# Different definitions of Borel sigma-algebra

Let $$X$$ be a locally compact Hausdorff space. Denote by $$\mathcal{B}(X)$$ the Borel $$\sigma$$-algebra (generated by all open sets in $$X$$) and by $$\mathcal{B}_c(X) \subseteq \mathcal{B}(X)$$ the collection of all relatively compact Borel sets. Then $$\mathcal{B}_c(X)$$ forms a $$\delta$$-ring.

For a ring $$\mathcal{R}$$ on some set $$S$$ denote by $$\mathcal{R}^{loc} := \{ E \subseteq S \mid E \cap A \in \mathcal{R} \textrm{ for all } A \in \mathcal{R} \}$$ the collection of all sets $$E \subseteq S$$ that are locally in $$\mathcal{R}$$. If $$\mathcal{R}$$ is a $$\delta$$-ring then $$\mathcal{R}^{loc}$$ is a $$\sigma$$-algebra that contains $$\mathcal{R}$$.

Back to the space $$X$$. The $$\sigma$$-algebra $$\mathcal{B}_c(X)^{loc}$$ contains all open sets of $$X$$ and therefore $$\mathcal{B}(X) \subseteq \mathcal{B}_c(X)^{loc}$$ [Dinculeanu, "Vector Measures", p. 291]. $$\mathcal{B}_c^{loc}(X)$$ is also sometimes referred to in the older literature as the $$\sigma$$-algebra of Borel sets. (It is useful in the context of vector measures, because the total variation measure of a vector measure can be defined on all of $$\mathcal{B}_c(X)^{loc}$$.)

I am searching for an example of a locally compact Hausdorff space $$X$$ such that $$\mathcal{B}(X) \subsetneq \mathcal{B}_c(X)^{loc}$$, i.e. these two definitions of a Borel $$\sigma$$-algebra are different. Equivalently, is there a locally compact Hausdorff space $$X$$ and a non-Borel measurable $$E \not\in \mathcal{B}(X)$$ but such that $$E \cap K \in \mathcal{B}_c(X)$$ (i.e. is Borel-measurable) for all compact $$K \subseteq X$$?

Edit: Just an idea: Consider the open ordinal space $$X = [0, \omega_1)$$ where $$\omega_1$$ is the first uncountable ordinal. Then $$X$$ is locally compact but not $$\sigma$$-compact. The sets $$[0, \alpha]$$, $$0 \leq \alpha < \omega_1$$ are compact sets and any compact set $$K \subseteq [0, \omega_1)$$ is contained in some $$[0, \alpha]$$. If $$E \in 2^{[0, \omega_1)} \setminus \mathcal{B}[0, \omega_1)$$ is a non-Borel measurable set then $$E \cap [0, \alpha]$$ is a Borel measurable subset of $$[0, \alpha]$$ for all $$\alpha < \omega_1$$ since $$\alpha$$ is a countable ordinal and $$\mathcal{B}[0, \alpha] = 2^{[0, \alpha]}$$. Hence $$E \cap [0, \alpha]$$ is a Borel measurable subset of $$[0, \omega_1)$$. So the question is: Is $$\mathcal{B}[0, \omega_1)$$ strictly smaller than the power set $$2^{[0, \omega_1)}$$?

As mentioned in the question, consider the first uncountable ordinal $$X = [0, \omega_1)$$. This is a locally compact Hausdorff space. The Borel $$\sigma$$-algebra $$\mathcal{B}([0,\omega_1))$$ is the collection of all sets $$A \subseteq [0, \omega_1)$$ such that $$A$$ or its complement $$A^c$$ contain a closed cofinite set [Fremlin, "Measure Theory", 4A3J]. In ZFC, one can show that there exists a subset $$E \subseteq [0, \omega_1)$$ that is not Borel-measurable (see Non-measurable subset of $\omega_1$). If $$K \subseteq [0, \omega_1)$$ is compact then $$K$$ is contained in some compact interval $$[0, \alpha] \subseteq [0, \omega_1)$$ for $$\alpha < \omega_1$$. Since $$\alpha$$ is a countable ordinal, the set $$[0, \alpha]$$ is countable and it follows that $$\mathcal{B}([0, \alpha]) = 2^{[0, \alpha]}$$. Moreover, $$\mathcal{B}_c([0, \omega_1)) = \bigcup_{K \subseteq [0, \omega_1) \textrm{ compact}} \mathcal{B}(K) = \bigcup_{\alpha < \omega_1} \mathcal{B}([0, \alpha]) = \bigcup_{\alpha < \omega_1} 2^{[0, \alpha]}$$ by [Dinculeanu, "Vector Measures", III.§14.1, Proposition 4]. Hence, $$E \cap K \in \mathcal{B}_c([0, \omega_1))$$ for all compact $$K \subseteq [0, \omega_1)$$. By [Dinculeanu, "Vector Measures", III.§14.2, Proposition 5] it follows that $$E \in \mathcal{B}_c([0, \omega_1))^{loc}$$, but $$E \not\in \mathcal{B}([0, \omega_1))$$.