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.
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.