This is from Sheldon Axler's text on Measure and integration: enter image description here

So we have defined the borel sigma algebra on the unit disk to be the pullback of the sigma algebra on $(-\pi, \pi]$ . However, the unit disk itself has a topology defined as the subspace topology on the complex plane, so if we take those open sets and generate the Borel sigma algebra from that, do we get the same Borel algebra as the pullback?

Also, it was not explicitly stated how we define a continuous function on the unit disk, but I assume it is defined using the subspace topology on the unit disk, so if that is the case, then these two different sigma algebras should coincide right?

  • $\begingroup$ Note in several places you say "disk" where you mean "circle"... $\endgroup$ Sep 9, 2022 at 22:05
  • $\begingroup$ Yes, they are the same. $\endgroup$
    – Mittens
    Dec 7, 2022 at 15:15

1 Answer 1


Yes it is. Let $\mathcal{F}$ denote the Borel $\sigma$-algebra as defined by Sheldon and $\mathcal{B}$ be the usual Borel $\sigma$-algebra defined as the minimal $\sigma$-algebra containing all open sets.

Now let $U\subseteq \partial D$ be open, and look at $\pi^{-1}(U)$ where $\pi:\mathbb{R}\rightarrow \partial D$ is the map $t\mapsto e^{it}$. Since $\pi$ is continuous, $\pi^{-1}(U)$ is open and therefore measurable in $\mathbb{R}$. We deduce that $U\in \mathcal{F}$ and so by minimality of the Borel $\sigma$-algebra we already have $\mathcal{B}\subseteq \mathcal{F}$.

To prove the other inclusion, let $E\in \mathcal{F}$ be measurable. This exactly means that $\pi^{-1}(E)$ is measurable in $\mathbb{R}$. Write $E = \bigcup E_n$ where $E_n = \pi^{-1}(E)\cap [\pi n, \pi(n+1)]$. Clearly, all the $E_n$'s are Borel measurable subsets of $\mathbb{R}$ (as intersection of Borel sets).

Now, let $\pi_n$ be the restriction of $\pi$ to $[\pi n,\pi(n+1)]$ is continuous and injective map from a compact Hausdorff space to a compact Hausdorff space (the image is a closed half hemisphere). Therefore, $\pi$ is an homeomorphism. In particular, $\pi_n^{-1}$ is continuous. As a result we see that $\pi_n(E_n)=(\pi_n^{-1})^{-1}(E_n)$ is a Borel subset of $\partial D$ (because the pre-image of a Borel set under a continuous function is Borel). Now since $E = \bigcup_n \pi(E_n)$, we deduce that $E$ is also Borel measurable. This completes the proof.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .