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The preimage of a Borel set under a continuous function is Borel, but the preimage of a Lebesgue measurable set need not be measurable.

Under what conditions is it true that the preimage of a general Lebesgue measurable set is measurable?

For example, is the preimage of a null set under a continuous function a null set?

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  • $\begingroup$ If $f(x)=0$ for all $x$ then t $f^{-1}\{0\}=R$. If $g(x)=x$ for $x<0,\;$ and $ g(x)=0$ for $x\in [0,1],$ and $g(x)=x-1$ for$ x>1$ then $g^{-1}\{0\}=[0,1].$ $\endgroup$
    – DanielWainfleet
    Mar 21, 2016 at 6:00

2 Answers 2

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Not sure if this is what you are looking for, but a homeomorphism that maps zero sets to zero sets in both directions will preserve Lebesgue Measurability.

Alternately, a homeomorphism which is Lipschitz in both directions (a so-called Lipeomorphism) preserves Lebesgue Measurability. This is because Lipschitz maps take zero sets to zero sets.

Finally, diffeomorphisms in general preserve Lebesgue Measurability. This is because they are locally Lipschitz. This follows from the $n-$dimensional Mean Value Theorem.

Also, it is not true that the continuous image of a zero set is a zero set. The Cantor Set is a zero set in $\mathbb{R}$, but yet there exists a continuous surjection of the Cantor Set onto any compact metric space.

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  • $\begingroup$ Space filling curves are another example. $\endgroup$
    – Behnam Esmayli
    Dec 25, 2021 at 17:58
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Let $E$ be the standard "middle-thirds" Cantor set, and $S$ a "fat" Cantor set where at stage $n$, instead of removing the middle third you remove the middle $2^{n}$'th. There is a continuous one-to-one function $F: [0,1] \to [0,1]$ that is a homeomorphism of $S$ onto $E$. Let $A$ be any non-measurable subset of $S$. Then $F(A)$ is a null set, but $F^{-1}(F(A)) = A$ is not measurable.

We may obtain $A$, for example, as the intersection of a Bernstein set with $S$.

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