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:


 A: 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).
A: 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...
