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As a follow up to Borel set with constant positive but not full measure in each interval.,

Let $\ell:\mathscr{B}(\mathbb{R})\to[0,\infty]$ be the Lebesgue measure defined over the Borel sigma algebra on $\mathbb{R}$. Consider for some $S\in\mathscr{B}(\mathbb{R})$ the function $f_s:\mathbb{R}\times\mathbb{R}\to[0,\infty]$ defined so: $$f_s(x,a) = \ell(S\cap(x,x+a])$$

Does there exist some $S\in\mathscr{B}(\mathbb{R})$ s.t. $\ell(S)=\infty=\ell(S^c)$ and $f_s(\bullet,a)$ is constant in $x$ for every choice of $a$?

In case my phrasing is too convoluted, what I'm ultimately aiming for is to show that for any $S\in\mathscr{B}(\mathbb{R})$ s.t. $\ell(S)=\infty=\ell(S^c)$ I can find two disjoint intervals of equal length satisfying $\ell(S\cap(x,x+a])\neq\ell(S\cap(y,y+a])$. Perhaps there's an easier way? Cheers.

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No. Let $\mu=\ell(S\cap (0,1])$ and note $0<\mu<1.$ For each $n$ the values $\ell(S\cap(k/n,(k+1)/n])$ for $k=0,\dots,n-1$ are equal and add up to $\mu,$ so they must all be $\mu/n.$ By Lebesgue's density theorem there is a density point $x\in(0,1]$ for $S,$ which implies that for large enough $n$ the interval $(k/n,(k+1)/n]$ satisfies $\ell(S\cap(k/n,(k+1)/n])>\mu/n,$ a contradiction.

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  • $\begingroup$ Thank you! Let me see if I'm parsing this correctly... (1) When you say $0<\mu<1$, this is a statement that holds WLOG and not per se. (2) Lebesgue's density theorem says $1>\mu=\frac{\ell(S\cap(k/n,(k+1)/n])}{1/n}\to1$, which is the contradiction we seek. Correct any mistakes I'm making if at all. tyty $\endgroup$ – procrastidigitation Feb 14 '18 at 3:33
  • $\begingroup$ @procrastidigitation: for (1) if $\mu=0$ then $S=\cup_n(S\cap(n,n+1])$ would have zero measure, but $\ell(S)=\infty$ is given. Similarly if $\mu=1$ then $S$ would have full measure contradicting $\ell(S^c)=\infty.$ $\endgroup$ – Dap Feb 14 '18 at 5:59
  • $\begingroup$ Ohhh that makes so much more sense. Fantastic! $\endgroup$ – procrastidigitation Feb 14 '18 at 6:24

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