I understand that that the Borel sigma algebra on the product topology of second countable metric spaces is the same as the sigma algebra generated from the product of their individual sigma algebras. In particular, $\mathscr B (\mathbb R^2) = \mathscr B (\mathbb R) \otimes \mathscr B (\mathbb R) $
Secondly, with their standard topologies, $ \mathbb C$ is isometrically homeomorphic to $\mathbb R^2$.
What is the relationship between $\mathscr B (\mathbb R^2) $ and $\mathscr B (\mathbb C) $ ? One reference I have says they are the same, but I think some form of isomorphism is more likely.
And, knowing that a function from a product Borel space (to another Borel space) is measurable if and only if its projections are measurable, how would one use that rigorously to show the same for the real and imaginary parts of a complex valued function ?
Update: the last part I think I have solved. Just use the continuity of the homeomorphism and the fact that compositions of continuous and Borel measurable functions are Borel measurable.
The standard homeomorphism $h$ between $\mathbb{R}^2$ and $\mathbb{C}$ ($h(x,y) =x+iy$) preserves all Borel sets, bijectively.
In fact if $X$ and $Y$ are both Polish spaces (i.e. completely metrisable separable spaces), there is a bijection $f$ from $X$ to $Y$ such that $f[A]$ is Borel for every Borel set $A$ of $X$, and every Borel set in $Y$ is of this form. All Borel $\sigma$-algebras of Polish spaces are isomorphic. See Wikipedia for example, or Kechris's book on descriptive set theory, or more classical texts in this vein.