I am struggling with this exercise. Can anyone please give me a hint?

Suppose f is Lebesgue-Measurable. Show that $f^{-1}(B)$ is Lebesgue- measurable for any borel set B.

I do know that both the borel and lebesgue measurable sets are $\sigma$-algebra. And that the borel sigma algebra is generated by all the open sets. I also know that the collection of borel-sets is a proper subset of the collection of the lebesgue measurable sets.

I know that what it means to be lebesgue measurable is that $f^{-1}(O)$ is lebesgue measurable, for any open set $O$.

But the problem is that $B$ may not be open. And it may even not be a countable union of open sets, if it were I would just use that the lebesgue measurable sets for a $\sigma$-algebra.


Hint: Show that the collection $\mathcal C_f = \{U : f^{-1}(U) \text{ is measurable}\}$ forms a $\sigma$-algebra. That is, show that this collection is closed under complements and countable unions.

Then, note that $\mathcal C_f$ is a $\sigma$-algebra containing the open sets.

  • $\begingroup$ Thanks that is very smart!, I wish I thought of that! $\endgroup$ – user119615 Sep 2 '14 at 20:09
  • $\begingroup$ I would appreciate it if whoever downvoted would explain his decision to do so, and perhaps tell me how my answer might be improved. $\endgroup$ – Omnomnomnom Sep 4 '14 at 13:54

If $f$ is Lebesgue-measurable, by the definition, it means that for each Lebesgue measurable set $X \subseteq R$, $f^{-1}(X)$ also is Lebesgue measurable. Since each Borel subset $B \subseteq R$ is Lebesgue measurable, we automatically get that $f^{−1}(B)$ is Lebesgue- measurable.

  • $\begingroup$ This is not the usual definition of lebesgue measurability $\endgroup$ – Omnomnomnom Sep 4 '14 at 13:56

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