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A quick question about the definition of the Borel $\sigma$-algebra. It was defined to me as the smallest $\sigma$-algebra generated by all open intervals.

What exactly does generated by refer to in rigorous terms?

I know an uncountable union of open sets is open, so it ought to be a Borel set, but does this follow from the definition (i.e. what is encompassed by generated by), or must some work be done to show that an uncountable union of open sets is a Borel set?

SSF
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    In rigorous terms, it means that the Borel $\sigma$-algebra equals the intersection of all $\sigma$-algebras containing the open intervals. Note that it also contains every open subset of $\mathbb R$ since an open set can be expressed as a countable (even pairwise disjoint, but that's not needed here) union of open intervals. See here, for example. –  Nov 13 '17 at 23:55
  • Alright. In that case I think some work must be done.

    Let A be an uncountable union of open sets. I want to show it is a Borel set, so I must show that it is in every $\sigma$-algebra containing the open intervals.

    Let X be a $\sigma$-algebra containing the open intervals. Then I claim ANY open set must be in X, so in particular A is in X.

    The claim follows because (as you say) any open set can be expressed as a countable union of open intervals, and is therefore in X. We're done.

    Thanks! I got the answer to my question -- the result doesn't follow directly from the definition.

     – SSF Nov 14 '17 at 00:01
  • An uncountable union of open sets is an open set. This is a topological fact that has nothing to do with $\sigma$-algebras. So no extra work is needed. If $A$ is an uncountable union of open sets, then it's an open set, hence it's a Borel set. –  Nov 14 '17 at 00:03
  • Oh, hmm. I would say that extra work has to be done to show that any open set can be expressed as a countable union of open intervals. No? – SSF Nov 14 '17 at 00:05
  • Yes, that's where the work is. The proof isn't too hard - see the link I gave in the first comment. It should also be in almost any elementary real analysis text. –  Nov 14 '17 at 00:06

2 Answers2

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Every open Set in R is the countable union of open intervals of the form (q-1/n,q+1/n) where q is rational and n is a positive interval .This show that every open set is a Borel set according to your definition .

user439545
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Let $\mathcal{O}$ be the $\sigma$-algebra containing all open intervals, then the Borel algebra $\mathcal{B}$ is defined by $$\mathcal{B}:=\langle \mathcal{O}\rangle=\bigcap_{I \in \mathcal{O}} I.$$ That is, $\mathcal{B}$ is the “smallest” $\sigma$-algebra containing the closed intervals.

Here $\langle {\mathcal{O}}\rangle$ is the set generated by the elements $I \in \mathcal{O}$.

James Garrett
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