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!

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    $\begingroup$ This is an EXCELLENT question $\endgroup$
    – Soapy Loaf
    Commented Oct 15, 2019 at 23:03
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    $\begingroup$ 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. $\endgroup$
    – Math1000
    Commented Oct 15, 2019 at 23:04
  • $\begingroup$ @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." $\endgroup$ Commented Oct 15, 2019 at 23:06

2 Answers 2


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 $f^{-1}(B^c)=A \setminus f^{-1}(B)$

  • $\begingroup$ Thank you!! I very much appreciate it $\endgroup$ Commented Oct 15, 2019 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.


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