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I want to prove:

If $f: \mathbb R \to \mathbb R $ is a function that send every Lebesgue measureable sets to Lebesgue measurable sets then it send measure zero sets to measure zero.

I do not know how to start to think. Can someone help me. Thanks

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  • $\begingroup$ The property that a function maps measure zero sets to measure zero sets is called "Lusin's (N)" condition. There is a result sometimes called the Rademacher-Ellis theorem that asserts for a measurable function that this condition is, in fact, equivalent to the property of mapping measurable sets to measurable sets. So your exercise proves one (easy) direction: don't forget to look up the converse proof. $\endgroup$ Commented Jan 6, 2016 at 19:32

1 Answer 1

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Suppose that $N \subset \mathbb R$ is a set of Lebesgue measure zero with the property that $f(N)$ has positive measure. Then $f(N)$ contains a nonmeasurable set $Z$. If you let $Y = N \cap f^{-1}(Z)$ then $Y$ has Lebesgue measure zero, so it is measurable, but $f(Y) = Z$ is not.

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    $\begingroup$ It's not obvious that every set of positive measure contains a nonmeasurable set. You might add a line or two... $\endgroup$ Commented Feb 17, 2015 at 0:45
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    $\begingroup$ Adding a line or two: en.wikipedia.org/wiki/Bernstein_set $\endgroup$
    – hot_queen
    Commented Feb 17, 2015 at 1:22
  • $\begingroup$ how do you know Y is measurable ? , i.e how do you know $f^{-1}(Z)$ is measurable. $\endgroup$
    – Monty
    Commented Nov 4, 2018 at 16:21
  • $\begingroup$ @Monty nobody said anything about $f^{-1}(Z)$ being measurable. To quote directly from the three line answer "$Y$ has Lebesgue measure zero, so it is measurable". $\endgroup$
    – Umberto P.
    Commented Nov 4, 2018 at 22:59
  • $\begingroup$ How do you know you can measure Y? i.e how can you claim it has measure zero therefore its measurable? surely you have to claim its measurable before you tell us its measure? $\endgroup$
    – Monty
    Commented Nov 5, 2018 at 17:28

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