# Prove $f^{-1}(B)\in\mathcal{A}$ where Borel set $B$, $\sigma$-algebra $\mathcal{A}$, $\mathcal{A}$-measurable function $f$.

I am trying to solve a homework problem in "Lebesgue integration" course.

And I found the theorem seems to be useful to my problem which is :

Let $\mathcal{A}$ be a $\sigma$-algebra of $\mathbb{R}$, $B$ be a Borel set of $\mathbb{R}$, and $f:\mathbb{R}\to\mathbb{R}$ be $\mathcal{A}$-measurable. Then, $f^{-1}(B)\in\mathcal{A}$.

I thought if I could express $B$ as the countable union of open intervals, then we have $$f^{-1}(B)=f^{-1}\left(\bigcup_{i=1}^\infty(a_i,b_i)\right)=\bigcup_{i=1}^\infty\left(f^{-1}((a_i,\infty))\cap f^{-1}((b_i,\infty))^c\right),$$ thus by the definition of measurable function, each $f^{-1}((x,\infty))\in\mathcal{A}$, and we conclude the proof.

But, I can not express $B$ as the countable union of open intervals. If it is impossible, more generally, I think considering not only the countable unions, but also countable intersections and the complement would give the same conclusion.

So, to deal with this problem, decomposing the structure of a Borel set with formal mathematical writing is necessary.

In my textbook, the Borel $\sigma$-algebra is defined by the $\sigma$-algebra generated by the set of all open sets and I know that all open sets can be expressed as the countable union of open intervals.

On the other hand, wikipedia define the Borel $\sigma$-algebra with the construction with three operations, countable union and intersection and the complement by the transfinite induction.

I do not want to use this definition since it is too complicate to me and it would be very straightforward work. Moreover, wikipedia's point of view is out of the scope of my course.

Anyway, so how can I prove this? How can I decribe the structure of an arbitrary Borel set just with the definition in my textbook?

• You are right: any open subset of $\mathbb{R}$ can be expressed as a countable union of disjoint open intervals. But, as you also pointed out, Borel sets are not necessarily open. The $\sigma$-algebra is generated by open sets, but it has sets in it that aren't open. – layman Oct 16 '14 at 17:11

## 1 Answer

Here's what you have to notice:

If $f$ is a measurable function, then $\{ A \subseteq \mathbb{R} \mid f^{-1}(A) \in \mathcal{A} \}$ is a $\sigma$-algebra. Can you prove this?

Once you know the above, all we need to do is check that the preimage of any open set is in $\mathcal{A}$, because the set above being a $\sigma$-algebra containing the open sets would imply that the set above contains all Borel sets, so all Borel sets would have preimage in $\mathcal{A}$.

So you need to be aware that if $\mathcal{B}$ is the Borel $\sigma$-algebra, then any $\sigma$-algebra containing all of the open sets will necessarily $\mathcal{B}$ since the Borel $\sigma$-algebra is the smallest $\sigma$-algebra containing the open sets. Do you understand this part?

Thus, to prove that the preimage of a borel set $B$ is in $\mathcal{A}$, all you need to do is prove the preimage of any open set is in $\mathcal{A}$, and for that, you can use the fact that open sets are countable unions of open intervals.

• Wow, I fully understand what you mean. I didn't consider the collection of subsets its preimage is measurable. Thanks! – Analysis Oct 16 '14 at 17:37
• @Analysis You're welcome! I'm glad the information was helpful. – layman Oct 16 '14 at 20:00