# Inverse image of Lebesgue set and Borel set

1. Mar 29, 2016

### JonnyG

1. The problem statement, all variables and given/known data
I need to prove that if $f: \mathbb{R} \rightarrow \mathbb{R}$ then the following two statements are NOT equivalent:

1) For every $E \subset \mathbb{R}$ that is Lebesgue-measurable, $f^{-1}(E)$ is Lebesgue-measurable.
2) For every $E \subset \mathbb{R}$ that is Borel-measurable, $f^{-1}(E)$ is Lebesgue-measurable.

2. Relevant equations

3. The attempt at a solution
Obviously 1) implies 2). But I am pretty sure that 2) does not imply 1). To find a counter example I was thinking that I could take a Lebesgue-measurable set and write it in the form $A \cup Z$ where $A$ is Borel-measurable and $Z$ is a zero-set. Then $f^{-1}(A \cup Z) = f^{-1}(A) \cup f^{-1}(Z)$ where $f^{-1}(A)$ is Lebesgue-measurable, by hypothesis. So a necessary condition would be to make sure that $f^{-1}(Z)$ is NOT Lebesgue-measurable.

Since the collection of open intervals generate the Borel-algebra, I'm thinking that I need to find a function such that the inverse image of open intervals is Lebesgue measurable, but the inverse image of some zero set is NOT Lebesgue-measurable. Coming up with such a function that is well-defined is proving to be more difficult than I thought.

Any thoughts?

2. Mar 29, 2016

### andrewkirk

Are you supposed to assume the Axiom of Choice (AC)?

If not then the propositions cannot be proven to be non-equivalent, as AC is required to prove the existence of non-Lebesgue-measurable sets. See here re Solovay's proof.

Conversely, if you are supposed to assume AC, you could use Vitali sets, which are not Lebesgue measurable (Vitali sets can only be proven to exist by using AC). The usual way to create counterexamples from pathological sets is to use the indicator function of such a set.

3. Mar 29, 2016

### JonnyG

Yes we are allowed to use the AC. I don't think we will be covering Vitali sets at all in this class, but I will read up ahead on them to answer this question. Thanks.

4. Mar 29, 2016

### andrewkirk

Well actually you don't need to specifically use Vitali sets - just any non-Lebesgue-measurable set. If you just assume the existence of at least non-Lebesgue measurable set S then you can use the indicator function of S as a counterexample.
Conversely, if one does not assume their existence, or AC (from which their existence can be proven, via Vitali for example) it cannot be proven that the two propositions are non-equivalent.

5. Mar 29, 2016

### JonnyG

I proved in my last homework assignment the existence of non-measurable sets...Now I feel like an idiot.

Thanks for the help!

6. Mar 29, 2016

### JonnyG

andrew, I just realized that the indicator function does not work. Remember, I have to assume that if a subset of the image is Borel measurable, then the inverse image is Lebesgue measurable. If S is non-measurable, and letting f denote the indicator function of S, then the image of f consists of the two points 0 and 1. Both of these are Borel-measurable, yet the inverse image of 1 is S, which is not Lesbesgue measurable.

7. Mar 30, 2016

### andrewkirk

Good observation Jonny. I hadn't noticed that. So it looks like we need a set that is Lebesgue but not Borel measurable, and a set that is not Lebesgue measurable.

This wiki link demonstrates a non-Borel set $A$. Since it does not appear to use AC, we conclude that it must be Lebesgue-measurable.
For the second we can use any non-Lebesgue-measurable set $N$, of which we assume there exists at least one such (which in turn requires AC).

If we can construct a function f that gives a number in $A$ if the input is in $N$ and a number outside $A$ otherwise, and f is surjective onto $\mathbb{R}$ (so that $f(N)=A$ and $f(\mathbb{R}-N)=\mathbb{R}-A$) then that will do the trick because setting $E=A$ falsifies (1) but not (2), and there doesn't appear to be any other Borel $E$ that would falsify (2). The challenge is to construct $f$. I expect we'll need AC.

If we take $N$ as a Vitali set then it's fairly easy to demonstrate that both $N$ and $\mathbb{R}-N$ have the same cardinality as $\mathbb{R}$, and hence at least as great cardinality as $A$ and $\mathbb{R}-A$. Hence there exist surjections from each of those former to each of the latter, and we can combine the two surjections to get our function f.

The bit I can't yet see how to do is to ensure that f can be constructed in such a way that there is no Borel set $E$ whose pre-image is non-Lebesgue measurable. The fact that $A$ and its complement are non-Borel does not guarantee that there won't be some other set $E$ that is Borel and has a non-Lebesgue pre-image.

8. Mar 30, 2016

### JonnyG

This seems to be the hardest part and I've made literally no progress on it. There are a lot more questions on this homework assignment, so I have to move on for now. If I still have time after I finish the rest of the questions then I will return to this question.

9. Mar 31, 2016

### micromass

Staff Emeritus