1 Outer Measure
1.1 Outer Measure
In measure theory an outer (exterior) measure is a function defined on all subsets of a given set taking values in the extended real numbers satisfying some additional technical conditions. The theory of outer measures was first introduced by Constantin Carathéodory to provide an abstract basis for the theory of measurable sets and countably additive measures.
An outer measure on non-empty \(X\) is a function \(\mu^*:2^X\to[0,+\infty]\) s.t.: 1. \(\mu^*(\emptyset)=0\) 2. \(A \subseteq B\implies \mu^*(A)\leq \mu^*(B)\) 3. \(\mu^*\left( \cup_{i=1}^\infty A_{i} \right)\leq \sum_{i=1}^\infty \mu^*(A_{i})\) (not necessarily disjoint).
Note that we often prove (2) and (3) simultaneously since they both hold iff if \(E \subseteq \cup_{i=1}^\infty A_{i}\) then \(\mu^*(E)\leq \sum_{i=1}^\infty \mu^*(A_{i})\).
The most common way of obtaining outer measures is to start with a family \(\mathcal{E}\) of elementary sets on which a notion of measure is defined (such as rectangles in the plane) and then to approximate arbitrary sets “from the outside” by countable unions of members of \(\mathcal{E}\). We state this more precisely in the following proposition.
Let \(\mathcal{E}\subset 2^X\) and \(\rho:\mathcal{E}\to[0,\infty]\) be such that \(\emptyset \in\mathcal{E}\), \(X\in\mathcal{E}\) and \(\rho(\emptyset)=0\). For any \(A\subset X\) we can define an outer measure by \[ \mu^*(A)=\inf\left\{ \sum_{j=1}^\infty \rho(E_{j}):E_{j}\in\mathcal{E},~A\subset \bigcup_{j=1}^\infty E_{j}\right\}. \]
PROOF: For any \(A \subset X\) there exists \(\{ E_{j} \}_{j=1}^\infty \subset\mathcal{E}\) such that \(A\subset \bigcup_{j=1}^\infty E_{j}\) (take \(E_{j}=X\) for all \(j\)) and so the definition of \(\mu^*\) makes sense. We proceed to show this satisfies the properties of outer measure.
(1) Firstly note that clearly \(\mu*(\emptyset)=0\) by taking \(E_{j}=\emptyset\) for all \(j\).
(2) Next for \(A\subset B\) we have that \(\mu^*(A)\leq \mu^*(B)\) since the set over which the infimum is taken in the definition of \(\mu^*(A)\) includes the corresponding set in the definition of \(\mu^*(B)\). (3) To prove countable subadditivity, suppose \(\{ A_{j} \}_{j=1}^\infty \subset2^X\) and fix \(\varepsilon>0\). For each \(j\) there exists \(\{ E_{j}^k \}_{k=1}^\infty \subset\mathcal{E}\) such that \(A_{j}\subset \bigcup_{k=1}^\infty E_{j}^k\) and \(\sum_{k=1}^\infty \rho(E_{j}^k)\leq \mu^*(A_{j})+\varepsilon{2}^{-j}\). But then, if \(A=\bigcup_{i=1}^\infty A_{j}\), we have \(A\subset \bigcup_{j,k=1}^\infty E_{j}^k\) and \(\sum_{j,k}\rho(E_{j}^k)\leq \sum_{j}\mu^*(A_{j})+\varepsilon\) whence \(\mu^*(A)\leq \sum_{j}\mu^*(A_{j})+\varepsilon\). Since \(\varepsilon\) is arbitrary, we are done. \(\square\)
Intuitively, we are defining the outer measure of the set \(A\) to be the infimum of all sums of collections of sets that cover \(A\). An important outer-measure in real analysis and Probability Theory is the Lebesgue outer-measure \(\lambda^*\).
The Lebesgue outer measure \(\lambda^*:2^\mathbb{R}\to [0,+\infty]\) is defined \[ \lambda^*(A)=\inf\left\{ \sum_{i=1}^\infty \lvert b_{i}-a_{i} \rvert :A \subseteq \bigcup_{i=1}^\infty (a_{i},b_{i}] \right\}. \]
Clearly from the previous result the function is an outer measure. We can also show: (1) \(\lambda^*((a,b])=b-a\), \(a\leq b\); and (2) \(\lambda^*\) is translation invariant. Importantly, it is not countable additive however, we will how that it becomes countably additive when restricted to “nice enough” sets.
Given any outer-measure \(\mu^*\), a set \(A \subseteq X\) is (outer) \(\mu^*\)-measurable if \(\forall E \subseteq X\) we have \[ \mu^*(E)=\mu^*(E \cap A)+\mu^*(E \cap A^c). \] We denote the set of outer-measurable sets \(\mathcal{F}_{\mu^*}:=\{ A\subseteq X:A\text{ is }\mu^*\text{-measurable} \}\). Intuitively, we say \(A\) breaks apart any set nicely.
Remark: Note that by countable subadditivity \(\leq\) holds for all \(A \subseteq X\) hence we need only show \(\geq\) for equality.
For any outer measure \(\mu^*\) if \(\mu^*(B)=0\) then \(B \in \mathcal{F}_{\mu^*}\).
PROOF: We very quickly see that \[ \mu^*(E)\geq \mu^*\left( E \cap B\right)+\mu ^*(E \cap B^c)=\mu^*(E\cap B^c). \] \(\square\)
The next key theory from Carathéodory shows that if we are given an outer-measure \(\mu^*\), then the set of outer-measurable sets \(\mathcal{F}_{\mu^*}\) is a \(\sigma\)-algebra and when restricted to this \(\sigma\)-algebra, \(\mu^*\) becomes a measure.