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Let $A$ be a subset of $\mathbb{R}$ such that its intersection with every finite segment is Lebesgue measurable. I am looking for an example of such an $A$ with the additional property that the function $\varphi(t)=\mu (A\cap\lbrack t,t+1])$ is strictly increasing in $t$, where $\mu$ is Lebesgue measure.

It is easy to see that such a set $A$ should have empty interior in $\mathbb{R}$.

Thanks.

  • 1
    So, $A = \bigcup_{n=1}^\infty A\cap [-n,n]$ is measurable itself?2011-10-10
  • 4
    Perhaps instead you should try to show that such a set cannot exist...2011-10-10
  • 0
    @Gortaur Sorry for this ambiguity, I was not sure about the terminology as I have seen somewhere that a set $A$ is called "measurable", if its intersections with all finite segments are measurable.2011-10-10
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    @GEdgar, after your remark (and some vain attempts) the nonexistence of such a set $A$ seems more likely to me. On the other hand, originally I was looking for TWO sets $A$ and $B$ such that the function $\frac{\varphi_{A}(t)} {\varphi_{A}(t)+\varphi_{B}(t)}$ is strictly increasing, where $\varphi _{X}(t)=\mu(X\cap\lbrack t,t+1])$, that gives much more freedom. It is clear that taking $B=$ $\mathbb{R}\backslash A$ we get the above question and I decided to "simplify" so the problem.2011-10-11

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