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.

Let $X$ be a locally compact (Hausdorff) space, and suppose I can write $X$ as the disjoint union of a family of open subsets $(X_i)_{i\in I}$ (so each $X_i$ is actually clopen). A typical example I have in mind is an uncountable disjoint union of copies of $\mathbb R$ (so I am certainly not wishing so assume $X$ is $\sigma$-finite, etc.)

Let's be careful: let $B$ be the Borel $\sigma$-algebra on $X$, so this is the intersection of all $\sigma$-algebras containing the open sets. For each $i$, treat $X_i$ as a locally compact space in its own right, and let $B_i$ be the Borel $\sigma$-algebra. Let $E\subseteq X$ be such that $E\cap X_i\in B_i$ for each $i$. Is $E\in B$?

The converse is true: if $\Omega=\{ E\subseteq X : \forall i, E\cap X_i\in B_i\}$ then $\Omega$ is a $\sigma$-algebra, and $\Omega$ contains all the open subsets of $E$, so $B\subseteq\Omega$. I want to show that $\Omega=B$, which seems much harder??

share|improve this question
    
Is there any other assumption on $X$? Something like second-countable (or that the decomposition into $X_i$ gives each $X_i$ second-countable)? –  Asaf Karagila May 18 '11 at 17:55
    
I'm certainly happy to just assume that $X$ is indeed an uncountable disjoint union of copies of $\mathbb R$; so each $X_i$ can be "nice", but there are a lot of them! –  Matthew Daws May 18 '11 at 19:05
add comment

1 Answer

up vote 3 down vote accepted

There are plenty of spaces for which (using your notation) $\Omega$ is strictly larger than $B$, and one such space is $\omega_1 \times \mathbb{R}$ (with the topology you described). To build a set in $\Omega \backslash B$, just paste a set of Borel rank $\alpha$ in the $\alpha^{\mathrm{th}}$ copy of $\mathbb{R}$.

share|improve this answer
    
Thanks-- I guess the key point is that $A$ is of Borel rank $\alpha$ in $X$ if and only if $A\cap X_i$ is Borel rank $\alpha$ in $X_i$; so if $A\cap X_i$ can have arbitrarily high rank, it's not possible for $A$ to be Borel (else $A$ would have countable rank...) Do you happen to know a reference for this, or is it just folklore? –  Matthew Daws May 18 '11 at 19:22
    
Sorry, I don't know of any reference for this. Seems pretty folklorical to me (or maybe something that would appear in the exercises of a standard text?). –  user92843 May 18 '11 at 19:58
add comment

Your Answer

 
discard

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.