2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ $\mathbb{C}$ is $\mathbb{R}^2$ as a set and topological space. Thus Borel sigma algebras are equal. Or am I missing something? $\endgroup$ – freakish Jul 7 '17 at 8:05
  • $\begingroup$ @freakish. I don't think they are "the same". They are different sets with an obvious bijection between them which is also a topological homeomorphism. Similarly their Borel sigma algebras are not the same but clearly have a strong equivalence which probably has a name and definition. $\endgroup$ – Tom Collinge Jul 7 '17 at 8:20
  • 1
    $\begingroup$ It actually depends on the definition of $\mathbb{C}$ whether they are isomorphic or the same. The first time I was introduced $\mathbb{C}$ it was defined as $\mathbb{R}^2$ with some weird multiplication and a nice addition. Further along, it was introduced as $\mathbb{R}[i]$, or $\mathbb{R}[X]/(X^2+1)$. But if you follow the first definition I received, then @freakish 's answer is correct : $\mathbb{C}$ is $\mathbb{R}^2$ as a set and a topological space. Otherwise, they are $C^\infty$-diffeomorphic, thus in particular homeomorphic, and their Borel algebras are isomorphic $\endgroup$ – Max Jul 7 '17 at 8:25
  • $\begingroup$ @Max That's an interesting point for sure. $\endgroup$ – Tom Collinge Jul 7 '17 at 8:32
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I guess that the correspondence between Borel sets follows from the continuity of the homeomorphism and its inverse and consequent measurability of these functions ? $\endgroup$ – Tom Collinge Jul 7 '17 at 10:42
  • $\begingroup$ Yes, and being open and bijective. For general Polish spaces we will just have a measurable bijection, which is weaker, but suffices.@TomCollinge $\endgroup$ – Henno Brandsma Jul 7 '17 at 10:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.