I was wondering how one can construct a Borel set that doesn’t have full measure on any interval of the real line but does have positive measure everywhere.

To be precise, if μ denotes Lebesgue measure, how would one construct a Borel set A⊂R such that

0<μ(A∩I)<μ(I)

for every interval I in R?Moreover, would such a set necessarily have to contain infinite measure?

**Answer**

If you got this from Rudin (it is Exercise 8, Ch. 2 in his *Real & Complex Analysis*), here is his personal answer (excerpted from *Amer. Math Monthly*, Vol. 90, No.1 (Jan 1983) pp. 41-42). He works with the unit interval [0,1], but of course this can be extended to R by doing the same thing in each interval (and by scaling these replications appropriately you can get the final set with finite measure). Anyways, here's how it goes:

"Let I=[0,1], and let CTDP mean compact totally disconnected subset of I, having positive measure. Let ⟨In⟩ be an enumeration of all segments in I whose endpoints are rational.

Construct sequences ⟨An⟩,⟨Bn⟩ of CTDP's as follows: Start with disjoint CTDP's A1 and B1 in I1. Once A1,B1,…,An−1,Bn−1 are chosen, their union Cn is CTD, hence In∖Cn contains a nonempty segment J and J contains a pair An,Bn of disjoint CTDP's. Continue in this way, and put

A=∞⋃n=1An.

If V⊂I is open and nonempty, then In⊂V for some n, hence An⊂V and Bn⊂V. Thus

0<m(An)≤m(A∩V)<m(A∩V)+m(Bn)≤m(V);

the last inequality holds because A and Bn are disjoint. Done.

The purpose of publishing this is to show that the highly computational construction of such a set in [another article] is much more complicated than necessary."

**Edit:** In his excellent comment below, @ccc managed to isolate the necessary components of my solution, and after incorporating his observation it has been greatly simplified. (Actually, after trimming the fat, I've realized that it is actually not entirely dissimilar from Rudin's.) Here it is:

Let {rn} be an enumeration of the rationals, let V1 be a segment of finite length centered at r1, and let Vn be a segment of length m(Vn−1)/3 centered at rn. Set

Wn=Vn−∞⋃k=1Vn+k,

and observe that

m(Wn)≥m(Vn)−∞∑k=1m(Vn+k)=m(Vn)−m(Vn)∞∑k=13−k=m(Vn)2.

In particular, m(Wn)>0.

For each n, choose a Borel set An⊂Wn with 0<m(An)<m(Wn). Finally, put A=⋃∞n=1An. Because An⊂Wn and the Wn are disjoint, m(A∩Wn)=m(An). That is to say,

0<m(A∩Wn)<m(Wn)

for every n. But every interval contains a Wn, so A meets the criteria, and has finite measure (specifically, m(A)≤∑nm(Vn)=2m(V1)<∞).

As a curiosity, here's my own "unnecessarily computational" way (though it's not quite as lengthy as that in the article Rudin was referring to), which I can't resist including because I slaved over it when I first came across this problem, before finding Rudin's solution:

Let {rn} be an enumeration of the rationals, and put

Vn=(rn−3−n−1,rn+3−n−1),Wn=Vn−∞⋃k=1Vn+k.

Observe that

m(Wn)>m(Vn)−∞∑k=1m(Vn+k)=m(Vn)−m(Vn)∞∑k=13−k=m(Vn)2.(1)

(We have strict inequality because there exist rationals ri, with i>n, in the complement of Vn.)

For each n, let Kn be a Borel set in Vn with measure m(Kn)=m(Vn)/2. Finally, put

An=Wn∩Kn,A=∞⋃n=1An.

To prove that A has the desired property, it is enough to verify that the inequalities

0<m(A∩Vn)<m(Vn)(3)

hold for every n. (This is because every interval contains a Vn.) For the left inequality, it is enough to prove that m(An∩Vn)=m(An)=m(Wn∩Kn)>0. This follows from the relations

m(Wn∪Kn)≤m(Vn)<m(Wn)+m(Kn)=m(Wn∪Kn)+m(Wn∩Kn),

the second inequality being a consequence of (1) and the fact that m(Kn)=m(Vn)/2.

For the right inequality of (3), observe that Vn⊂Wci for i<n, and that therefore

m(A∩Vn)=m(∞⋃k=0An+k∩Vn)≤∞∑k=0m(Kn+k∩Vn)

<∞∑k=0m(Kn+k)=∞∑k=0m(Vn+k)2=∞∑k=0m(Vn)2k+1=m(Vn).

The strict inequality above follows from three observations: (i) m(Ki)>0 for every i; (ii) Ki⊂Vi; and (iii) there exist neighborhoods Vi, with i>n, that are contained entirely in the complement of Vn.

So A meets the criteria (and also has finite measure).

**Attribution***Source : Link , Question Author : user1736 , Answer Author : Nick Strehlke*