Annales Academiæ Scientiarum Fennicæ Series A. I. Mathematica Volumen 18, 1993, 77–91

NULL SETS FOR DOUBLING AND DYADIC DOUBLING MEASURES

Jang-Mei Wu

University of Illinois at Urbana-Champaign, Department of Mathematics 273 Altgeld Hall, MC-382, 1409 West Green Street, Urbana, IL 61801, U.S.A.

Abstract. In this note, we study sets on the real line which are null with respect to all doubling measures on R, or with respect to all dyadic doubling measures on R. We give some sufficient conditions for the former, a test for the latter, and some examples.

Our work is motivated by a characterization of dyadic doubling measures by Fefferman, Kenig and Pipher [5], and by a result of Martio [8] on porous sets and sets of total 𝒜 -harmonic measure zero for certain class of nonlinear 𝒜 -operators.

A measure μ on R is said to have the doubling property with constant λ if, whenever I and J are two neighboring intervals of same length then μ(I) ≤ λμ(J); denote by 𝒟(λ) the collection of all doubling measures with constant λ, and 𝒟 = ∪_{λ≥1}𝒟(λ). A measure μ on R has the dyadic doubling property with constant λ if μ(I) ≤ λμ(J) whenever I and J are two dyadic neighboring intervals of same length and I ∪ J is also a dyadic interval; denote by 𝒟_{d}(λ) and 𝒟_{d} the corresponding collections of dyadic doubling measures.

Given {a_{n}}, 0 < α_{n} < 1, a set E ⊆ R is called {α_{n}}-porous if there exists a sequence of coverings ℰ_{n} = {E_{n,j}} of E, by intervals with mutually disjoint interiors, so that each E_{n,j} \ E contains an interval J_{n,j} of length ≥ α_{n}|E_{n,j}|, ∪_{ℰ_{n+1}} E_{n+1,k} is contained in ∪_{ℰ_{n}} (E_{n,j} \ J_{n,j}) and sup |E_{n,j}| → 0 as n → ∞. The Cantor ternary set is \{\frac{1}{3}\}-porous. The porous sets studied by Martio [8] are {α}-porous for some α > 0.

Theorem 1. If 0 < α_{n} < 1, ∑_{1}^{∞} α_{n}^{K} = ∞ for all K ≥ 1, and E is {α_{n}}-prorous, then E is null for all doubling measures on R.

Corollary. There exist sets of Hausdorff dimension one which are null for all doubling measures.

The condition given in Theorem 1 cannot be improved:

Theorem 2. If 0 < α_{n} < \frac{1}{4} is a decreasing sequence satisfying ∑_{n} α_{n}^{K} < ∞ for some K ≥ 1, then there exists a perfect set which is {α_{n}}-porous, but carries a positive measure for some μ ∈ 𝒟.

1991 Mathematics Subject Classification: Primary 28A12.

Research partially supported by the National Science Foundation.

Let E be a closed set in [0, 1] and λ > 1. In Theorem 3, we give a deterministic procedure of testing whether E is 𝒟_{d}(λ)-null. In this process, an optimal measure μ_{E,λ} among 𝒟_{d}(λ) is selected for E. The precise statement is given in Section 2.

Denote by 𝒩 the collection of null sets for doubling measures {E : μ(E) = 0 for all μ ∈ 𝒟}, and 𝒩_{d} its dyadic counterpart {E : μ(E) = 0 for all μ ∈ 𝒟_{d}}. Clearly 𝒩_{d} ⊆ 𝒩 and 𝒩 is translation invariant. The assertion that 𝒩_{d} ≠ 𝒩 is not suprising, however it requires a lot of work.

Theorem 4. There exists a perfect set S ⊆ [0, 1] which is in 𝒩 \ 𝒩_{d}. And corresponding to this S, there exists a set T of dimension one, so that t + S ∈ 𝒩_{d} for each t ∈ T.

It would be interesting to know whether a pair of sets S, T can be chosen to satisfy length (T) > 0 in addition to the properties in Theorem 4.

Theorem 5. Let t be any number whose binary expansion has infinitely many zeros and infinitely many ones. Then there exists a perfect set S_{t} so that S_{t} ∈ 𝒩 \ 𝒩_{d} but t + S_{t} ∈ 𝒩_{d}.

Finally, in Section 4, we shall comment on relations betwen sets in 𝒩 and null sets of the harmonic measures with respect to the p-Laplacians in the upper half plane.

The author would like to thank A. Hinkkanen for his joint contribution on Theorem 1, and R. Kaufman for many conversations.

1. Proofs of Theorems 1 and 2

We first state a useful lemma.

Lemma 1. Let μ be a dyadic doubling measure on [0, 1] and I be any subinterval. Then there exists K > 1 depending on the dyadic doubling constant λ only, so that

4|I|^{1/K}μ([0, 1]) ≥ μ(I) ≥ \frac{1}{4}|I|^{K}μ([0, 1]).

Proof. Let I_{1} and I_{2} be two adjacent dyadic closed intervals in [0, 1] such that I ⊆ I_{1} ∪ I_{2} and 2|I| ≥ |I_{1}| + |I_{2}|. Then

μ(I) ≤ μ(I_{1}) + μ(I_{2})

≤ ((\frac{λ}{1 + λ})^{− log_{2} |I_{1}|} + (\frac{λ}{1 + λ})^{− log_{2} |I_{2}|})μ([0, 1])

≤ 2(2|I|)^{log_{2}((1+λ)/λ)}μ([0, 1]) ≤ 4|I|^{log_{2}((1+λ)/λ)}μ([0, 1]).

If

|I| ≥ 1 − 16^{−(log_{2}((1+λ)/λ))^{−1}} ≡ A,

then

μ([0, 1] \ I) ≤ 8(1 − |I|)^{log_{2}((1+λ)/λ)}μ([0, 1]) ≤ \frac{1}{2}μ([0, 1]).

Hence μ(I) ≥ \frac{1}{2}μ([0, 1]). If |I| < A, let J be the largest dyadic interval contained in I. Therefore |J| ≥ |I|/4 and

μ(I) ≥ μ(J) ≥ (\frac{1}{1 + λ})^{− log_{2} |J|}μ([0, 1]) ≥ (|I|/4)^{log_{2}(1+λ)}μ([0, 1])
≥ μ([0, 1])|I|^{(log_{2}(1+λ))(1−2(log_{2} A)^{−1})}.

Proof of Theorem 1. Assume E ⊆ [0, 1], and let ℰ_{n} = {E_{n,j}} be the coverings of E and {J_{n,j}} be the subintervals of E_{n,j} \ E in defining {α_{n}}-porosity. Let μ ∈ 𝒟, it follows from Lemma 1 that μ(J_{n,j}) ≥ \frac{1}{4}α_{n}^{K}μ(E_{n,j}) for some K ≥ 1 depending on μ only. Thus

μ(E_{n,j} \ J_{n,j}) ≤ (1 − \frac{1}{4}α_{n}^{K})μ(E_{n,j}).

Summing over j, we obtain

∑_{k} μ(E_{n+1,k}) ≤ (1 − \frac{1}{4}α_{n}^{K}) ∑_{j} μ(E_{n,j}).

Therefore

μ(E) ≤ ∏_{n}(1 − \frac{1}{4}α_{n}^{K})μ([0, 1]) = 0.

Proof of Theorem 2. Let N_{n} be a rapidly increasing sequence of odd integers with N_{1} = 1, N_{n} ≥ α_{n−1}^{−1} for n ≥ 2. After replacing α_{n} by a number which is at most twice its size, we may assume that α_{n} = m_{n}N_{n+1}^{−1} for some odd integer m_{n}. The construction of E resembles that of the Cantor set. First we remove the open interval which constitutes the middle α_{1} position of [0, 1], and subdivide the two remaining closed intervals into subintervals of equal length N_{2}^{−1} , call this collection of subintervals 𝒮_{1} . This subdivision is possible due to the modification on α_{n} ’s. On each interval in 𝒮_{1} , remove the open interval which constitutes its middle α_{2} portion, and subdivide the remaining intervals into subintervals of equal length (N_{2}N_{3})^{−1} , call this new collection of subintervals 𝒮_{2} . Continue the process indefinitely and let

E = ⋂_{n} ( ⋃_{I∈𝒮_{n}} I).

Clearly E is {α_{n}}-porous.

It remains to choose N_{n} so that μ(E) > 0 for some μ ∈ 𝒟. Our idea comes from Ahlfors and Beurling [2; Theorem 3]. First, we construct a function h which plays the role of 1 + λcosine in [2].

Lemma 2. Given 0 < α < \frac{1}{4} , K > 2, there exists a function h continuous on R, of period 1, monotonic in [0, \frac{1}{2}] and in [\frac{1}{2}, 1] respectively, which satisfies ∫_{0}^{1} h(x) dx = 1,

h(x) = \{ \begin{matrix}α^{K−1} & on [\frac{1}{2}(1 − α), \frac{1}{2}(1 + α)], \\ 1 + \sqrt{α} & on [0, \frac{1}{2} − \sqrt{α}] ∪ [\frac{1}{2} + \sqrt{α}, 1], \end{matrix}

and h(x) dx is in 𝒟(B^{K}) for some absolute constant B > 2.

As an example, we may choose

h(x) = α^{K−1} + α^{−K}(x − \frac{1 + α}{2})^{2K−1}

on

[\frac{1 + α}{2}, \frac{1 + α}{2} + \frac{α}{4}^{(2K−1)/(4(K−1))}],

piecewise linear on

[\frac{1 + α}{2} + \frac{α}{4}^{(2K−1)/(4(K−1))}, \frac{1}{2} + \sqrt{α}]

with derivatives between 1/4\sqrt{α} and 4/\sqrt{α}, and h(x) = h(1 − x) for x in [\frac{1}{2} − \sqrt{α}, \frac{1}{2}(1 − α)], so that the continuity, monotonicity and ∫_{0}^{1} h(x) dx = 1 are satisfied. For this h, h dx ∈ 𝒟(B^{k}) for some absolute constant B > 2.

In the hypothesis ∑ α_{n}^{K} < ∞, we may assume K > 2. Corresponding to each pair (α_{n}, K), we fix a function h_{n} which satisfies properties in Lemma 2 with α = α_{n}. Denote by

A_{n} = ⋃_{k=−∞}^{∞} ([k, k + \frac{1}{2} − \frac{1}{2}α_{n}]) ∪ ([k + \frac{1}{2} + \frac{1}{2}α_{n}, k + 1]),

F_{k} = ⋃_{I∈𝒮_{k}} I, M_{n} = ∏_{1}^{n} N_{k}, and f_{n}(x) = ∏_{1}^{n} h_{k}(M_{k}x).

We shall choose N_{n} inductively so that N_{n+1} ≫ N_{n} and that f_{n}(x) is “nearly constant” on each interval of length M_{n+1}^{−1} .

Recall that N_{1} = 1 and assume that odd integers N_{2}, N_{3}, …, N_{n} have been chosen so that

(1.1)  ∫_{F_{k}} f_{k}(x) dx ≥ ∏_{j=1}^{k}(1 − 2α_{j}^{K})  (1 ≤ k ≤ n),

and whenever |x − x′| ≤ M_{k}^{−1} , (2 ≤ k ≤ n),

(1.2)  \frac{k − 1}{k} < h_{k−1}(M_{k−1}x)/h_{k−1}(M_{k−1}x′) < \frac{k + 1}{k},

and

(1.3)  \frac{k − 1}{k} < f_{k−1}(x)/f_{k−1}(x′) < \frac{k + 1}{k}.

Note that ∫_{0}^{1} h_{k}(x) dx = 1, ∫_{[0,1]∩A_{k}} h_{k}(x) dx = 1 − α_{k}^{K} and that f_{k}(x) is uniformly continuous on R for each k ≥ 1. Let F ⊆ [0, 1] be any measurable set. Then χ_{F} f_{n} is the pointwise a.e. and L^{1} limit of an increasing sequence of simple functions, each of which has the form ∑ a_{j}χ_{I_{j}}, where {a_{j}} are constants and {I_{j}} are finitely many mutually disjoint open intervals with rational end points. Therefore,

∫_{0}^{1} χ_{F} (x)f_{n}(x)χ_{A_{n+1}}(Mx)h_{n+1}(Mx) dx → (1 − α_{n+1}^{K}) ∫_{F} f_{n}(x) dx

as M → ∞. Thus a large odd integer N_{n+1} can be found so that N_{n+1} > α_{n}^{−1} , and (1.1), (1.2) and (1.3) hold with k = n + 1. Here we have used the fact that F_{n+1} = {x ∈ F_{n} : M_{n+1}x ∈ A_{n+1}}.

Let μ be a weak limit point of f_{n}(x) dx. Then μ(E) > 0 in view of (1.1) and ∑ α_{n}^{K} < ∞. To verify that μ is a doubling measure we consider two neighboring intervals I and I′ satisfying

M_{n+1}^{−1} ≤ |I| = |I′| ≤ M_{n}^{−1}.

In view of (1.3)

(\frac{n − 1}{n})^{2} ≤ \frac{f_{n−1}(x)}{f_{n−1}(x′)} ≤ (\frac{n + 1}{n})^{2}

whenever x ∈ I and x′ ∈ I′. We note that f_{m}(x)/f_{n}(x) has period M_{n+1}^{−1} if m ≥ n + 1, and that

\frac{n}{n + 1} < \frac{h_{n}(M_{n}x)}{h_{n}(M_{n}x′)} < \frac{n + 2}{n + 1}

whenever |x − x′| ≤ M_{n+1}^{−1} . Writing

f_{m}(x) = f_{n−1}(x)h_{n}(M_{n}x)\frac{f_{m}(x)}{f_{n}(x)}, for m ≥ n + 1,

we deduce from the fact h_{n}(M_{n}x) dx ∈ 𝒟(B^{K}) that

(CB^{K})^{−1} ≤ ∫_{I} f_{m}(x) dx/ ∫_{I′} f_{m}(x) dx ≤ CB^{K}

for every m ≥ n + 1. Therefore μ ∈ 𝒟.

82 Jang-Mei Wu

2. A test for 𝒟_{d}(λ)-null sets Given a closed set E in [0, 1], we shall develop a procedure to test whether

E is in 𝒟_{d}(λ) for some λ > 1.

Let I_{n,j}, 1 ≤ j ≤ 2^{n}, be the dyadic closed intervals in [0, 1] of length 2^{−n}, I_{n,j}^{r} be the closed interval which forms the right half of I_{n,j} and let I_{n,j}^{l} = I_{n,j} \ I_{n,j}^{r}. Let

h_{n,j}(x) = \{ \begin{matrix}1 & on I_{n,j}^{r} \\ −1 & on I_{n,j}^{l} \\ 0 & on R \ I_{n,j}. \end{matrix}

For a fixed integer n ≥ 2, define

f_{n}^{(n)} ≡ ∏_{j=1}^{2^{n}} (1 + δ(n, j)τh_{n,j}),

and dν_{n}^{(n)} ≡ f_{n}^{(n)}dx, where τ = (λ − 1)/(λ + 1) and

δ(n, j) = \{ \begin{matrix}1, & E ∩ I_{n,j}^{r} ≠ ∅, \\ −1, & E ∩ I_{n,j}^{r} = ∅. \end{matrix}

Denote by E_{n} = ∪{I_{n+1,j} : I_{n+1,j} ∩ E ≠ ∅, 1 ≤ j ≤ 2^{n+1}}. Let

f_{n−1}^{(n)} ≡ ∏_{j=1}^{2^{n−1}} (1 + δ(n − 1, j)τh_{n−1,j}),

and dν_{n−1}^{(n)} ≡ f_{n−1}^{(n)}dν_{n}^{(n)} , where

δ(n − 1, j) = \{ \begin{matrix}1, & ν_{n}^{(n)}(I_{n−1}^{r} ∩ E_{n}) ≥ ν_{n}^{(n)}(I_{n−1}^{l} ∩ E_{n}), \\ −1, & otherwise. \end{matrix}

After defining f_{k}^{(n)} and dν_{k}^{(n)} , we let

f_{k−1}^{(n)} ≡ ∏_{j=1}^{2^{k−1}} (1 + δ(k − 1, j)τh_{k−1,j}),

and dν_{k−1}^{(n)} ≡ f_{k−1}^{(n)}dν_{k}^{(n)} , where

δ(k − 1, j) = \{ \begin{matrix}1, & ν_{k}^{(n)}(I_{k−1,j}^{r} ∩ E_{n}) ≥ ν_{k}^{(n)}(I_{k−1,j}^{l} ∩ E_{n}), \\ −1, & otherwise. \end{matrix}

Continue this process until we arrive at ν_{1}^{(n)} . Define μ_{n} ≡ ν_{1}^{(n)} .

Notice that

(2.1)  ν_{k}^{(n)}(I_{k,j}) = 2^{−k}  (1 ≤ k ≤ n)

with the understanding that I_{0,j} ≡ I_{0,1} ≡ [0, 1]; and that from dν_{k}^{(n)} to dν_{k−1}^{(n)} , total measure in each I_{k−1,i} is kept unchanged, but the total measures of I_{k−1,i}^{r} and I_{k−1,i}^{l} are redistributed in the most advantageous way.

Repeat for each n ≥ 2, to obtain a sequence of measures {μ_{n}}. Let μ_{E,λ} be a weak limit point of {μ_{n}}, extended to R with period 1.

Theorem 3. Among all the measures in 𝒟_{d}(λ) which have mass one on [0, 1], μ_{E,λ} has the maximum measure on E. In particular, E is 𝒟_{d}(λ)-null if and only if μ_{E,λ}(E) = 0.

Proof. It is clear that μ_{E,λ}([0, 1]) = 1 and μ_{E,λ} ∈ 𝒟_{d}(λ).

Let ω be any measure in 𝒟_{d}(λ) with ω([0, 1]) = 1. We claim, in fact, that

(2.2)  ω(E_{n}) ≤ μ_{n}(E_{n}).

Let m be the largest integer in [1, n], if it exists, such that there exists at least one interval I_{m,j} on which

(2.3)  \frac{ω(I_{m,j}^{r})}{ω(I_{m,j}^{l})} ≠ \frac{μ_{n}(I_{m,j}^{r})}{μ_{n}(I_{m,j}^{l})}.

(If such m does not exist, then ω(E_{n}) = μ_{n}(E_{n}).) We shall redistribute the measure ω on these I_{m,j} ’s and keep ω unchanged elsewhere. Denote by ℐ_{m} = {I_{m,j} : (2.3) holds on I_{m,j}}; and let ω_{m} = ω on [0, 1] \ ∪_{ℐ_{m}} I_{m,j}, and

(2.4)  dω_{m} = \{ \begin{matrix}\frac{ω(I_{m,j})μ_{n}(I_{m,j}^{r})}{ω(I_{m,j}^{r})μ_{n}(I_{m,j})} dω & on I_{m,j}^{r}, \\ \frac{ω(I_{m,j})μ_{n}(I_{m,j}^{l})}{ω(I_{m,j}^{l})μ_{n}(I_{m,j})} dω & on I_{m,j}^{l}, \end{matrix}

for each I_{m,j} ∈ ℐ_{m}. Clearly, if I_{m,j} ∈ ℐ_{m}, then

\frac{ω_{m}(I_{m,j}^{r})}{ω_{m}(I_{m,j}^{l})} = \frac{μ_{n}(I_{m,j}^{r})}{μ_{n}(I_{m,j}^{l})},

and ω_{m} is dyadic on I_{m,j} with constant ≤ λ. Actually, ω_{m} is in 𝒟_{d}(λ) by the following lemma.

Lemma 3. Let I be a dyadic subinterval of [0, 1], and μ and ν be dyadic doubling measures on [0, 1] and I respectively, satisfying μ(I) = ν(I). Then the new measure ω defined by ω ≡ ν on I, ≡ μ on [0, 1] \ I is dyadic doubling on [0, 1] with constant bounded by the maximum of those of μ and ν.

First, we shall verify that ω(E_{n}) ≤ ω_{m}(E_{n}). To show this, it is enough to prove

(2.5)  ω(E_{n} ∩ I_{m,j}) ≤ ω_{m}(E_{n} ∩ I_{m,j})

for each I_{m,j} ∈ ℐ_{m}.

Fix I_{m,j} ∈ ℐ_{m}, clearly (2.5) holds when m = n. Thus we assume m < n and note from the definition of m that

\frac{ω(I_{k,i}^{r})}{ω(I_{k,i})} = \frac{μ_{n}(I_{k,i}^{r})}{μ_{n}(I_{k,i})},  and  \frac{ω(I_{k,i}^{l})}{ω(I_{k,i})} = \frac{μ_{n}(I_{k,i}^{l})}{μ_{n}(I_{k,i})}

for m + 1 ≤ k ≤ n. Therefore

\frac{ω(I_{m,j}^{r} ∩ E_{n})}{ω(I_{m,j}^{r})} = \frac{μ_{n}(I_{m,j}^{r} ∩ E_{n})}{μ_{n}(I_{m,j}^{r})}

and

\frac{ω(I_{m,j}^{l} ∩ E_{n})}{ω(I_{m,j}^{l})} = \frac{μ_{n}(I_{m,j}^{l} ∩ E_{n})}{μ_{n}(I_{m,j}^{l})}.

Moreover, from the construction of μ_{n},

(2.6)  \frac{μ_{n}(I_{n+1,l})}{μ_{n}(I_{k,i})} = \frac{ν_{k}^{(n)}(I_{n+1,l})}{ν_{k}^{(n)}(I_{k,i})},

if 1 ≤ k ≤ n and I_{n+1,l} ⊆ I_{k,i}. Thus by (2.1) and the above identities,

(2.7)  \frac{ω(I_{m,j}^{r}) ∩ E_{n}}{ω(I_{m,j}^{r})} = ν_{m+1}^{(n)}(I_{m,j}^{r} ∩ E_{n})2^{m+1}

and

(2.8)  \frac{ω(I_{m,j}^{l}) ∩ E_{n}}{ω(I_{m,j}^{l})} = ν_{m+1}^{(n)}(I_{m,j}^{l} ∩ E_{n})2^{m+1}.

Writing I in place of I_{m,j} for the rest of this paragraph, we obtain

ω(E_{n} ∩ I) = [\frac{ω(E_{n} ∩ I^{r})}{ω(I^{r})} \frac{ω(I^{r})}{ω(I)} + \frac{ω(E_{n} ∩ I^{l})}{ω(I^{l})} \frac{ω(I^{l})}{ω(I)}]ω(I)

= 2^{m+1}ω(I)[ν_{m+1}^{(n)}(E_{n} ∩ I^{r})\frac{ω(I^{r})}{ω(I)} + ν_{m+1}^{(n)}(E_{n} ∩ I^{l})\frac{ω(I^{l})}{ω(I)}]

≤ 2^{m+1}ω(I)(ν_{m+1}^{(n)}(E_{n} ∩ I^{r})A + ν_{m+1}^{(n)}(E_{n} ∩ I^{l})(1 − A)) where A = \frac{1}{2}(1+τ) if ν_{m+1}^{(n)}(E_{n}∩I^{r}) ≥ ν_{m+1}^{(n)}(E_{n}∩I^{l}), and A = \frac{1}{2}(1−τ) otherwise. From the definition of ν_{m}^{(n)} , (2.6), (2.7), and (2.8), it follows that

\begin{aligned}ω(E_{n} ∩ I) & ≤ 2^{m+1}ω(I)[ν_{m+1}^{(n)}(E_{n} ∩ I^{r})\frac{ν_{m}^{(n)}(I^{r})}{ν_{m}^{(n)}(I)} + ν_{m+1}^{(n)}(E_{n} ∩ I^{l})\frac{ν_{m}^{(n)}(I^{l})}{ν_{m}^{(n)}(I)}] \\ & = ω(I)[\frac{ω(E_{n} ∩ I^{r})}{ω(I^{r})} \frac{μ_{n}(I^{r})}{μ_{n}(I)} + \frac{ω(E_{n} ∩ I^{l})}{ω(I^{l})} \frac{μ_{n}(I^{l})}{μ_{n}(I)}] = ω_{m}(E_{n} ∩ I). \end{aligned}

This proves (2.5) and hence ω(E_{n}) ≤ ω_{m}(E_{n}).

We proceed to make modifications of ω_{m} on each dyadic interval I_{m−1,i} of size 2^{−m+1} on which (2.3) holds with m, j, ω replaced by m − 1, i and ω_{m} respectively, according to the rule (2.4) adapted for m − 1, i and ω_{m}; call this new measure ω_{m−1} . Continue to modify ω_{m−1} on dyadic intervals of size 2^{−m+2} if necessary to obtain ω_{m−2}, . . .. Finally we arrive at a measure ω_{1} , and obtain

ω(E_{n}) ≤ ω_{m}(E_{n}) ≤ ω_{m−1}(E_{n}) ≤ ⋯ ≤ ω_{1}(E_{n})

and

\frac{ω_{1}(I_{m,j}^{r})}{ω_{1}(I_{m,j}^{l})} = \frac{μ_{n}(I_{m,j}^{r})}{μ_{n}(I_{m,j}^{l})},

for all 1 ≤ m ≤ n, 1 ≤ j ≤ 2^{m}. Therefore ω_{1}(E_{n}) = μ_{n}(E_{n}) and (2.2) is proved.

We note that E = ∩_{m}E_{m}. Therefore for any ε > 0 and sufficiently large m and n with m > m(ε) and n > n(ε, m), we have

\begin{aligned}μ_{E,λ}(E) & ≥ μ_{E,λ}(E_{m}) − ε ≥ μ_{n}(E_{m}) − 2ε ≥ μ_{n}(E_{n}) − 2ε \\ & ≥ ω(E_{n}) − 2ε ≥ ω(E) − 2ε. \end{aligned}

This shows that ω(E) ≤ μ_{E,λ}(E).

3. Proofs of Theorems 4 and 5

Given a, ε, δ ∈ (0, 1), εa < δ < ε, we choose a sequence of integers {n_{k}} satisfying n_{k} ≥ 4 and

(3.1)  n_{n+1} > n_{k} + [ε log_{2} k].

For k ≥ 2 + [2^{1/δ}] and 0 ≤ j ≤ 2^{n_{k}} − 1, denote by

L_{k,j} = [\frac{j}{2^{n_{k}}} , \frac{j + 1}{2^{n_{k}}}],

I_{k,j} = [\frac{j}{2^{n_{k}}} , \frac{j}{2^{n_{k}}} + \frac{1}{2^{n_{k}} k̇^{δ}} ],

and

J_{k,j} = [\frac{j + 1}{2^{n_{k}}} − \frac{1}{2^{n_{k}} k̇^{ε}} , \frac{j + 1}{2^{n_{k}}}],

where k̇^{δ} = 2^{[δ log_{2} k]} , k̇^{ε} = 2^{[ε log_{2} k]} and [ ] is the greatest integer function. Note that intervals L’s, I ’s and J ’s are dyadic,

(3.2) |J_{k,j}|/|I_{k,j′}| = O(k^{δ−ε}) = o(1) as k → ∞,

and

(3.3)  |J_{k,j}|/|L_{k+1,j′}| = 2^{n_{k+1}−n_{k}−[ε log_{2} k]} > 2.

The construction of a set S ∈ 𝒩 \ 𝒩_{d} is similar to the Cantor set; collections of nested intervals from {J_{k,j}} are used. The measure μ in 𝒟_{d} to be produced with μ(S) > 0 will satisfy

\frac{μ(J_{k,j})}{μ(L_{k,j})} = (\frac{|J_{k,j}|}{|L_{k,j}|})^{a}

on infinitely many J_{k,j} ’s.

Let {K_{i}} be an increasing sequence of integers with K_{0} ≡ 2+[2^{1/δ}] and some other properties to be specified later. Let S_{0} = [0, 1], 𝒞_{1+K_{0}}^{I} be the collection of all I_{1+K_{0},j} ⊆ S_{0} and 𝒞_{1+K_{0}}^{J} be the collection of all J_{1+K_{0},j} ⊆ S_{0} . After 𝒞_{k}^{I} and 𝒞_{k}^{J} have been defined for some k, 1 + K_{0} ≤ k ≤ K_{1} − 1, we let 𝒞_{k+1}^{I} = 𝒞_{k}^{I} ∪ {I_{k+1,j} ⊆ S_{0} : I_{k+1,j} is not contained in any interval in 𝒞_{k}^{I} ∪ 𝒞_{k}^{J}},

𝒞_{k+1}^{J} = 𝒞_{k}^{J} ∪ {J_{k+1,j} ⊆ S_{0} : J_{k+1,j} is not contained in any interval in 𝒞_{k}^{I} ∪ 𝒞_{k}^{J}}; and let

S_{1}^{I} = union of all intervals in 𝒞_{K_{1}}^{I},
S_{1} = union of all intervals in 𝒞_{K_{1}}^{J}.

Next let 𝒞_{1+K_{1}}^{I} be the collection of all I_{1+K_{1},j} ⊆ S_{1} and 𝒞_{1+K_{1}}^{J} be the collection of all J_{1+K_{1},j} ⊆ S_{1} . And define for each k, 1 + K_{1} ≤ k ≤ K_{2} − 1, 𝒞_{K+1}^{I} = 𝒞_{k}^{I} ∪ {I_{k+1,j} ⊆ S_{1} : I_{k+1,j} is not contained in any interval in 𝒞_{k}^{I} ∪ 𝒞_{k}^{J}},

𝒞_{K+1}^{J} = 𝒞_{k}^{J} ∪ {J_{k+1,j} ⊆ S_{1} : J_{k+1,j} is not contained in any interval in 𝒞_{k}^{I} ∪ 𝒞_{k}^{J}},

S_{2}^{I} = union of all intervals in 𝒞_{K_{2}}^{I},

and

S_{2} = union of all intervals in 𝒞_{K_{2}}^{J}.

Clearly S_{2}^{I} ⊆ S_{1} and S_{2} ⊆ S_{1} .

Continue this procedure to obtain 𝒞_{K_{3}}^{I}, 𝒞_{K_{3}}^{J}, S_{3}^{I} and S_{3} , …, and so on, and let

S = ⋂_{1}^{∞} S_{m}.

To construct μ ∈ 𝒟_{d} with μ(S) > 0, we shall use scale invariant versions of Lemma 3 and the following lemma repeatedly.

Lemma 4. Given a, α, β ∈ (0, 1) with α^{a} + β < 1/16 and c_{1}, c_{2} ∈ (\frac{1}{2}, 2), there exists a measure μ ∈ 𝒟_{d}(10^{1/a}), which satisfies μ([0, 1]) = 1, μ([0, α]) = c_{1}α^{a}, and μ([1 − β, 1]) = c_{2}β.

As an example, we may choose

μ([0, t]) = \{ \begin{matrix}c_{1}t^{a}, & 0 ≤ t ≤ t_{0} ≡ (\frac{1}{8})^{1/a}, \\ \frac{1}{8}c_{1} + (1 − \frac{1}{8}(c_{1} + c_{2}))(t − t_{0})/(\frac{7}{8} − t_{0}), & t_{0} ≤ t ≤ \frac{7}{8}, \\ c_{2}t + 1 − c_{2}, & \frac{7}{8} ≤ t ≤ 1. \end{matrix}

Then extend μ periodically to R with period 1.

All measures μ_{k} defined below are periodic with period 1. Choose μ_{1+K_{0}} ∈ 𝒟_{d}(10^{1/a}) so that

\begin{aligned}μ_{1+K_{0}}(L_{1+K_{0},j}) & = |L_{1+K_{0},j}|, \\ μ_{1+K_{0}}(I_{1+K_{0},j}) & = |L_{1+K_{0},j}|(1 + K_{0})^{−δ}, \\ μ_{1+K_{0}}(J_{1+K_{0},j}) & = |L_{1+K_{0},j}|(1 + K_{0})^{−εa} \end{aligned}

for each 0 ≤ j ≤ 2^{1+K_{0}} − 1. After μ_{k} is selected for some k, 1 + K_{0} ≤ k ≤ K_{1} , we choose μ_{k+1} ∈ 𝒟_{d}(10^{1/a}), so that μ_{k+1} = μ_{k} on each interval in 𝒞_{k}^{I} ∪ 𝒞_{k}^{J}, and μ_{k+1} is a redistribution of μ_{k} on each L_{k+1,j} which is not contained in any interval in 𝒞_{k}^{I} ∪ 𝒞_{k}^{J}:

(3.4)  μ_{k+1}(L_{k+1,j}) = μ_{k}(L_{k+1,j}),

(3.5)  μ_{k+1}(I_{k+1,j}) = (1 + k)^{−δ}μ_{k+1}(L_{k+1,j}),

(3.6)  μ_{k+1}(J_{k+1,j}) = (1 + k)^{−εa}μ_{k+1}(L_{k+1,j}).

The measure μ_{K_{1}} so chosen has the properties that

μ_{K_{1}}(S_{1}^{I} ∪ S_{1}) = 1 − ∏_{1+K_{0}}^{K_{1}}(1 − k^{−εa} − k^{−δ})

and

\begin{aligned}μ_{K_{1}}(S_{1}) & ≥ μ_{K_{1}}(S_{1}^{I} ∪ S_{1}) inf_{1+K_{0}≤k≤K_{1}} \frac{k^{−εa}}{k^{−εa} + k^{−δ}} \\ & ≥ (1 − ∏_{1+K_{0}}^{K_{1}}(1 − k^{−εa}))(1 − K_{0}^{εa−δ}), \end{aligned}

because εa < δ.

Next choose μ_{1+K_{1}} ∈ 𝒟_{d}(10^{1/a}) so that μ_{1+K_{1}} = μ_{K_{1}} on S_{0} \ S_{1} , and on each L_{1+K_{1},j} ⊆ S_{1} it is a redistribution of μ_{K_{1}} satisfying (3.4), (3.5) and (3.6) with k = K_{1} . After μ_{k} is constructed for some k, 1 + K_{1} ≤ k < K_{2} , build μ_{k+1} from μ_{k} following the same steps as in the case 1 + K_{0} ≤ k ≤ K_{1} . The dyadic doubling measure μ_{K_{2}} so obtained belongs to 𝒟_{d}(10^{1/a}), moreover

μ_{K_{2}}(S_{2}^{I} ∩ S_{2}) = (1 − ∏_{1+K_{1}}^{K_{2}}(1 − k^{−εa} − k^{−δ}))μ_{K_{1}}(S_{1}),

and

μ_{K_{2}}(S_{2}) ≥ μ_{K_{2}}(S_{2}^{I} ∪ S_{2}) inf_{1+K_{1}≤k≤K_{2}} \frac{k^{−εa}}{k^{−εa} + k^{−δ}}

≥ μ_{K_{2}}(S_{2}^{I} ∪ S_{2})(1 − K_{1}^{εa−δ})

≥ (1 − ∏_{1+K_{0}}^{K_{1}}(1 − k^{−εa}))(1 − ∏_{1+K_{1}}^{K_{2}}(1 − k^{−εa}))(1 − K_{0}^{εa−δ})(1 − K_{1}^{εa−δ}).

Whenever μ_{K_{m}} is constructed, keep μ_{1+K_{m}} = μ_{K_{m}} on S_{0} \ S_{m}, redistribute the mass on each L_{1+K_{m},j} ⊆ S_{m} according to (3.4), (3.5) and (3.6) with k = K_{m}, and keep the dyadic doubling constant bounded by 10^{1/a}. Continue this indefinitely. Thus, we obtain a sequence of measures μ_{K_{m}} ∈ 𝒟_{d}(10^{1/a}), with μ_{K_{m}}([0, 1]) = 1 and

μ_{K_{m}}(S_{m}) ≥ ∏_{i=0}^{m−1} ((1 − K_{i}^{εa} − δ)(1 − A_{i}))

where A_{i} = ∏_{1+K_{i}}^{K_{1+i}}(1 − k^{−εa}). Let μ be a weak limit point of {μ_{K_{m}}}. Clearly μ ∈ 𝒟_{d}(10^{1/a}).

Since εa < 1, it is possible to choose {K_{i}} so that

(3.7)  ∑_{i=1}^{∞} K_{i}^{εa−δ} + ∑_{i=1}^{∞} A_{i} < +∞.

With respect to this choice of {K_{i}}, we have μ(S) > 0, hence μ ∉ 𝒩_{d}.

It remains to show that S ∈ 𝒩 . Let ν ∈ 𝒟. Recall that J_{k,j} and I_{k,j+1} have the common boundary point (j + 1)/(2^{n_{k}}); by the doubling property

ν(J_{k,j} ∪ I_{k,j+1}) ≥ A^{(ε−δ) log_{2} k−5}ν(J_{k,j})

for some A > 1 depending only on the doubling constant of ν. For m ≥ 2, intervals in 𝒞_{K_{m}}^{J} ∪ {I_{k,j+1} : J_{k,j} ∈ 𝒞_{K_{m}}^{J}} (≠ 𝒞_{K_{m}}^{J} ∪ 𝒞_{K_{m}}^{I}) may meet in their interiors; however, because of (3.3), every point in [0, 1] is covered by at most three such intervals. Therefore

\begin{aligned}3ν([0, 1]) & ≥ ∑_{J_{k,j}∈𝒞_{Km}^{J}} ν(J_{k,j} ∪ I_{k,j+1}) ≥ A^{(ε−δ) log_{2} K_{m−1}−5}ν(S_{m}) \\ & ≥ A^{(ε−δ) log_{2} K_{m−1}−5}ν(S). \end{aligned}

Hence ν(S) = 0. Therefore S ∈ 𝒩 \ 𝒩_{d}.

Let

T = {t = ∑_{n=1}^{∞} t_{n}2^{−n}, where t_{n} = 0 or 1,
but t_{n_{k}+[δ log_{2} k]+1} = 1 and t_{n_{k}+[δ log_{2} k]+2} = 0
for each integer k > K_{0}}.

In view of (3.1), it has Hausdorff dimension 1.

Fix t ∈ T and ν ∈ 𝒟_{d} and let J_{k,j} be any interval in 𝒞_{K_{m}}^{J}. We note that

\frac{p + \frac{1}{2}}{2^{n_{k}} k̇^{δ}} < t + \frac{j + 1}{2^{n_{k}}} < \frac{p + \frac{3}{4}}{2^{n_{k}} k̇^{δ}}

for some integer p, because

q + \frac{1}{2} < t2^{n_{k}} k̇^{δ} < q + \frac{3}{4}

for some integer q.

Therefore t + J_{k,j} is contained in the middle half of some dyadic interval

M_{k,j} = [\frac{p}{2^{n_{k}} k̇^{δ}} , \frac{p + 1}{2^{n_{k}} k̇^{δ}} ].

Recall that the interval I_{k,j+1} shares an end point (j + 1)/2^{n_{k}} with J_{k,j} and has length 1/2^{n_{k}} k̇^{δ}. Therefore

|(t + I_{k,j+1}) ∩ M_{j,k}| > \frac{1}{4} \frac{1}{2^{n_{k}} k̇^{δ}} .

The dyadic doubling property of ν, (3.2) and Lemma 1,

ν(t + (J_{k,j} ∪ I_{k,j+1}) ∩ M_{k,j}) ≥ c(k, ν)ν(t + J_{k,j})

with c(k, ν) → ∞ as k → ∞. Summing over all J_{k,j} in 𝒞_{K_{m}}^{J} and reasoning as before, we obtain

3ν([0, 1]) ≥ c(K_{m−1}, ν)ν(t + S_{m}) ≥ c(K_{m−1}, ν)ν(t + S).

Letting m → ∞, we have ν(t + S) = 0. This completes the proof of Theorem 4.

It would be interesting to characterize those t’s so that t + S is in 𝒩_{d}. However this seems difficult.

To prove Theorem 5, we note that in the binary expansion of t, the event that a digit 1 is followed immediately by a digit 0 occurs infinitely often. Choose ε, δ and a as in Theorem 4, and {n_{k}} depending on t, so that (3.1),

t_{n_{k}+[δ log_{2} k]+1} = 1  and  t_{n_{k}+[δ log_{2} k]+2} = 0

hold for each k > k_{0} . Let S_{t} ≡ S in Theorem 4 associated with this sequence {n_{k}}. Then S_{t} ∈ 𝒩 \ 𝒩_{d}. The proof of t + S_{t} ∈ 𝒩_{d} is similar to that in Theorem 4.

4. Null sets for p-harmonic measures

Consider the p-Laplace equation (1 < p < ∞)

div (|∇u|^{p−2}∇u) = 0

in the half plane Ω ≡ {x ∈ R^{2} : x_{2} > 0}. For the definition and properties of p-harmonic measure (the harmonic measure for p-Laplacian) see [6; Chapter 10].

Let E be a compact set on ∂Ω which has positive p-harmonic measure for some p. Then there exists a nonconstant solution u (0 ≤ u ≤ 1) of the p-Laplacian in Ω, with continuous boundary value 0 on ∂Ω \ E.

Following [1], we may apply a linearization technique in [7] or an approximation technique in [4], and Theorem 4.5 in [3], to write

u(x) = ∫_{∂Ω} K(x, y)f(y) dω(y),

where K is a limit of kernel functions and ω is a weak limit of harmonic measures at a fixed point, corresponding to a sequence of uniformly elliptic operators of nondivergence form in Ω, with ellipticity constants depending only on p. Moreover ω has the doubling property and u has nontangential limit on ∂Ω ω -a.e.

Because u has zero boundary value on ∂Ω \ E, f(y) dω(y) is supported in E. This implies that ω(E) > 0. Therefore, we have

Theorem 6. Compact sets in 𝒩 are null sets for any p-harmonic measure with respect to the half plane {x ∈ R^{2} : x_{2} > 0}.

Remark. Martio has defined a version of porosity and proved that a porous set on {x_{2} = 0} has zero 𝒜 -harmonic measure with respect to all those nonlinear operators 𝒜 on {x_{2} > 0} considered in [8]; p-Laplacians are examples of such operators. We do not know whether Theorem 6 can be extended to all such 𝒜 -operators. However a compact set E on {x_{2} = 0} is 𝒜 -harmonic measure null for all such 𝒜 if it satisfies a stronger {a_{n}}-porous condition for some {a_{n}} (∑ α_{n}^{K} = ∞ for all K > 1), namely, in defining {a_{n}}-porosity, E ∩ E_{n,j} is required to lie in the middle 1 − 2α_{n} portion of E_{n,j} for each n and j. Proof follows by combining the original proof of Martio and that of Theorem 1.

Null sets for doubling and dyadic doubling measures 91

References

[1] Avilés, P., and J. Manfredi: On null sets of p-harmonic measures. - Preprint.

[2] Beurling, A., and L. Ahlfors: The boundary correspondence under quasiconformal mappings. - Acta Math. 96, 1956, 125–142.

[3] Carffarelli, L., E. Fabes, S. Mortola, and S. Salsa: Boundary behavior on nongative solutions of elliptic operators in divergence form. - Indiana Univ. Math. J. 30, 1981, 621–640.

[4] Fabes, E., N. Garofalo, S. Marín-Malave, and S. Salsa: Potential theory and Fatou theorems for some nonlinear elliptic operators. - Rev. Mat. Iberoamericana 4, 1988, 227–252.

[5] Fefferman, R.A., C.E. Kenig, and J. Pipher: The theory of weights and the Dirichlet problem for elliptic equations. - Ann. Math. 134, 1991, 65–124.

[6] Heinonen, J., J. Kilpeläinen, and O. Martio: Nonlinear potential theory of solutions of degenerate elliptic equations. - Oxford University Press (to appear).

[7] Manfredi, J., and A. Weitsman: On the Fatou-theorem for p-harmonic functions. - Comm. Partial Differential Equations 13:6, 1988, 651–688.

[8] Martio, O.: Harmonic measures for second order nonlinear partial differential equations. - In: Function spaces, differential operators and nonlinear analysis, L. Päivärinta (editor), Longman Scientific and Technical Publishers, 1989.

Received 17 December 1991