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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 semiring 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 semiring 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.

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..
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..