2.5 Corollary. A function $f: X \rightarrow \mathbb{C}$ is $\mathcal{M}$ -measurable iff $\operatorname{Re} f$ and $\operatorname{Im} f$ are $\mathcal{M}$-measurable.

Proof. This follows since $\mathcal{B}_{\mathbb{C}}=\mathcal{B}_{\mathbb{R}^{2}}=\mathcal{B}_{\mathbb{R}} \otimes \mathcal{B}_{\mathbb{R}}$ by Proposition $1.5 .$

In Corollary 2.5, why are we allowed to say that $\mathcal{B}_{\mathbb{C}}=\mathcal{B}_{\mathbb{R}^{2}}$? How could a set of complex numbers be in $\mathcal{B}_{\mathbb{R}^2}$? Is this a different notion of equality used here?

The following are referenced propositions:

enter image description here

  • $\begingroup$ Did you mean Borel when you typed bore and boreal in the title? $\endgroup$ Jul 21, 2021 at 18:57
  • $\begingroup$ @J.W.Tanner Thanks! I typed that on an iPad $\endgroup$ Jul 21, 2021 at 18:58
  • $\begingroup$ "Boreal" could be interesting, too! :) $\endgroup$ Jul 21, 2021 at 19:14

2 Answers 2


As metric spaces, $\mathbb C$ and $\mathbb R^2$ are isomorphic (via $a+bi\mapsto(a,b)$). If we denote the isomorphism by $\varphi:\mathbb C\to\mathbb R^2$, then the corollary really says $\mathcal B_{\mathbb C}=\{\varphi(S)|S\in\mathcal B_{\mathbb R^2}\}$.

But such isomorphisms (surjective isometries? I don't know of a better name) preserve properties of metric spaces, so usually isomorphic things are identified (as an abuse of notation).

  • $\begingroup$ Hmm, I know not everything is preserved via isomorphisms (Like boundedness, etc.). I wonder if anything is not preserved under isometric isomorphisms between metric spaces? $\endgroup$
    – Alan
    Jul 21, 2021 at 19:10
  • $\begingroup$ I meant isometric isomorphism in this answer. What do you mean exactly by (non-isometric) isomorphisms of metric spaces? $\endgroup$ Jul 21, 2021 at 19:13
  • $\begingroup$ Bleh, I meant homeomorphisms (That preserve topology but not everything), not isometries. $\endgroup$
    – Alan
    Jul 21, 2021 at 19:14
  • $\begingroup$ Oh, I see. In that case boundedness, completeness, etc. won't be preserved. On the other hand isometric isomorphisms should be good enough for metric spaces. $\endgroup$ Jul 21, 2021 at 19:18
  • $\begingroup$ You can say more: $\mathbb C$ is exactly equal to $\mathbb R^2$ as sets. $\endgroup$
    – zhw.
    Jul 21, 2021 at 19:34

The more substantive assertion is not really about real versus complex numbers, but, for example, that a function $f(x)=(f_1(x),\ldots,f_n(x)):X\to \mathbb R^n$ is measurable if and only if each of the component functions $f_j$ is measurable. (Where each of these maps $X\to\mathbb R$, as suggested by the notation.)

This follows by the same argument as in your cited source.

The specific relevance to complex-valued functions is via the identification of $\mathbb C$ with $\mathbb R^2$ by $x+iy\to \mathbb R^2$, which is a homeomorphism. Since it's a homeomorphism, it will map Borel sets in the one to the Borel sets in the other...


You must log in to answer this question.

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