = \mu_1^{-1} \mu_2 q_2 \]

\[ \Rightarrow q_1 = \mu ' q_2 \leftrightarrow q_1 \sim q_2 \] but \(q_1,q_2 \in \Phi\), contradiction!

\[ \therefore \mu_1 \Phi \cap \mu_2 \Phi = \phi \] so now we can make use of the (3) property of our proportion function - \[ P_r(S) = P_r(\bigcup \limits_{\mu \in T} \mu \Phi) = \sum_{\mu \in T} P_r(\mu \Phi) \]

\[ \Rightarrow 1 = P_r(S) = \sum_{\mu \in T} P_r(\Phi) \] (as \(\Phi \cong \mu \Phi\))

claim: The set of all the equivalence classes i.e. \(\Phi\) is uncountable.

proof of claim: Suppose not then: \[ S = \bigcup \limits_{i=1}^{\infty} E_i, \] thus, \(\forall~ i, E_i \subset S \rightarrow S\) is countable but \(S = \{ e^{2\pi i r} : r \in \mathbb{R} \}\) which is clearly not countable (otherwise we would able to find a bijection between \(\mathbb{N}\) and \(\mathbb{R}\)) therefore, \(\Phi\) is uncountable.

and since we have \(P_r:2^{S} \rightarrow [0,1], \sum_{\mu \in T} P_r(\Phi)\) is either 0 or \(\infty\), in any case, not 1. \[ \therefore 1 = \sum_{\mu \in T} P_r(\Phi) \neq 1 \] thus the properties (1),(2) and (3) are logically inconsistent.