Let $f$ be a continuous real-valued function on $\mathbb{R}$. Show that the inverse image with respect to $f$ of an open set is open, of a closed set is closed, and of a Borel set is Borel.
I got open and closed pretty easily by following the definition. But Borel? A Borel set is formed from open sets by complement or countable union (or repeating these operations). Doesn't seem very concrete. How can I deal with its inverse image?
Best Answer
You might use the following result:
Let $(X, \mathcal{S})$ and $(Y, \mathcal{T})$ be measurable spaces. Let $T$ be a transformation from $X$ into $Y$. Also, let $\mathcal{G}$ be a class of subsets of $Y$ that generates $\mathcal{T}$. Then $T$ is measurable if $T^{-1}(G) \in \mathcal{S}$ for all $G \in \mathcal{G}$.
What is one example of a class of subsets of $\mathbb{R}$ that generates the Borel sets of $\mathbb{R}$? Which you might easily be able to take advantage of using a fact about continuous functions?