This is a homework question, so please just very small hints.

I am trying to prove that the following two statements are not equivalent:

Let $f: \mathbb{R} \rightarrow \mathbb{R}$.

1) $f^{-1}(E)$ is Lebesgue-measurable for every Lebesgue-measurable $E \subset \mathbb{R}$.

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

Obviously 1) implies 2), but I am having trouble proving that 2) does not imply 1). I can't seem to come up with a counter example. I was thinking, since the Lebesgue sets are the completion of the Borel sets then every Lebesgue measurable set can be written in the form $A \cup Z$, where $A$ is Borel and $Z$ is a zero set (quick question - is this even true?). So then I will have $f^{-1}(A \cup Z) = f^{-1}(A) \cup f^{-1}(Z)$, so by assumption of statement 2), $f^{-1}(A)$ will be Lebesgue-measurable, which must mean that a necessary condition is that $f^{-1}(Z)$ is NOT Lebesgue measurable. Is this the right train of thought to continue along?

  • 1
    $\begingroup$ Everything you've said so far is correct. $\endgroup$ – Omnomnomnom Mar 29 '16 at 3:16
  • $\begingroup$ It's at this point in my train of thought where I get stuck...any ideas? $\endgroup$ – Jonathan Gafar Mar 29 '16 at 3:31
  • $\begingroup$ Every continuous function satisfy $2)$, but not always $1)$, see the first answer (with the Devil's staircase) to this question (math.stackexchange.com/questions/479441/…) $\endgroup$ – Charles Madeline Nov 6 '16 at 16:39

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