Extending a pre measure

[fishturtle1]

Homework Statement

Consider on $\mathbb{R}$ the family $\sum$ of all Borel sets which are symmetric w.r.t. the origin. Show that $\sum$ is a $\sigma$ algebra. Is it possible to extend a pre measure $\mu$ on $\sigma$ to a measure on $\mathcal{B}(\mathbb{R})$? If so, is this extension unique?

Relevant Equations

Definition: A semi-ring is a family $\mathcal{S} \subset \mathcal{P}(X)$ with the following properties:

1) $\emptyset \in \mathcal{S}$

2) $S, T \in \mathcal{S} \Rightarrow S \cap T \in \mathcal{S}$

3) For $S, T \in \mathcal{S}$ there exists finitely many disjoint $S_1, S_2, \dots, S_M \in \mathcal{S}$ such that $S \setminus T = \bigcup_{i=1}^M S_i$.

Theorem: (Carathéodory) Let $\mathcal{S} \subset \mathcal{P}(X)$ be a semi-ring and $\mu:[0, \infty]$ a pre measure, i.e. a set function with

i) $\mu(\emptyset) = 0$

ii) $(S_n)_{n\in\mathbb{N}} \subset \mathcal{S}$, disjoint and $S = \bigcup_{n\in\mathbb{N}}S_n \in \mathcal{S}\Rightarrow \mu(S) = \sum_{n\in\mathbb{N}} \mu(S_n)$.

Then $\mu$ has an extension to a measure $\mu$ on $\sigma(\mathcal{S})$. Moreover, if $\mathcal{S}$ contains an exhausting sequence $(S_n)_{n\in\mathbb{N}}$, $S_n \uparrow X$ such that $\mu(S_n) < \infty$ for all $n\in\mathbb{N}$, then the extension is unique.
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Attempt at solution: We have $\sum = \lbrace B \subset \mathbb{R} : b \in B \Rightarrow -b \in B \rbrace$. Clearly, $\emptyset \in \sum$. Let $B \in \sum$ and $a \in B^c$. Then $a \not\in B$ which implies $-a\not\in B$. So $-a \in B^c$ i.e. $B^c \in \sum$. Lastly, for any $(B_n)_{n\in\mathbb{N}} \subset \sum$, if $b \in \bigcup_{n\in\mathbb{N}}B_n$, then $b \in B_k$ for some $k$ which implies $-b \in B_k \subset \bigcup_{n\in\mathbb{N}}B_n$. This shows $\sum$ is a $\sigma$ algebra.

However, I don't think we can extend a measure on $\sum$ to $\mathcal{B}(\mathbb{R})$ only using the above theorem. My reasoning is because $\sigma(\sum) \neq \mathcal{B}(\mathbb{R})$. My questions are this:

1) How can I prove that $\sigma(\sum) \neq \mathcal{B}(\mathbb{R})$? I guessed this is true because I don't see how to show $[1, 2) \in \sigma(\sum)$.

2) The book's solution says that we can extend a pre measure on $\sum$ to a measure on $\mathcal{B}(\mathbb{R})$, but in theorem 6.1. we require $\sigma(\sum) = \mathcal{B}(\mathbb{R})$ for such an extension to exist? I think I am missing something..

 

[member 587159]

Since $\Sigma$ is a $\sigma$-algebra, we have $\sigma(\Sigma)=\Sigma$.

Thus trivially $\sigma(\Sigma)\subsetneq B(\mathbb{R})$.

 

[fishturtle1]

Since $\Sigma$ is a $\sigma$-algebra, we have $\sigma(\Sigma)=\Sigma$.

Thus trivially $\sigma(\Sigma)\subsetneq B(\mathbb{R})$.

Thank you, that makes sense.

I think things are cleared up. Carathéodory's theorem (the one in OP) gives us sufficient but not necessary conditions to extend a pre measure. So we don't necessarily have to meet the requirement of the theorem in order to extend a pre measure to a measure. And this answers my 2nd question, I think.