Construct a measurable set $E\subset [0,1]$ such that for any non-empty open sub-interval $I$ in $[0,1]$, both sets $E\cap I$ and $E^c\cap I$ have positive measure.

[Stein's Hint: For the first part, consider a Cantor-like set of positive measure, and add in each of the intervals that are omitted in the first step of its construction, another Cantor-like set. Continue this procedure indefinitely.]

So I tried using the hint to construct the desired set, but I must be misunderstanding something because I do not think the construction works.

My Thoughts:

I believe the basic idea is that we generate a Cantor-like set $C_1$ by removing repeatedly open intervals of some appropriate length at each stage of the construction starting from $[0,1]$.

Then as said in the hint, at the first step in the construction we removed centrally some open interval $I_1$ from $[0,1]$. Then we generate another Cantor-like set $C_2$ from $I_1$. During the first stage of $C_2$'s construction we removed $I_2$ from $I_1$. Repeat indefinitely...

My problem with this is that during the second stage of the construction of $C_1$, we removed $2$ open intervals, call one of them $\mathcal{I}$, from $[0,1]\setminus I_1$. If we take the union of all these Cantor-like sets, denote it as $E$, then wouldn't $E\cap \mathcal{I}=\emptyset$?

If true, then it seems we would have to apply this to all the open intervals that $C_1$ removes and generate a collection of Cantor-like sets from them. Each of which would need the same procedure done to them.

This doesn't seem correct to me and we may need to define $E$ as the countable union of countable unions of Cantor-like sets.

Any help thinking correctly about this problem would be much appreciated.

Note: It's proven (I've proved) that the Cantor-like sets Stein is talking about have positive measure (Exercise 4 in Stein).


marked as duplicate by Tim kinsella, Community Feb 17 at 10:50

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I typed this up a while ago -- I hope it helps.

First I claim that if $I\subset \mathbb{R}$ is any nonempty open interval, there exists a closed set $S\subset I$ such that $S$ has positive Lebesgue measure and $S$ has empty interior. For instance we may construct a "fat Cantor set" (sometimes called a "thick Cantor set" or a "Smith-Voltera Cantor set") inside the interval. Doing the same construction inside any open interval $J\subset I-S$ we may find another closed set, $T$, with positive measure and empty interior such that $S\cap T=\emptyset$.

Now the set of "rational intervals" (that is, intervals with rational center and rational radius) is countable. Thus we may enumerate them: $I_1, I_2, I_3,...$. Now, as above, we may find a pair of closed disjoint positive-measure empty-interior sets $$S_1,~~T_1\subset I_1.$$ Now $ I_2- S_1-T_1$ contains a nonempty open interval $J_2$ since $S_1,T_1$ are closed and their union cannot be $I_2$ since $S_1\cup T_1$ has empty interior, since $S_1$ and $T_1$ have empty interior. By working inside $J_2$, we may find closed sets $S_2, T_2\subset I_2$ with positive measure and empty interior and such that $S_1,T_1,S_2,T_2$ are pairwise disjoint. Then working inside an open interval in $$I_3- S_1-T_1-S_2-T_2$$ we may find closed sets $S_3, T_3\subset I_3$ with positive measure and empty interior and such that $S_1,T_1,S_2,T_2, S_3, T_3$ are pairwise disjoint. Continuing in this way we get a sequence of pairwise disjoint sets with positive measure $$S_1, T_1, S_2, T_2,S_3,T_3,...$$ such that every rational interval contains one (in fact infinitely many) of the pairs $S_j, T_j$.

Now set $E:= \cup_{k=1}^\infty S_k$ and let $I$ be an arbitrary nonempty interval of positive length. Then $I$ contains a rational interval, and therefore $I$ contains a pair $S_j, T_j$. Then since $E\cap T_j=\emptyset$ and $S_j\subset E$, we have

$$T_j\subset I-E \subset I-S_j.$$ Therefore

$$0<\mu(T_j)<\mu(I-E)\leq \mu (I-S_j)= \mu (I)-\mu(S_j) <\mu(I).$$ where in the first and last inequalities we used the fact that the sets $S_j, T_j$ have positive measure, and in the equality we used the fact that $S_j\subset I$. We have proved that $E$ possesses the property.


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