21/10/24
Reference: Lecture notes 3.3, 3.4
Lecture 7 Caratheodory theorem II.pdf
Carathéodory theorem allows us to extend an outer measure to a measure defined on a $\sigma$-algebra of sets. As we discussed, outer measures are easy to construct: just pick a collection of sets $\mathcal{S} \subset P(X)$ and a function $\rho \colon \mathcal{S} \to [0,\infty]$, then define for any $A \subset X$
$\mu^*(A) = \inf \{ \sum_{n=1}^\infty \rho(E_n) \ | \ E_n \in \mathcal{S} \text{ and } A \subset \bigcup_{n=1}^\infty E_n \}$
The set $\mathcal{A}$ of $\mu^$-measurable sets is a $\sigma$-algebra and the restriction $\mu$ of $\mu^$ to $\mathcal{A}$ is a measure. However, in general the sets in the original collection $\mathcal{S}$ are not $\mu^$-measurable and it is not necessarily true that $\mu^(A)= \rho(A)$ for all $A \in \mathcal{S}$.
Example. Let $X=\{a,b\}$ and $\mathcal{S} = \{\emptyset, \{a\}, \{a,b\}\}$. Define
$\rho(\emptyset)=0, \ \rho(\{a\})=2, \ \rho(\{a,b\}) = 1$
Then $\mu^(\{a\})=1 \neq \rho(\{a\})$ because $\{a\}$ is covered by $\{a,b\}$. Similarly, $\mu^(\{b\})=1$. This implies that $\{a\}$ is not $\mu^*$-measurable because
$\mu^(\{a,b\}) \neq \mu^(\{a\}) + \mu^*(\{b\})$.
We see from this example what the problem is. In the above construction we imposed no constraints on $\rho$, so that it can be very far from what we imagine a measure is. In particular, in this case $\rho(\{a,b\})<\rho(\{a\})$ even though $\{a\} \subset \{a,b\}$. The solution is to impose more conditions of $\mathcal{S}$ and $\rho$ that are more compatible with the properties of measures.
Definition. A collection of sets $\mathcal{S}\subset P(X)$ is a semiring if
$\emptyset\in\mathcal{S}$,
If $A,B\in\mathcal{S}$, then $A\cap B\in\mathcal{S}$,
If $A,B\in\mathcal{S}$, then there exist pairwise disjoint $D_1,\ldots, D_n \in \mathcal{S}$ such that
$A\setminus B = \bigcup_{n=1}^ND_n$.
Example. Let $X=\R^n$. The collection of all half-open $n$-dimensional cubes
$Q = [a_1,b_1) \times \cdots [a_n,b_n)$
is a semi-ring. Verify this as an exercise.
Example. In our previous example $X=\{a,b\}$, the collection $\mathcal{S} = \{\emptyset, \{a\}, \{a,b\}\}$ is not a semiring.