Inverse of the function $f(\theta) =\exp(i\theta)$ is Borel measurable

Question

Let $\Lambda$ denotes the circle in the Complex plane, that is $\Lambda:=\{\exp(i\theta): \theta \in \mathbb R\}$. Now consider the function $f:[0,2\pi) \to \Lambda$ defined by the assignment $$f(\theta)=\exp(i\theta),~~\text{for all }~ \theta \in [0,2\pi).$$ Now notice that $f$ is continuous and also bijective. Let $f^{-1}$ denotes the inverse of $f$. I want to show that $f^{-1}$ is Borel measurable. Please help me to solve this. Means, how can I show that inverse image of an open set under $f^{-1}$ is Borel. Thank you.

Answer 1

Let $C$ be a closed subset of $[0,2\pi)$ and $C_n =C \cap [0,2\pi-\frac 1 n]$. Then $C_n$ is compact and $C =\bigcup_n C_n$. Since $f$ is continuous it follows that $f(C)=\bigcup_n f(C_n)$ is a countable union of compact sets, hence a Borel set. For any open set $U$ in $[0,2\pi)$ we have $(f^{-1})^{-1} (U)=f(U)=[f(U^{c})]^{c}$ which is Borel since $U^{c}$ is closed in $[0,2\pi)$.