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Let $F_n\subset [0,1]$ be a Fat Cantor Set (so that $[0,1]\setminus F_n$ is dense) of Lebesgue measure $1 - 1/n$, and let $F = \bigcup_n F_n$. Does there exist a probability measure $\mu$ on $[0,1]$ such that $\mu(F + x) = 0$ for all $x\in [0,1]$? Here $+$ is a cyclic shift, so $0.5 + 0.6 = 0.1$.

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If I am not mistaken, Fubini's theorem shows that this is impossible, as follows \begin{eqnarray*} 0 & = & \int_{0}^{1}\mu\left(F+x\right)\,{\rm d}\lambda\left(x\right)\\ & = & \int_{0}^{1}\int1_{F+x}\left(y\right)\,{\rm d}\mu\left(y\right)\,{\rm d}\lambda\left(x\right)\\ & = & \int_{0}^{1}\int_{0}^{1}1_{y-F}\left(x\right)\,{\rm d}\lambda\left(x\right)\,{\rm d}\mu\left(y\right)\\ & = & \int_{0}^{1}\lambda\left(y-F\right)\,{\rm d}\mu\left(y\right)\\ & = & \int_{0}^{1}\lambda\left(F\right)\,{\rm d}\mu\left(y\right)=\lambda\left(F\right)\cdot\mu\left(\left[0,1\right]\right)=\lambda\left(F\right)=1. \end{eqnarray*} Here, I assumed that $\lambda$ is the Lebesgue measure and $\mu$ is a measure as you want to have it.

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  • $\begingroup$ Nice, I'll have to double check, but apparently you just got a bounty for another question of mine :) $\endgroup$
    – SBF
    Commented Nov 13, 2015 at 22:53
  • $\begingroup$ Please check this question. If there are no answers before the bounty expires, I think it should go to you, so don't forget to post an answer there. Maybe, just a reference to this question if you have nothing else to add. $\endgroup$
    – SBF
    Commented Nov 15, 2015 at 14:23

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