0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ what is the question? $\endgroup$ – Masacroso Oct 2 '18 at 16:44
0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Because a continuous function may not satisfy this condition, for example the Cantor function. $\endgroup$ – TNT Oct 2 '18 at 16:46
  • $\begingroup$ What is your definition of measurable functions btw? $\endgroup$ – Kyle Oct 2 '18 at 20:15
  • $\begingroup$ 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. $\endgroup$ – Kyle Oct 2 '18 at 20:27
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.