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 $2^{[0, \omega_1)}$?
Best Answer
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 $E \not\in \mathcal{B}([0, \omega_1))$.