The inverse of a Borel set

Question

Let $f$ be a continuous function and $B$ a Borel set, show that $f^{-1}(B)$ is a Borel set. (Hint, the collection $\mathcal{M}=\{E: f^{^{-1}}(E)\ \ $is Borel$ \}$ is $\sigma$- algebra).

Solution: I am struggling with showing that $\mathcal{M}$ is $\sigma$ algebra. Clearly, $\emptyset \in \mathcal{M}$ since by the continuity of $f$, we have $f^{-1}(\emptyset)=\emptyset$ is open and $\emptyset$ is Borel. Is $\mathbb{R}\in \mathcal{M}$ and how to prove it is closed under the union?

Thanks for any hint.

Answer 1

Let $\mathcal{O}$ denote the set of open sets and $\mathcal{B}$ the set of Borel sets.

Since $f$ is continuous we have $\mathcal{O} \subset \mathcal{M}$. Therefore

$\mathcal{B}= \sigma (\mathcal{O}) \subset \sigma (\mathcal{M})=\mathcal{M}$.

Answer 2

Use the facts that $f^{-1}(\cup_i A_i)=\cup_i f^{-1} (A_i)$ and $f^{-1}(A^{c})=(f^{-1}(A))^{c}$ to shwo that $\mathcal M$ is a sigma algebra. Since it contains open sets it must contains all Borel sets and this is what we want to prove. Note: the set theoretic identities I have stated hold always; they hold for any function and any sets.