As the hint suggests, it is easy to show that for all :

A.

B.

Recall that a sigma algebra is a set closed under compliments and countable unions; and a Borel sigma algebra is the "smallest" sigma algebra containing every open set; thus, you have to show 3 things.

1. If B is open or closed then so is . This follows from continuity of f.

2. If B = union of open and/or closed sets, then is Borel. This follows from 1 and A.

3. If B is a compliment of a set from 1 or 2, then is borel. This follows from 1/2 and B.