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I was asked to verify whether the next claim is true or false:

Let $f:\mathbb{R}^n\rightarrow\mathbb{R}^n$ be a homeomorphism and define $\mu(A)=m_n(f(A))$, where $m_n$ is the $n$-dimensional Lebesgue measure. If every $m_n$-measurable set $A\subset\mathbb{R}^n$ is also $\mu$-measurable, then $\mu\ll m_n$.

I think this is false, if we restrict the $\sigma$-algebra to be all Borel sets and let $f:[0, 1]\rightarrow [0, 2]$, $f(x)=\Psi(x)+x$, where $\Psi:[0, 1]\rightarrow [0, 1]$ is the Cantor function. Then $f$ is homeomorphism and because of that every $m_1$-measurable set is also $\mu$-measurable. Moreover $m_1(C)=0$ but $\mu(C)=m_1(f(C))=1$, where $C$ is the Cantor ternary set, so $\mu$ is not absolutely continuous w.r.t. $m_n$.

My question is, can we do this without restricting the $\sigma$-algebra, or without restricting the domain and codomain of $f$? And above all, is my reasoning correct? Thanks in advance.

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  • $\begingroup$ A set is $\nu$-measurable if it is measurable with respect to the completion of $\nu$. $\endgroup$ – Michael Greinecker Nov 3 '17 at 19:11
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If there were some $A$ with Lebesgue measure zero such that $\mu(A)>0$, then every subset of $A$ would be Lebesgue measurable and hence $\mu$-measurable. This would imply that $f(A)$ is a set of positive Lebesgue measure that has no subset that is not Lebesgue measurable. This is not possible.

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  • $\begingroup$ I don't understand the line "f(A) has no subset that is not Lebesgue measurable", would you clarify this? so the claim is true then? $\endgroup$ – peastick Nov 4 '17 at 8:19
  • $\begingroup$ The claim is true. Every subset of $f(A)$ is of the form $f(B)$ for some $B\subseteq A$. Since every subset of $A$ is $\mu$-measurable, $B$ is $\mu$-measurable and therefore $f(B)$ Lebesgue measurable. $\endgroup$ – Michael Greinecker Nov 4 '17 at 10:54
  • $\begingroup$ okay, thanks, now I got it $\endgroup$ – peastick Nov 4 '17 at 15:27

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