Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is it possible to define a Borel $\sigma$-algebra on a topological space which is not second countable, i.e. one which does not have a countable base?

I am trying to learn measure theory and my intuition says that this is NOT possible, because if the space is not second countable, then there are "too many" open sets to be able to get closure taking countable unions (etc.) of open sets.

(The book by Bogachev (and other books) did not help me understand this any better, which is why I am asking here.)

share|improve this question
Isn't the Borel $\sigma$-algebra on a topological space $X$ just the $\sigma$-algebra generated by all open sets? That is, the intersection by $\sigma$-algebras containing the open sets. Why wouldn't that work if the space is not second countable? There is always one of those, namely the power set of $X$. If the space is not second countable, it might not be so useful for other reasons. –  Jonas Teuwen Oct 26 '11 at 19:43
add comment

1 Answer 1

up vote 4 down vote accepted

Yes, it is possible, no problems.

Let $X$ be a space with topology $T$. Let $\{A_\alpha\}_\alpha$ be any (possibly uncountable) collection of $\sigma$-algebras on $X$.

Consider $\hat A := \bigcap_\alpha A_\alpha$. It's easy to check this again forms a $\sigma$-algebra.

In particular, consider the collection $\mathcal{C}$ of all $\sigma$-algebras containing $T$. Surely the power set $\mathcal{P}(X)$ is a $\sigma$-algebra containing $T$, so $\mathcal{C}$ is not empty.

We now define the Borel $\sigma$-algebra on $(X,T)$ to be $\mathcal{B} := \bigcap_{C \in \mathcal{C}} C$.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.