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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.

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What do we know about the topology of $X$? – T. Eskin Jun 11 '12 at 10:04
The X is a Locally Compact Hausdorff Space. – Sandra Jun 11 '12 at 11:04
I couldn't find this Lemma from Royden's book. What page is it? – T. Eskin Jun 11 '12 at 17:39
Found it now. And it's true that this is very questionable since no proof is given. If $X$ is $\sigma$-compact then the Lemma is true, as $E$ is a countable union of Borel sets. Let me see if I'm able to do something with the locally compact Hausdorff case. – T. Eskin Jun 11 '12 at 18:09
Thanks Thomas. I tried making a counterexample but seem not able to do one. Perhaps someone or you might be able to show me one. I anticipate your response. – Sandra Jun 11 '12 at 18:20
up vote 2 down vote accepted

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.

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I changed your link to the webpage of the author of the note you were referring to, I hope that's fine with you. There is indeed a nice counterexample to be found on page 5. Its construction starts on the line beginning with A bad apple and relies on the "explicit" transfinite construction of the $\sigma$-algebra generated by the open sets on $\omega_1 \times \mathbb R$, where $\omega_1$ is discrete. – Sam Jun 11 '12 at 21:12
@SamL.: Thank you. – T. Eskin Jun 11 '12 at 21:13

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.

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