# Caratheodory continuation theorem: existence of the sequence for the definition of the extended measure

I was trying to understand the proof for the Caratheodory's continuation theorem: any measure on an algebra $$R$$ has a unique continuation to a measure on $$\sigma(R)$$.

The proof suggested an extension of a measure $$\mu$$ defined on $$R$$ as $$\lambda(A) = \inf \left( \sum_{n \in \mathbb{N}} \mu(A_n)\right)$$ defined on $$P(\Omega)$$, which is the power set of $$\Omega$$, the set of outcomes, and $$A_n$$ is a sequence in $$R$$ satisfying $$A \subset (\cup A_n)$$.

I failed to understand why such sequence of $$A_n$$ exists, because $$A_n$$ is supposed to be elements of $$R$$ while $$P(\Omega)$$ is the power set of $$\Omega$$. For example, if $$\Omega$$ is the outcome of dice roll: $$\Omega = {1,2,3,4,5,6}$$, and I take $$R = \{ \emptyset, \{1, 3,5\}, \{2,4,6\}, \Omega \}$$, then $$A = \{1,3\}$$ is a valid subset of $$P(\Omega)$$ but $$A$$ cannot be constructed by taking unions of sets in $$R$$.

Maybe I have misunderstood something but it is unclear for me why most proofs work with power sets, not with e.g., $$\sigma(R)$$.

• In the case where $$\mathcal{R}$$ is an algebra, for any $$A\subset \Omega$$ there is always sequence $$\{A_n:n\in\mathbb{N}\}\subset\mathcal{R}\}$$ covering $$A$$ as $$\Omega\in\mathcal{R}$$.

• The situation that arises in your posting may appear when $$\mathcal{R}$$ is a ring or a semiring, and $$\mu$$ is a finitely additive and countably semi additive function on $$\mathcal{R}$$ . If there is no sequence $$\{A_n:n\in\mathbb{N}\}\subset\mathcal{R}$$ that covers $$A$$, then recall that $$\inf\emptyset:=\infty$$, in which case, $$\lambda(A)=\inf\{\sum_n\mu(A_n):\,A_n\in\mathcal{R},\,A\subset\bigcup_nA_n\}=\infty$$

• Here, additive means that if $$A,B\in\mathcal{R}$$, $$A\cap B=\emptyset$$, and $$A\cup B\in\mathcal{R}$$, then $$\mu(A\cup B)=\mu(A)+\mu(B)$$; countably sub additivity means that if $$(A_n:n\in\mathbb{N})\subset\mathcal{R}$$, $$A_n\cap A_m=\emptyset$$ if $$n\neq m$$, and $$\bigcup_nA_n\in\mathcal{R}$$, then $$\mu(\bigcup_nA_n)\leq \sum_n\mu(A_n)$$; $$\mathcal{R}$$ is a semiring if $$I,J\in\mathcal{R}$$ implies that $$I\cap J\in\mathcal{R}$$, and $$I\setminus J$$ is the finite union of elements in $$\mathcal{R}$$. A careful study of the outer measure $$\mu^*(A)= \inf\{\sum_n\mu(A_n):\,A_n\in\mathcal{R},\,A\subset\bigcup_nA_n\}$$ shows that $$\mu$$ can be extended to a measure $$\mu'$$ defined on a $$\sigma$$-algebra that contains $$\sigma(\mathcal{R})$$. In fact, any other extension $$\nu$$ of $$\mu$$ satisfies $$\nu\leq \mu'$$ on $$\sigma(\mathcal{R})$$. But this is a matter for other discussions.

• Rings and semiring appear naturally in applications. For example, in the construction of the Lebesgue integral, consider $$\mathcal{R}$$ to be the collection of finite union of intervals $$\{(a,b]:a,b\in\mathbb{R}, a\leq b\}$$, $$\mu((a,b])=b-a$$. Here $$\mathcal{R}$$ is neither an algebra not a ring but a semiring. The method of Cartheodory produces the Lebesgue Measure here.

• One last example: Suppose $$\Omega=\{1,2,3,4,5,6,7\}$$, $$\mathcal{R}=\{\emptyset,\{1,3,5\},\{2,4,6\}\}$$, and $$\mu(\emptyset)=0$$, $$\mu(\{2,4,6\})=\frac12=\mu(\{1,3,5\}).$$ $$\mathcal{R}$$ is a semiring and generates the $$\sigma$$-algebra $$\sigma(\mathcal{R})=\{\emptyset,\{1,3,5\},\{2,4,6\},\{1,3,5,7\},\{2,4,6,7\},\{1,2,3,4,5,6\},\{7\},\Omega\}$$. The function $$\mu$$ is of course additive and countably subadditive on $$\mathcal{R}$$. The extension $$\mu'$$ of $$\mu$$ using Caratheodory's construction gives $$\mu'(A)=\infty$$ for any set $$A\in\sigma(\mathcal{R})$$ that contains $$7$$. The extension $$\nu$$ of $$\mu$$ that assigned $$\nu(\{7\})=0$$ yields a probability measure on $$\sigma(\mathcal{R})$$ The measure provided by Charatheodory is unique in the sense that is maximal: If extension $$\nu$$ of $$\mu$$ from $$\mathcal{R}$$ to $$\sigma(\mathcal{R})$$ satisfies $$\nu\leq \mu'$$ on $$\sigma(\mathcal{R})$$. But this is a matter for other discussions

• Super! I didn't expect that the measurement will be $\infty$ for sets that cannot be "constructed" from the union of sets in the algebra $\mathcal{R}$ May 18, 2021 at 10:22
• That is what you get from the Caratheodory construction, which gives you a maximal measure (any other extension $\nu$ of $\mu$ satisfies $\nu\leq\mu'$. notice that in the last example, there are uncountably many ways to extend $\mu$ to a measure: $\nu_a(\{7\})=a$ with $0\leq a\leq\infty$. $\nu_{\infty}$ is the one that Caratheidory gives. May 18, 2021 at 14:33

By definition any algebra on $$\Omega$$ contans $$\Omega$$. So $$A_n=\Omega$$ for each $$n$$ gives one such choice.