TITLE: Show that $\{(r\cos t,r\sin t)\mid 0\leq t<\theta,r\geq 0\}$ is a borel set where $\theta$ is fixed
QUESTION [1 upvotes]: Let $0<\theta \leq 2\pi$. Define the set $A=\{(r\cos t,r\sin t)\mid 0\leq t<\theta,r\geq 0\}$. I want to show that it is a borel set. I have drawn a figure of the set, but I am not sure if it is correct (it looks like that the domain lies in the circular sector depending on the angle $t$). At first I thought about finding an open ball that lies in $A$, but the restriction of $t$ confuses me. Is there a way of rewriting this set in terms of the intersection or union of open/closed sets?

REPLY [0 votes]: $A$ is an $F_{\sigma}$ set.  Let $S_q=\{(r\cos t,r\sin t): t\in [0,q]\land r\geq 0\}.$ 
Every $S_q$ is closed and $A=\cup \{S_q:q\in \Bbb Q \cap [0,\theta)\}.$
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