Introdution: A Non-Measurable Set
Can we define a measure on \(\mathbb{R}\), such that it works on all the subsets of \(\mathbb{R}\)?
(i) \(\lambda: \mathfrak{P}(\mathbb{R}) \to \mathbb{R}_+ \cup \{\infty\}\)
(ii) \(\lambda((a,b)) = \lambda([a,b]) = b-a\)
(iii) \(\lambda(A+x) = \lambda(\{a_i + x, a_i \in A\}) = \lambda(A)\)
(iv) \(\lambda(\cup_{j\ge 1} A_j) = \sum_{j \ge 1} \lambda (A_j)\)
It is impossible to define such a measure on all the subsets.
We define \(x \sim y \quad \iff \quad y - x \in \mathbb{Q}\).
Define \([x] = \{ y \in \mathbb{R}, y - x \in \mathbb{Q}\}\).
then \(\Lambda = \mathbb{R} \vert_{\sim}\).
then using the axiom of choice, from each family in \(\Lambda\), we choose one representitive to form \(\Omega\), it is possible to make \(\Omega \subseteq \mathbb{R}\) and \(\Omega \subseteq (0,1)\). For example, \(\Omega = \{\dfrac{1}{2}, \dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{3}}{2}, \cdots\}\).
Then it is easy to show if \(p, q \in \mathbb{Q}\) and \(p \ne q\), we have \((\Omega + p) \cap (\Omega + q) = \emptyset\)
For set \(\sum_{q\in \mathbb{Q}, -1 < q < 1} (\Omega + q)\)
first, \(\sum_{q \in \mathbb{Q}, -1< q <1} (\Omega + q) \subseteq (-1,2)\), then \(\lambda(\Omega) = 0\).
second, \((0,1) \subseteq \sum_{q\in \mathbb{Q}, -1<q<1} (\Omega + q)\), then \(\lambda(\Omega) \ne 0\).
Set Operations
The following set operations are valid, or shown as identity equation
\(A \cup B = (A^c \cap B^c)^c\)
\(\cap_{\alpha \in I} A_{\alpha}\), where \(I\) is any index set (e.g., it can be uncountable)