# Question on Borel Measurable Functions and Borel Sigma Algebras.

On an assignment, we have been given a the following setup: Let $$f:A\rightarrow \bar{\mathbb{R}}$$ be a borel measurable function. prove that if $$f$$ is Borel measurable and $$B$$ is a Borel set, then $$f^{-1}(B)$$ is a Borel set.

The definition of Borel Measurability we were given is as follows: "A function $$f : \mathbb{R} \to \mathbb{R}$$ is said to be Borel measurable provided its domain A ⊆ R is a Borel set and for each c, the set {$$x ∈ A : f (x) < c$$} is a Borel set.

We were not given any description of where this set lives, I'm assuming $$B \subset \mathbb{R}$$ but this may be incorrect. I figure we need to show that the set {$${B \subset \mathbb{R}:f^{-1}(B)}$$ is a Borel set} is a sigma algebra but I am unsure how to do this. I know its a little silly because all you need to do is check that the definitions of a sigma algebra are met but showing those things is proving more difficult than i anticipated. We may also use the fact that borel measurable functions are Lebesgue measureable. Any help would be greatly appreciated!

• This is an EXCELLENT question – Sacha L'Heveder Oct 15 '19 at 23:03
• I don't understand - that is the definition of a Borel-measurable function, that for every Borel set $B$, $f^{-1}(B)$ is a Borel set. There is nothing to prove. – Math1000 Oct 15 '19 at 23:04
• @Math1000 the definition we were give was as follows, i will edit the question to be more clear. "A function f : R → R is said to be Borel measurable provided its domain A ⊆ R is a Borel set and for each c, the set {x ∈ A : f (x) < c} is a Borel set." – Dominic Petti Oct 15 '19 at 23:06

Note that for $$b \in \Bbb{R}$$

$$f^{-1}((-\infty,b])=A \setminus f^{-1}(b,+\infty)$$ which is a Borel set.

The family of sets of the form $$(-\infty,b]$$ with the empty set,generate the Borel sigma algebra.

So if you show that $$A=\{B \subseteq R: f^{-1}(B) \text{is Borel}\}$$ is a sigma algebra then you are done since this sigma algebra contains the sets of the form $$(-\infty,b]$$ and the empty set,so it will contain the Borel sigma algebra by definition.

To show that $$A$$ is sigma algebra,just use the fact that the inverse image of a union is the union of the inverse images and

if $$B \subseteq \Bbb{R}$$ then $$f^{-1}(B^c)=A \setminus f^{-1}(B)$$

• Thank you!! I very much appreciate it – Dominic Petti Oct 15 '19 at 23:34

The Borel $$\sigma$$-algebra of some topological space $$A$$ is defined as the smallest $$\sigma$$-algebra that contains all the open sets of $$A$$.

When we says that a function $$f:A\to\overline{\Bbb R}$$ is Borel measurable we are assuming the standard topology in $$\overline{\Bbb R}$$ and that if $$C\subset \overline{\Bbb R }$$ is a Borel set then $$f^{-1}(C)$$ is Borel in the induced Borel $$\sigma$$-algebra of $$A$$.

By the properties of $$f^{-1}$$ respect to set operations it can be shown that it is enough to say that $$f^{-1}(C)$$ is Borel in $$A$$ for any $$C$$ of the form $$[-\infty,a)$$, as is generally explained in any analysis textbook that cover an introduction to Lebesgue integration theory.