True if sigma-compact Halo to readers of this post, I would like to enquire from the book of Royden's Real Analysis book, there is a lemma which I believe is wrong but am not certain. Perhaps someone can how me a counter-example?

Let $X$ a locally compact Hausdorff space and $E$ be a subset of $X$ such that $E \bigcap K$ is a Borel Set for each compact set $K$. Then $E$ is a Borel set.

I think this is only true if $X$ is sigma-compact. Please correct me if I am wrong.
 A: I think $\omega_1$ should supply a counterexample.  Compact sets are countable, so every set $E$ has the property that $E \cap K$ is Borel for all compact $K$.  But I'm pretty sure $\omega_1$ contains non-Borel sets (assuming enough choice), though I don't know a proof off the top of my head.
A: As with Nate, I'm leaning towards the side that Royden's Lemma is false.
Look at page 5 in the Notes on measure and integration in locally compact spaces written by William Arveson from Berkeley. The document can be found on his homepage by following the link "papers" and scrolling down to the section titled "Lecture notes and snippets". Here is the direct download link.
It seems that they refer to Royden's Lemma and say that it is misleading, providing seemingly a counter-example for someone (with more knowledge than me) to verify it if possible.

However, a weaker result does hold, which I believe Royden might have assumed to imply the above result since the Borel $\sigma$-algebra is generated by closed sets.
Namely, that if $X$ is a locally compact Hausdorff space and $F\subset X$ is a set for which $F\cap K$ is closed for all compact $K\subset X$, then $F$ is closed. I believe that it can be shown with the following argument:
Assume the contrary that $F$ is not closed. Then we find $x\in\bar{F}\setminus F$ where $\bar{F}$ denotes the closure of $F$ in $X$. In particular, this means that $U\cap F\neq \emptyset$ for all open neighborhoods $U$ of $x$. Since $X$ is locally compact, we find an open neighborhood $U_{x}$ of $x$ so that $\bar{U_{x}}$ is compact. Thus $V:=\bar{U_{x}}\cap F\neq\emptyset$ and $V$ is closed by our assumption. Since $x\notin F\supset V$ then $x\notin V$, and since $V$ is closed its complement is open. Thus we find an open neighborhood $W$ of $x$ so that $x\in W\subset V^{c}$. We then take $G=U_{x}\cap W$ as an open neighborhood of $x$, whence
\begin{align*}
G=U_{x}\cap W\subset U_{x}\cap V^{c}=U_{x}\cap(\bar{U_{x}}\cap F)^{c}=U_{x}\cap(\bar{U_{x}}^{c}\cup F^{c})=U_{x}\cap F^{c}.
\end{align*}
But this implies that $G\cap F\subset U_{x}\cap F^{c}\cap F=\emptyset$, which is a contradiction since $x\in\bar{F}$ and $G$ is an open neighborhood of $x$. Thus $F$ must be closed.
