Let $m$ be Lebesgue measure on $\mathbb{R}$.
Let $f(x)=x^{3}$,$f: \mathbb{R} \to \mathbb{R}$.
For every measurable set $A \subset \mathbb{R}$ we define a measure: $$\nu(A)=m(f^{-1}(A))$$
Is it true that $\nu<<m$ ?
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Well, I thought this is true mainly because I couldn't find any proper counterexample (perhaps some variation of Cantor set?).
Let $A \subset \mathbb{R}$ be a measurable set so $m(A)=0$. Obviously Lebesgue measure is not invariant under $\sqrt[3]{.}$ (also $f$ is invertible, of course), but it might be the case for sets of zero measure.
I tried using the fact that: $$ \nu(A)=m(f^{-1}(A))=inf\Big\{\sum_{j=1}^\infty(b_j-a_j) : f^{-1}(A) \subset\bigcup_{j=1}^\infty(a_j,b_j) \Big\} =inf\Big\{\sum_{j=1}^\infty(\sqrt[3]{b_j}-\sqrt[3]{a_j}) : A \subset \bigcup_{j=1}^\infty(a_j,b_j) \Big\} $$
While knowing that: $$m(A)=inf\Big\{\sum_{j=1}^\infty(b_j-a_j) : A \subset\bigcup_{j=1}^\infty(a_j,b_j) \Big\}=0$$
But no progression from here...
Thanks in advance!