Thread: Borel-Sigma Algebras

Let f: R --> R be a continuous function, and let = {A C R: there exists B with where are the Borel subsets of the range space R. Show that , the Borel subsets of the domain space R.

Okay, so I know f being continuous means that is open for all open v in R.

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

I am somewhat confused as to what you are asking. I assume you want to show that is a subset of the Borel sigma algebra of the domain space.

Take any , then there exists a B in the Borel sigma algebra of the range, such that . 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 . Now , and as f is cts, each is open in the domain. Hence A is a Borel set.

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 .

You definitely can't say that. Borel subsets can be much more complicated than intersections of open sets.

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

The proof goes this way: let

.

Then is a subset of , it contains the open subsets, and it is easy (but you really should write it down) to prove that is a -algebra. Therefore, by definition of the borelian -algebra, , hence . This is exactly what we needed: for any , .