1. ## Borel-Sigma Algebras

Let f: R --> R be a continuous function, and let $\mathcal{A}$ = {A C R: there exists B $\epsilon \mathcal{B}$ with $A=f^{-1}(B)$ where $\mathcal{B}$ are the Borel subsets of the range space R. Show that $\mathcal{A} C \mathcal{B}$, the Borel subsets of the domain space R.

Okay, so I know f being continuous means that $f^{-1}(v)$ is open for all open v in R.

I'm not exactly sure what I need to show here. Is it just that $\mathcal{A}$ is open in R? Then wouldn't that show it is in $B$? And wouldn't this be true because f inverse is open? So, all the A's in $\mathcal{A}$ would be open. Am I on the right track? Thank you!

2. I am somewhat confused as to what you are asking. I assume you want to show that $\mathcal{A}$ is a subset of the Borel sigma algebra of the domain space.

Take any $A \in \mathcal{A}$, then there exists a B in the Borel sigma algebra of the range, such that $A=f^{-1}(B)$. As B is a Borel set, it is the intersection of some open sets (countable union of open sets is also open, all we are missing is the countable intersection), so $B=\bigcap G_n$. Now $A=f^{-1}\left(\bigcap G_n \right)=\bigcap f^{-1}(G_n)$, and as f is cts, each $f^{-1}(G_n)$ is open in the domain. Hence A is a Borel set.

3. Originally Posted by Focus
As B is a Borel set, it is the intersection of some open sets (countable union of open sets is also open, all we are missing is the countable intersection), so $B=\bigcap G_n$.
You definitely can't say that. Borel subsets can be much more complicated than intersections of open sets.

The Borel $\sigma$-algebra is defined as the smallest $\sigma$-algebra containing the open subsets. That's the only property we can use to prove that a set is in the Borel $\sigma$-algebra. Hence the proof can't be direct. We know that $f^{-1}(O)\in\mathcal{B}$ for any open subset $O$ (because $f^{-1}(O)$ is open, too). And we want to generalize this to any borelian subset instead of just open subsets.

The proof goes this way: let

$\mathcal{C}=\{B\in\mathcal{B}|f^{-1}(B)\in\mathcal{B}\}$.

Then $\mathcal{C}$ is a subset of $\mathcal{B}$, it contains the open subsets, and it is easy (but you really should write it down) to prove that $\mathcal{C}$ is a $\sigma$-algebra. Therefore, by definition of the borelian $\sigma$-algebra, $\mathcal{B}\subset\mathcal{C}$, hence $\mathcal{C}=\mathcal{B}$. This is exactly what we needed: for any $B\in\mathcal{B}$, $f^{-1}(B)\in\mathcal{B}$.