1 Measure
A measurable space is a set \(X\) with a σ-algebra \(\mathcal{F}\) containing all measurable subsets of \(X\). Upon a measurable space we are able to define a measure, a function assigning elements of \(\mathcal{F}\) a measure of size.
Given a measurable space \((X,\mathcal{F})\) a measure \(\mu:\mathcal{F}\to[0,+\infty]\) is a mapping frm the collection of measurable sets to \([0,+\infty]\) such that: 1. \(\mu(\emptyset)=0\) 2. \(\{ E_{i} \}_{i=1}^\infty \subseteq \mathcal{F}~\&~\text{disjoint}\implies \mu(\cup_{i=1}^\infty E_{i})=\sum_{i=1}^\infty \mu(E_{i})\).
We call the triple \((X, \mathcal{F}, \mu)\) a measure space.
Example (Dirac Measure): For \((X,\mathcal{F})=(X, 2^X)\) we fix \(x_{0}\in X\) and define \[ \mu(A)=\delta_{x_{0}}(A)=\begin{cases} 1&x_{0}\in A \\ 0& \text{o.w.} \end{cases} \] known as the Dirac measure.
For any measure space \((X,\mathcal{F},\mu)\) and \(A,B\in \mathcal{F}\), \(\{ A_{i} \}_{i=1}^n\subseteq \mathcal{F}\) we have the following properties: 1. \(A \subseteq B\implies \mu(A)\leq \mu(B)\) (Monotonicity) 2. \(A \subseteq B\implies \mu(B\setminus A)=\mu(B)-\mu(A)\) 3. \(\mu(\cup_{i=1}^\infty A_{i})\leq \sum_{i=1}^\infty \mu(A_{i})\) (Subadditivity) 4. \(A_{i}\subseteq A_{i+1},~\forall i\implies \mu(\cup_{i=1}^\infty A_{i})=\lim_{_{i \to \infty}}\mu(A_{i})\) (Continuity from below) 5. \(A_{i}\supseteq A_{i+1},~\forall i\implies \mu(\cap_{i=1}^\infty A_{i})=\lim_{_{i \to \infty}}\mu(A_{i})\) (Continuity from above)
PROOF: (1) + (2) Assume there exists \(A\subseteq B\) then we can partition \(B=A\cup(B\setminus A)\) and so \[ \mu(B)=\mu(A)+\mu(B\setminus A)\geq \mu(A). \] (3) Similarly, we can define \(A_{1}=B_{1}\) and \(B_{n}=A_{n}\setminus \cup_{i=1}^{n-1}A_{i}\) then we have that \[ \mu(\cup_{i=1}^\infty A_{i})=\mu(\cup_{i=1}^\infty B_{i})=\sum_{i=1}^\infty \mu(B_{i})=\sum_{i=1}^\infty \mu(A_{i})-\mu(\cup_{i=1}^{n-1}A_{i})\leq \sum_{i=1}^\infty \mu(A_{i}). \] (4) Defining \(B_{1}=A_{1}\) and \(B_{i}=A_{i}\setminus A_{i-1}\) for \(i \geq 2\). By definition \(\cup_{i=1}^n B_{i}=A_{n}\) hence \[ \cup_{i=1}^\infty B_{i}=\cup_{n=1}^\infty A_{n}\implies \mu(A_{n})=\mu(\cup_{i=1}^nB_{i})=\sum_{i=1}^n\mu(B_{i}). \] Consequently taking the limit as \(n \to \infty\) we find \[ \lim_{n \to \infty}\mu(A_{n})=\sum_{i=1}^\infty \mu(B_{i})=\mu\left( \bigcup_{i=1}^\infty B_{i} \right) =\mu\left(\bigcup_{i=1}^n A_{i}\right) \] (5) Define \(B_{i}=A_{1}\setminus A_{i}\). By constraint \(B_{i}\subseteq B_{i+1}\) and so \[ \begin{align} \bigcup_{i=1}^\infty B_{i} & =\bigcup_{i=1}^\infty(A_{1}\setminus A_{i})=\bigcup_{i=1}^\infty(A_{i}\cap A_{i}^c)=A_{1}\cap\left( \bigcup_{i=1}^\infty A_{i}^c \right) \\ & = A_{1}\cap \left( \bigcap_{i=1}^\infty A_{i} \right)^c=A_{1}\setminus \left( \bigcap_{i=1}^\infty A_{i} \right). \end{align} \] Therefore we have that \[ \begin{align} \mu(A_{1}) & =\mu\left( A_{1}\cap\left( \bigcap_{i=1}^\infty A_{i} \right) \right)+\mu\left( A_{1}\setminus \bigcap_{i=1}^\infty A_{i} \right) =\mu\left( \bigcap_{i=1}^\infty A_{i} \right)+\lim_{i \to \infty} \mu(B_{i}) \\ & = \mu\left( \bigcap_{i=1}^\infty A_{i} \right)+\lim_{i \to \infty}\mu(A_{i})-\mu(A_{i} ) = \mu\left( \bigcap_{i=1}^\infty A_{i} \right)+\mu(A_{1})-\lim_{i \to \infty}\mu(A_{i}), \end{align} \] and rearranging gives the result. \(\square\)
Given a measure space \((X,\mathcal{F}, \mu)\) we define the following terminology: 1. the measure is finite if \(\mu(X)<+\infty\) 2. the measure is \(\sigma\)-finite if \(\exists \{ E_{i} \}_{i=1}^\infty \subseteq \mathcal{F}\) such that \(\cup_{i=1}^\infty E_{i}=X\) and \(\mu(E_{i})< \infty\) for all \(i \in \mathbb{N}\) 3. \(E \in S^X\) is a null set (of \(\mu\)) if \(E\in \mathcal{F}\) and \(\mu(E)=0\) 4. a property holds (\(\mu\)-) almost everywhere if the set of points where it fails is a null set 5. a measure whose domain includes all subsets of null sets is called complete.
The next theorem defines a complete measure space, a measure space in which every subset of every null set is measurable (with measure zero).
Suppose that \((X,\mathcal{F}, \mu)\) is a measure space. Let \(\mathcal{N}=\{ N \in \mathcal{F}:\mu(N)=0 \}\) and \(\overline{\mathcal{F}}=\{ E \cup F:E \in\mathcal{F},~F\subset N~\text{for some }N \in\mathcal{N} \}\). Then \(\overline{\mathcal{F}}\) is a \(\sigma\)-algebra known as the completion of \(\mathcal{F}\) with respect to \(\mu\), and there is a unique extension \(\overline\mu\) of \(\mu\) to a complete measure on \(\overline{\mathcal{F}}\), known as the completion of \(\mu\). The resulting measure space \((X,\overline{F}, \overline{\mu})\) is called a complete measure space.
PROOF: Firstly, we show that \(\overline{\mathcal{F}}\) is a \(\sigma\)-algebra. Note that since \(\mathcal{F}\) and \(\mathcal{N}\) are both closed under countable unions, so to is \(\overline{\mathcal{F}}\) and further, since \(\emptyset \in \mathcal{F},\mathcal{N}\) then we have that \(\emptyset \in\overline{\mathcal{F}}\), hence we need only show that \(\overline{\mathcal{F}}\) is closed under compliments.
\(\square\)
Naturally you might ask whether there exists a measure \(\mu\) on the measurable real space \((\mathbb{R}, \mathcal{B}_{\mathbb{R}})\) (i.e. the couple of the real numbers and the Borel σ-Algebra) where \(\mu((a,b))=b-a\) and \(\mu\) is translation invariant. To prove existence we must first consider outer-measures, functions similar to measures but that are defined on all collections of subsets.