A function is a Borel function if and only if for any c, the set $f^{-1}((c,\infty))$ is a Borel set

Question

Prove that a function $f:\mathbb{R}\rightarrow\mathbb{R}$ is a Borel function if and only if for any $c$, the set $f^{-1}((c,\infty))$ is a Borel set.

Recall that a Borel set is obtained from open and closed sets using the combination of complement, countable unions, and countable intersections. From this definition, a Borel set is measurable but a measurable set may not be a Borel set. A function $f:\mathbb{R}\rightarrow\mathbb{R}$ is called a Borel function if the preimage of any Borel set $B$ is a Borel set (i.e. $f^{-1}(B)$ is a Borel set whenever $B$ is a Borel set).

We define a Borel function by saying $f^{-1}(B)$ is Borel whenever $B$ is Borel; but we do not define a measurable function by saying that the preimage of a measurable set is measurable because a continuous function may not satisfy this condition.

This is what I have so far, but I'm not sure about the forward direction.

($\Leftarrow$) Suppose for any $c$, the set $f^{-1}((c,\infty))=\{x\in\mathbb{R}:f(x)>c\}$ is a Borel set. We know that any open set can be written as a countable union of disjoint open intervals. The collection of all sets formed this way is the Borel collection where a set in the collection is a Borel set. So, every open set is a Borel set. Let $B=\{x\in \mathbb{R}:x\in (c,\infty)\}$. By definition, a function is a Borel function if the preimage of any Borel set $B$ is a Borel set. Hence $f$ is a Borel function.

Answer 1

To prove it, note $f^{-1}$ preserves unions, intersections and complements. And Borel sets=$\sigma(\{(c,\infty),c\in \mathbb{R}\})$.

Also, why would functions such that "the preimage of a measurable set is measurable" not a measurable function?

Answer 2

How to prove the backward direction: Let $\mathcal{C} = \{A\subset\mathbb{R} : f^{-1}(A) \in \mathcal{B}\}$ where $\mathcal{B}$ is the collection of Borel sets. Show that $\mathcal{C}$ is a $\sigma$-algebra that contains the open intervals. Hence $\mathcal{B} \subset \mathcal{C}$ since $\mathcal{B}$ is the smallest $\sigma$-algebra that contains the open intervals. But this means for any borel set $B\in\mathcal{B}$, we have $f^{-1}(B)\in\mathcal{B}$.