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Let $f:[0,1]\to [0,1]^2$ be some continuous surjective function, and $\mu_n$ - the canonical Lebesgue measure on $\Bbb R^n$. So, on $[0,1]^2$ we can define two probability measures: $\mu_2$ and $\nu_2:=f_*\mu_1$ is the pushforward measure: $\nu_2(A) = \mu_1(f^{-1}(A))$ for each Borel-measurable set $A$.

My guess would be that they are mutually singular for all $f$, and that it shold not be too hard to find example sets on which each of them takes $0$ values while the other assigns positive measure, however I have not practiced measure theory for a while, so can't think of anything simple.

Question: I would like to know whether $\mu_2$ and $\nu_2$ are indeed mutually singular for any $f$ as above.

SBF
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    Could you just remind me how you define $f*\mu_1$ please ? – charmd Jun 05 '18 at 08:48
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    @charMD: added definition to the OP. – SBF Jun 05 '18 at 21:22
  • The answer depends on the choice of $f$. If $f: [0, 1] \to [0, 1]^2$ is the Hilbert or the Peano space filling curve, then $f$ is ($\mu_1-\mu_2$) measure preserving -- An interesting corollary is that there are continuous functions from $[0, 1]$ to $[0, 1]$ which are neither one-one nor constant on any set of positive measure. You can also get singular measures by composing the Cantor-Lebesgue function with Hilbert/Peano curve. A good reference for these facts is "Space filling curves" by Hans Sagan. – hot_queen Jun 06 '18 at 11:36
  • @hot_queen: I see; actually, should not have come as a surprise, given that any uncountable Polish space is Borel isomorphic to $[0, 1]$. You can post your comment as an answer – SBF Jun 09 '18 at 08:32

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The answer depends on the choice of $f$. If $f: [0, 1] \to [0, 1]^2$ is the Hilbert or the Peano space filling curve, then it is measure preserving. You can also get singular measures by composing the Cantor-Lebesgue function with Hilbert/Peano curve. A good reference for these facts is "Space filling curves" by Hans Sagan.

hot_queen
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