I am studying Simon & Reed's Methods of Modern Mathematical Physics, Volume 1, Functional Analysis.
In chapter 1, exercise 14, it is asked to show that if $f$ is a Borel function, then $f^{-1}(B) \in \mathcal{B}$ for any $B \in \mathcal{B}$. Simon & Reed define Borel sets and Borel functions in the following way:
The Borel sets of $\mathbb{R}$, $\mathcal{B}$, is the smallest family of subsets of $\mathbb{R}$ with the following properties:
- The family is closed under complements.
- The family is closed under countable unions.
- The family contains each open interval.
A function $f$ is called a Borel function if and only if $f^{-1}[(a,b)]$ is a Borel set for all $a,b$.
This is my first introduction to the subject and I am not that comfortable with that. I feel this should not be too hard but at the same time, I have no idea how to do this. Any help would be much appreciated.
This argument is the kind of thing one usually learns in a first graduate-level measure theory course. I'll outline it and let you fill in the details.
We're given that $f^{-1}(I)$ is a Borel set whenever $I$ is an open interval, and we want to show that $f^{-1}(B)$ is a Borel set whenever $B$ is an arbitrary Borel set. So let us define $\mathcal{F}$ to be the family of all subsets $E \subseteq \mathbb{R}$ such that $f^{-1}(E)$ is Borel. Since the family of Borel sets is defined to be the smallest family satisfying conditions 1,2,3, if we show that $\mathcal{F}$ also satisfies 1,2,3 then it will follow that $\mathcal{B} \subseteq \mathcal{F}$, which is what we want.
That $\mathcal{F}$ satisfies condition 3 is exactly the assumption on our function $f$. I will leave it to you to verify that $\mathcal{F}$ also satisfies 1 and 2.