DECOMPOSITION OF THE LINE INTO COUNTABLY–MANY

MEASURE–THEORETIC DENSE SETS

Joseph Bennish

California State University, Long Beach

Like the warp and woof of a piece of cloth, two sets may be thoroughly intermingled. But how intermingled can disjoint sets be? With this in mind we ask the following question:

Can ℝ^{n} be decomposed into countably–many (or even just two) disjoint Lebesgue measurable sets such that the intersection of any one of these sets with any continuous (non–constant) curve has positive one–dimensional Hausdorff measure (or, at least, positive Hausdorff dimension)? (For the definition of Hausdorff measure and Hausdorff dimension, see, e.g., [1].)

In this note we show that the one–dimensional case is true, that is, we show that the real line ℝ can be decomposed into countably–many Lebesgue measurable sets such that the intersection of any of these sets with any open interval has positive measure. (Decomposition of ℝ into two such sets was posed as a problem in [2, p. 59].)

We call a measurable subset E of an interval I an m–dense set (with respect to I) if for any open interval I_{1} ⊂ I we have 0 < m(E ∩ I_{1}), where m represents one–dimensional Lebesgue measure. The one–dimensional problem is then whether ℝ can be expressed as the union of mutually–disjoint m–dense sets. It suffices to carry out the construction of countably–many m–dense sets on the unit interval [0, 1), then extend these sets to be periodic of period one. Suppose the following statement is true:

Any set B which is m-dense with respect to the unit interval is the

(1)

union of two disjoint m-dense sets C and D with m(D) = m(B)/2.

Then, starting with the unit interval and iterating this decomposition, we obtain

[0, 1) = ⋃_{1}^{n} A_{i} ∪ B_{n}

for all n ≥ 1, where B_{0} = [0, 1), and B_{n}, n ≥ 0, is the union of disjoint m–dense sets A_{n+1} and B_{n+1} such that m(B_{n+1}) = m(B_{n})/2. Then, [0, 1) = ∪_{1}^{∞}A_{i} ∪ B_{∞}, where B_{∞} has measure 0 and so can be incorporated into any of the other sets. Thus, it suffices to prove statement (1).

First, given a set B of positive measure contained in an interval I of length L, we define a generalized Cantor set E = E(B, I, μ), 0 < μ < 1. Given δ, 0 < δ < L, choose {δ_{n}} so that

(2)  L = δ_{0} > δ_{1} > δ_{2} > ⋯, δ_{n} → δ.

Put E_{0} = Ī. For n ≥ 0, E_{n} is constructed so that E_{n} is the union of 2^{n} disjoint closed intervals, each of length 2^{−n}δ_{n}. Delete an open interval in the center of each of these 2^{n} intervals, so that each of the remaining 2^{n+1} intervals has length 2^{−n−1}δ_{n+1} and let E_{n+1} be the union of these 2^{n+1} intervals. Then E_{1} ⊃ E_{2} ⊃ ⋯, m(E_{n}) = δ_{n}, and the generalized Cantor set E = ∩_{1}^{∞}E_{n} has measure δ. Here δ is chosen so that m(E ∩ B) = (1 − μ)m(B) (a continuity argument shows that this can be done). Thus, F = F (B, I, μ), the complement of E with respect to I, is a dense open subset of I satisfying m(F ∩ B) = μm(B). (Note that E and F depend on the choice of the sequence {δ_{n}}.)

We now give the proof of statement (1). Choose positive numbers μ_{i} < 1, i = 1, 2, …, such that ∏_{1}^{∞} μ_{i} = 1/2. We define a decreasing sequence of open dense sets D_{n} contained in the unit interval such that

(3)  m(B ∩ D_{n}) = ∏_{1}^{n} μ_{i} · m(B).

The set D_{1} = F (B, [0, 1), μ_{1}) is the union of disjoint open intervals I_{j}^{(1)}, j ≥ 1. Let

D_{2} = ⋃_{j=1}^{∞} F (B ∩ I_{j}^{(1)}, I_{j}^{(1)}, μ_{2}).

Then, equation (3) is satisfied for n = 2, and D_{2} ⊂ D_{1}. D_{2} is open so it is also the union of disjoint open intervals I_{j}^{(2)}. Continuing this process we obtain the prescribed sequence of open sets D_{n}.

Let D = B ∩ (∩_{1}^{∞}D_{n}). We have m(D) = lim m(D_{n} ∩ B) = m(B)/2, since D_{n} ∩ B form a decreasing sequence. Next, we show that the sets C = B\D and D are m–dense subsets of the unit interval provided that in (2) the term δ_{1} is always chosen larger than L/2. With this proviso the subintervals I_{j}^{(n)} of D_{n} have length less than 2^{−n}. Since D_{n}, n ≥ 1, is a dense subset of [0, 1), any interval I ⊂ [0, 1) contains an interval of the form I_{j}^{(n)} for n sufficiently large. But, by the construction,

m(D ∩ I) ≥ m(D ∩ I_{j}^{(n)}) = ∏_{n+1}^{∞} μ_{i} · m(B ∩ I_{j}^{(n)}) > 0.

The above equality shows that m(D ∩ I_{j}^{(n)}) < m(I_{j}^{(n)}) so that m(C ∩ I) > 0. This completes the proof of statement (1). Note that the decomposition of the unit interval into m–dense sets A_{i}, the measure of A_{i}, i ≥ 2, can be made arbitrarily small by taking ∏ μ_{i} slightly less than 1. A simple argument then shows that the real line can be decomposed into m–dense sets, all but one of which has measure less than an arbitrarily assigned positive number.

\underline{References}

1. K. J. Falconer, The Geometry of Fractal Sets. Cambridge University Press, Cambridge, 1985.

2. W. Rudin, Real and Complex Analysis, 2nd ed., McGraw–Hill Book Co., New York, 1974.