# Can one divide a set into subintervals [duplicate]

Sorry that the question title is unclear, I didn't know how to ask it.

Take set $A \subseteq [0,1]$, measurable. Does there exist a sequence $x_1,x_2,\dots$ such that $\forall x_i$,

\begin{align*} \displaystyle\frac{\mu\left([x_i,x_{i+1}]\cap A \right)}{\mu\left([x_i,x_{i+1}]\right)} \in \{0,1\} \end{align*}

where $\mu$ is the Lebesgue measure?

• are you requiring $x_i$ distinct? or $x_1 < x_2 < ...$? – user66081 Jan 26 '16 at 0:50
• So you ask: is there a measurable set $A$ such that for every interval $[a,b]$ with $a<b$ both $[a,b]\cap A$ and $[a,b]\setminus A$ have positive measure? yes. – GEdgar Jan 26 '16 at 0:53
• Sorry, I'm sure I'm being unclear. Take the measurable set $A$ as given. Does there exist a collection of intervals, $[a_i,b_i]$ such that $\{$ $\mu\left([a_i,b_i]\right)=\mu\left([a_i,b_i]\right)$ or $\mu\left([a_i,b_i]\right)=0$ $\}$ and $\cup\limits_{i=1}^\infty [a_i,b_i] = [0,1]$. – user257254 Jan 26 '16 at 0:57
• A countable list of intervals? Doesn't the fat Cantor set violate that? – Brian Tung Jan 26 '16 at 1:01
• Do the $x_i$ have to be in $A$? If so, it's true if $A$ contains an interval (just choose an increasing sequence of points in the interval). What if $A$ has positive measure but contains no intervals, such as a fat Cantor set? – Ian Jan 26 '16 at 1:18

No, this is not always possible, and in fact it can fail in an extreme way: we can have Lebesgue measurable (even Borel!) $A$ such that both $A$ and the complement of $A$ have positive-measure intersection with every nontrivial interval. For such an $A$ and any points $x_1< x_2$, we will never have $${\mu([x_1, x_2]\cap A)\over \mu([x_2,x_1])}\in\{0, 1\}.$$ See Construct a Borel set on R such that it intersect every open interval with non-zero non-"full" measure, and also Construction of a Borel set with positive but not full measure in each interval.