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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:

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  • $\begingroup$ 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

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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).

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  • $\begingroup$ 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
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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...

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