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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 $\mathbb{R}$.


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So, $A = \bigcup_{n=1}^\infty A\cap [-n,n]$ is measurable itself? – Ilya Oct 10 '11 at 17:32
Perhaps instead you should try to show that such a set cannot exist... – GEdgar Oct 10 '11 at 17:47
@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. – t22 Oct 10 '11 at 17:48
@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. – t22 Oct 11 '11 at 5:57

Here is how to construct such a set $A$.

It is well-known that the interval $[0,1)$ contains two disjoint measurable subsets $X$ and $Y$ such that for any $0\leq x < y \leq 1$, both $\mu(X\cap (x,y))>0$ and $\mu(Y\cap(x,y))>0$. See, for example, the very short proof of this by Rudin. It is straightforward to extend Rudin's argument to show that $[0,1)$ contains countably many disjoint sets which all intersect any such interval $(x,y)$ in positive measure.

Let $\ldots,X_{-1},X_0,X_1,X_2,\ldots$ be such a partition of $[0,1)$. Let $T_x$ be the map which translates subsets of $\mathbb{R}$ right by $x$ and define \[ A = \bigcup_{\stackrel{m,n\in\mathbb{Z}}{m \leq n}} T_n(X_m), \] so the intersection of $A$ with $[n,n+1)$ is the union of appropriate translates of $X_m$ for all $m \leq n$.

Now suppose $k\in\mathbb{Z}$ and $x,y\in\mathbb{R}$ satisfy $k\leq x<y\leq k+1$. Let $s = x-k$ and $t = y-k$, so $0\leq s<t \leq 1$. Then \[ \begin{split} \phi(y)-\phi(x) & = \mu(A\cap [y,y+1]) - \mu(A\cap [x,x+1]) \\ &= \mu(A\cap (x+1,y+1]) - \mu(A\cap [x,y)) \\ & = \mu(A\cap(x+1,y+1)) - \mu(A\cap (x,y)) \\ & = \sum_{l\leq k+1} \mu(X_l\cap (s,t)) - \sum_{l\leq k} \mu(X_l\cap (s,t)) \\ & = \mu(X_{k+1}\cap(s,t)) > 0. \end{split} \] By transitivity of inequality $\phi(x) < \phi(y)$ for all $x < y$ in $\mathbb{R}$.

To see where the construction above came from, you can prove that any $A$ with the properties specified in the original post is of the form constructed above, with two modifications. First, there may be an additional $X_{-\infty}$ disjoint from all the $X_k$ translated into every $A\cap[n,n+1)$. Second, the whole construction may be modified by any measure zero set.

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What a clever trick!! So, when you "shift" from $x$ to $y$, assuming $x$ and $y$ are "close", you are swapping the interval $[x,y)$ with $(x+1,y+1]$ whose intersection with $A$ is strictly bigger (in measure), because $(x+1,y+1)$ intersects a translation of $X_{k+1}$ as well. Of course, the boundaries of the intervals don't interfere since the measure is Lebesgue. – André Caldas Sep 19 '13 at 15:42

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