1
$\begingroup$

I'm trying to prove this very precisely.

Let $\Sigma_n$ be the sigma-algebra of n-dimensional Lebesgue measurable sets.

Let $F:\mathbb{R}^n\times\mathbb{R}^m \rightarrow \mathbb{R}^{n+m}$ be a function such that $F(x,y)=(x_1,\cdots,x_n,y_1,\cdots,y_m)$.

(This function is meaningful since $\mathbb{R}^n\times\mathbb{R}^m \neq \mathbb{R}^{n+m}$ precisely.)

I'm trying to prove the following statement:

Define $H=\{F(A):A\in \Sigma_n \otimes \Sigma_m\}$

Then, $\mathscr{B}_{\mathbb{R}^{n+m}}\subset H \subset \Sigma_{n+m}$

(Of course, $\mathscr{B}$ denotes Borel algebra)

Is it possible to prove this not invoking Borel hierarchy?

Let $S\subset P(\mathbb{R}^n\times\mathbb{R}^m)$

It's easy to prove $\sigma(\{F(A) : A\in S\})\subset \{F(A) : A\in \sigma(S)\}$ but i think it's impossible to prove the converse not invoking Borel hierarchy.

This process would be clear, but it is not if one does not know Borel hierarchy (just like me). I don't understand why people say this so trivially.

Anyway, is there a easy way to prove this?

$\endgroup$
1
$\begingroup$

You don't need the Borel hierarchy here, all you need is that $F$ is bijective. Thus the induced maps between the power sets are also bijective, and commute with complements, unions, and intersections, and hence for every $\sigma$-algebra $\mathcal{A}$ on $\mathbb{R}^n\times \mathbb{R}^m$, the family

$$F_\ast(\mathcal{A}) = \left\{ F(A) : A\in\mathcal{A}\right\}$$

is a $\sigma$-algebra on $\mathbb{R}^{n+m}$, and for every $\sigma$-algebra $\mathcal{B}$ on $\mathbb{R}^{n+m}$, the family

$$F^\ast(\mathcal{B}) = \left\{ F^{-1}(B) : B\in\mathcal{B}\right\}$$

is a $\sigma$-algebra on $\mathbb{R}^n\times\mathbb{R}^m$, and $F^\ast$ and $F_\ast$ are inverses of each other, $F^\ast(F_\ast(\mathcal{A})) = \mathcal{A}$, and $F_\ast(F^\ast(\mathcal{B})) = \mathcal{B}$.

So $F_\ast(\sigma(S))$ is a $\sigma$-algebra containing $F(A)$ for every $A\in S$, hence $\sigma(F_\ast(S)) \subset F_\ast(\sigma(S))$ ($F_\ast$ and $F^\ast$ are defined as above for all families of subsets). Conversely, $F^\ast(\sigma(F_\ast(S)))$ is a $\sigma$-algebra containing $F^\ast(F_\ast(S)) = S$, whence $\sigma(S) \subset F^\ast(\sigma(F_\ast(S)))$ and consequently $F_\ast(\sigma(S)) \subset F_\ast(F^\ast(\sigma(F_\ast(S)))) = \sigma(F_\ast(S))$, altogether,

$$F_\ast(\sigma(S)) = \sigma(F_\ast(S)).$$

$\endgroup$
  • $\begingroup$ Wonderful!! Thank you! $\endgroup$ – John. p Feb 20 '14 at 19:16

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.