Show that inverse image of a Lebesgue measurable function is Lebesgue-measurable 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.
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.