Measurable subset of $\mathbb{R}$ with a specific property

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.

Solution 1:

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 $\mathbb{R}$.