# Function that sends every Lebesgue measureable sets to a lebesgue measurable set. Then it sends measure zero sets to measure zero sets.

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

• 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. Commented Jan 6, 2016 at 19:32

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.
• how do you know Y is measurable ? , i.e how do you know $f^{-1}(Z)$ is measurable. Commented Nov 4, 2018 at 16:21
• @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". Commented Nov 4, 2018 at 22:59