I remember hearing some time ago that there is a locally compact Hausdorff space $X$ and a non-Borel subset $E$ which intersects every compact set in a Borel set. (This would contradict Lemma 13.9 of Royden, Real Analysis 3rd edition 1988, which is stated without proof).


Is there a reference for this? Can this happen if the space $X$ is perfectly normal?



non-Borel set which intersects every compact in a Borel set
If I understand the question, then you are correct, there is such a space. I'll sketch what I hope is a correct argument.


Take $X=\coprod\_AY$ for some fixed locally compact Hausdorff space $Y$ and some index set $A$. As long as $Y$ is sufficiently complicated (probably $Y=\mathbb R$ would work) and $A$ is sufficiently large ($\left|A\right|\geq\aleph\_1$ is enough), then you can find a collection of Borel sets $\{B\_\alpha\subseteq Y\}\_{\alpha\in A}$ which are not all contained in any countable "stage" towards the Borel $\sigma$-algebra of $Y$ (I'm sorry I don't know the standard terminology for this; using the notation of [the wikipedia entry on Borel sets](http://en.wikipedia.org/wiki/Borel_set), I mean that for any countable ordinal $m$, not all of the $B\_\alpha$ are contained in $G^m$). Now the subset:
$$\coprod\_{\alpha\in A}B\_\alpha\subseteq\coprod\_{\alpha\in A}Y$$
is not a Borel set. If it were, it would be contained in some $G^m(X)$ for some countable ordinal $m$, but this would imply that every $B\_\alpha$ is in $G^m(Y)$, a contradiction.


If we take $Y$ metrizable, then $X$ will be metrizable as well, and hence perfectly normal.


