# Borel $\sigma$-algebra on non- second countable topological space

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.)

-
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

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 $\mathcal{B} := \bigcap_{C \in \mathcal{C}} C$.

-

I found this post because I wondered something related, and I thought I'd post an observation here:

Yes, defining the Borel $\sigma$-algebra goes through trivially. But I think certain fundamental, "nice" theorems don't. For example, theorem 1.8 of big Rudin, page 11, says that if $u,v: X\to\mathbb{R}$ are measurable then so is $u\times v: \mathbb{R}^2$. The argument is as follows.

Given an open set $\Omega\subset\mathbb{R}^2$, the inverse image of any rectangle $(a_1,b_1)\times (a_2,b_2)$ is just $u^{-1}((a_1,b_1))\times v^{-1}((a_2,b_2))$, which is measurable in $X$. Any open $\Omega$ can be written as a countable union of rectangles (here the second countability is used) and so $(u\times v)^{-1}(\Omega)$ is the countable union of these measurable sets, hence is measurable.

This proof goes through precisely the same, replacing $\mathbb R$ with any second countable space. My sense of urgency tells me not to try to construct a counterexample for the long line to say that second countability is needed, but that might be an interesting question.

-