# Borel sets and clopen decompositions

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??

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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

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 $\mathbb{R}$.
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