# Exercise 36 Ch 1 in Stein's Real Analysis [duplicate]

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.

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

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.