# How is it that Borel sets of $\mathbb{C}$ equal the Borel sets of $\mathbb{R}^2$

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:

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

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).

• 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?
– Alan
Jul 21, 2021 at 19:10
• I meant isometric isomorphism in this answer. What do you mean exactly by (non-isometric) isomorphisms of metric spaces? Jul 21, 2021 at 19:13
• Bleh, I meant homeomorphisms (That preserve topology but not everything), not isometries.
– Alan
Jul 21, 2021 at 19:14
• 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. Jul 21, 2021 at 19:18
• You can say more: $\mathbb C$ is exactly equal to $\mathbb R^2$ as sets.
– 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...