Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\mathbb{R}$.


share|improve this question
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
add comment

1 Answer

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.

share|improve this answer
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
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.