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

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.

• what is the question? – Masacroso Oct 2 '18 at 16:44

## 2 Answers

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?

• Because a continuous function may not satisfy this condition, for example the Cantor function. – TNT Oct 2 '18 at 16:46
• What is your definition of measurable functions btw? – Kyle Oct 2 '18 at 20:15
• I think you are confusing the idea preimage with image. The Cantor function is cts but not necessarily maps a measurable set onto a measurable set. Also you might wanna specify the sigma algebra(Borel or Lebesgue measurable sets?) equipped with the space. – Kyle Oct 2 '18 at 20:27

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 $$f^{-1}(B)\in\mathcal{B}$$.