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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., $\sigma(R)$.

Grateful for your input.

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2 Answers 2

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  • 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

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  • $\begingroup$ 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 $\mathcal{R}$ $\endgroup$ May 18, 2021 at 10:22
  • $\begingroup$ 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. $\endgroup$
    – Mittens
    May 18, 2021 at 14:33
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By definition any algebra on $\Omega$ contans $\Omega$. So $A_n=\Omega$ for each $n$ gives one such choice.

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