Annales Academiæ Scientiarum Fennicæ Series A. I. Mathematica Volumen 19, 1994, 115–132

APPLICATIONS OF GEOMETRIC MEASURE THEORY TO THE STUDY OF GAUSS– WEIERSTRASS AND POISSON INTEGRALS

Neil A. Watson

University of Canterbury, Department of Mathematics Private Bag 4800, Christchurch, New Zealand; naw@math.canterbury.ac.nz

Abstract. The behaviour as t → 0 of t^{δ(n−q)}u(x, t) is studied, where u is either the Gauss–Weierstrass or Poisson integral of a signed measure μ on R^{n}, q ∈ [0, n], and δ is \frac{1}{2} for a Gauss–Weierstrass integral but 1 for a Poisson integral. Such behaviour is used to characterize rectifiable subsets of R^{n}, positive sets for μ, and sets to which the restriction of μ is absolutely continuous with respect to q -dimensional Hausdorff measure.

1. Introduction

For any locally finite, signed Borel measure μ on R^{n}, the convolution

u(x, t) = W μ(x, t) = ∫_{R^{n}} W (x − y, t) dμ(y),

where W (x, t) = (4πt)^{−n/2} exp(−∥x∥^{2}/4t) for all (x, t) ∈ R^{n}×]0, ∞[, is called the Gauss–Weierstrass integral of μ. If u(x_{0}, t_{0}) exists and is finite, then u(x, t) is finite whenever (x, t) ∈ R^{n}×]0, t_{0}[, and u is a solution of the heat equation there. We shall always assume implicitly that u is finite on R^{n}×]0, a[ for some a > 0.

Similarly, the convolution

w(x, t) = P μ(x, t) = ∫_{R^{n}} P (x − y, t) dμ(y),

where P (x, t) = 2s_{n+1}^{−1}t(∥x∥^{2} + t^{2})^{−(n+1)/2} for all (x, t) ∈ R^{n}×]0, ∞[, and s_{n+1} is the surface area of the unit sphere in R^{n+1} , is called the (half-space) Poisson integral of μ. If w(x_{0}, t_{0}) exists and is finite, then w(x, t) is finite whenever (x, t) ∈ R^{n}×]0, ∞[, and w is harmonic there. We shall always assume implicitly that w is finite on R^{n}×]0, ∞[.

It is well-known that these two integrals have much in common, and here we study them together, writing u = Kμ with K either W or P. We are particularly concerned with the limiting behaviour as t → 0 of t^{δ(n−q)}u(x, t), where q ∈ [0, n], and δ = \frac{1}{2} if K = W, δ = 1 if K = P. Such behaviour is used to characterize rectifiable subsets of R^{n}, positive sets for μ, and sets to which the restriction of μ is absolutely continuous with respect to the q -dimensional Hausdorff measure m_{q} (defined in [7, p. 7]). Our main tool is the link with geometric measure theory in R^{n} which is provided by the inequalities

1991 Mathematics Subject Classification: Primary 31B10, 35J05, 35K05; Secondary 28A15, 28A78, 35C15, 35B05.

lim inf_{r→0} \frac{μ(B(x, r))}{ν(B(x, r))} ≤ lim inf_{t→0} \frac{u(x, t)}{v(x, t)}

≤ lim sup_{t→0} \frac{u(x, t)}{v(x, t)} ≤ lim sup_{r→0} \frac{μ(B(x, r))}{ν(B(x, r))},

where u = Kμ, v = Kν with ν positive, and B(x, r) = { y ∈ R^{n} : ∥x − y∥ ≤ r }. These inequalities lead to a very precise form of the relative Fatou theorem, whose applications are found throughout this paper; the principal ones we now describe.

In Theorem 4 we show that, if q is an integer and Z is a rectifiable set (as defined in Section 4) with σ -finite m_{q} -measure then

lim_{t→0} t^{δ(n−q)}u(x, t) = cf(x)

for m_{q} -almost all x ∈ Z, where c is a specific constant, and f is the Radon– Nikodým derivative of the restriction of μ to Z with respect to m_{q}. Conversely, if μ is positive, m_{q}(Z) > 0, and

0 < lim_{t→0} t^{δ(n−q)}u(x, t) < ∞

for m_{q} -almost all x ∈ Z, then q is an integer and Z is a rectifiable set of σ -finite m_{q} -measure. Thus the classical Fatou theorem and its converse are generalized. The proof depends essentially on the considerable measure theoretic achievements of Federer [8], Marstrand [9], and Preiss [10].

In Section 5, we consider the problem of determining when a given Borel subset Z of R^{n} is a positive set for the signed measure μ. For example, Theorem 8 shows that, if Z has σ -finite m_{q} -measure, then Z is positive for μ if and only if

lim inf_{t→0} t^{δ(n−q)}u(x, t) ≥ 0

for μ-almost all x ∈ Z. The case q = n, Z = R^{n}, is well-known. This result depends on measure theory due to Wallin [12].

In Section 6, we determine when the restriction of μ to Z is absolutely continuous with respect to m_{q}, where Z is a Borel set which is σ -finite with respect to m_{q}. Theorem 9 shows that this is the case if and only if

lim sup_{t→0} t^{δ(n−q)}|u(x, t)| < ∞

for μ-almost all x ∈ Z. Here the measure theory involved is mainly due to Besicovitch [3], [4].

Given a signed measure μ and a Borel set Z, we denote by μ_{Z} the restriction of μ to Z, and by μ^{+} , μ^{−}, and |μ| the positive, negative, and total variations of μ. The term ‘positive’ is used in the wide sense.

Remark. The referee has pointed out that the results in [3], [4] which are used below, can all be found in [8, Chapters 2.8 and 2.9].

2. Preliminaries

Let u = Kμ and v = Kν, where μ is signed but ν is positive. Consider the inequalities

(1)

lim inf_{r→0} \frac{μ(B(x, r))}{ν(B(x, r))} ≤ lim inf_{t→0} \frac{u(x, t)}{v(x, t)}

≤ lim sup_{t→0} \frac{u(x, t)}{v(x, t)} ≤ lim sup_{r→0} \frac{μ(B(x, r))}{ν(B(x, r))}.

For the case K = W, it was proved in [13, Theorem 1] that (1) holds at any point x such that ν(B(x, r)) > 0 for all r > 0. Subsequently, in [1, Theorem 1], Armitage proved that (1) holds when K = P, but only at those points x for which

(2)  lim_{t→0} t^{−1}v(x, t) = ∞.

He did not discuss the necessity of this more stringent condition, but it is essential. For example, if n = 1, x = 0, l ∈]0, ∞[, and

dν(s) = l|s|^{−3} exp(−s^{−2}) ds,

then for every r > 0 we have ν(B(0, r)) = l exp(−r^{−2}), so that, by a formula in [1, p. 241],

lim_{t→0} t^{−1}v(0, t) = 2lπ^{−1} ∫_{0}^{∞} s^{−3} exp(−s^{−2}) ds = lπ^{−1}.

Taking μ to be the unit mass at the point 1, it is obvious that

lim_{r→0} \frac{μ(B(0, r))}{ν(B(0, r))} = 0,

but

\frac{u(0, t)}{v(0, t)} ∼ \frac{πP (−1, t)}{lt} → \frac{1}{l}

as t → 0.

The inequalities (1) are essential for this paper, and the extra condition (2) in the harmonic case is not a real problem because it is satisfied by ν -almost all points x. We introduce a notation which masks this difference in the two cases. Given a positive measure ν such that W ν is finite at some point, we put

X(ν) = X_{W} (ν) = { x ∈ R^{n} : ν(B(x, r)) > 0 for all r > 0 }.

Clearly ν is concentrated on X(ν), and (1) holds for all x ∈ X(ν). If ν is a positive measure such that v = P ν is finite, we put

X(ν) = X_{P} (ν) = { x ∈ R^{n} : lim_{t→0} t^{−1}v(x, t) = ∞ }.

It was shown in [1, p. 241] that lim_{t→0} t^{−1}v(x, t) exists for every x ∈ R^{n}, and therefore lim_{t→0} v(x, t) = 0 for every x ∈ R^{n} \ X_{P} (ν), so that ν(R^{n} \ X(ν)) = 0 by a result of Brelot [5]. Thus ν is again concentrated on X(ν), and (1) holds for all x ∈ X(ν).

In particular, we have the following result, which is given in [14] for the case K = W.

Lemma 1. Let u = Kμ and v = Kν, where μ is signed and ν is positive. If μ and ν are mutually singular, then

u(x, t) = o(v(x, t)) as t → 0

for ν -almost all x.

Proof. We can suppose that μ is positive. By [3, Theorem 3],

lim_{r→0} \frac{μ(B(x, r))}{ν(B(x, r))} = 0

for ν -almost all x. Since (1) also holds for ν -almost all x, the result follows.

Another set of inequalities, which can be derived from (1), will also be needed. If u = Kμ and q ∈ [0, n], then for all x ∈ R^{n}

(3)

lim inf_{r→0} r^{−q}μ(B(x, r)) ≤ c_{n,q} lim inf_{t→0} t^{δ(n−q)}u(x, t)

≤ c_{n,q} lim sup_{t→0} t^{δ(n−q)}u(x, t) ≤ lim sup_{r→0} r^{−q}μ(B(x, r)),

where δ is as defined in Section 1, and

(4)  c_{n,q} = \{ \begin{matrix}π^{n/2}2^{n−q}/Γ(\frac{1}{2}q + 1) & if K = W, \\ π^{(n+1)/2}/Γ(\frac{1}{2}q + 1)Γ(\frac{1}{2}(n + 1 − q)) & if K = P. \end{matrix}

These inequalities were proved for K = W in [13], and for K = P in [1]. Their proofs are valid for all q ∈ [0, ∞[ and all q ∈ [0, n + 1[ respectively, but the extended ranges are not useful here. In particular, it follows from (3) that if lim_{r→0} r^{−q}μ(B(x, r)) = λ then lim_{t→0} t^{δ(n−q)}u(x, t) = c_{n,q}^{−1}λ. For Theorem 4 below, we require the reverse implication under the Tauberian condition that μ is positive. For K = W this was proved in [13], for all λ, q ∈ [0, ∞[ (despite the statement of [13, Theorem 4]). For K = P a different Tauberian condition was used in [1], while that of positivity was used by Rudin in [11] for the case λ ∈ [0, ∞[, q = n only. We now extend the result of [11, Theorem A] to an arbitrary q ∈ [0, n + 1[.

Theorem 1. Let u = P μ with μ positive, let q ∈ [0, n + 1[, and suppose that

(5)  lim_{t→0} t^{n−q}u(x, t) = λ < ∞

for some x. Then

lim_{r→0} r^{−q}μ(B(x, r)) = c_{n,q}λ,

where c_{n,q} is given by (4).

Proof. The case q = 0 is simple, because it is always true that lim_{r→0} μ(B(x, r)) = μ({x}), and therefore that lim_{t→0} t^{n}u(x, t) = μ({x})c_{n,0}^{−1} by (3).

The proof when q > 0 is similar to that given for q = n in [11], so we give only an outline, in which all unexplained notations have the same meanings as in [11], and all omitted calculations are similar to the corresponding ones there. Note that our Poisson integral differs from Rudin’s by a factor of π^{n/2}Γ(\frac{1}{2}n + 1).

If v(t) = t^{n−q}u(x, t), then (5) implies that

lim_{r→0}(H_{q} ∗ v)(r) = λ.

Furthermore, it follows from (3) that, if μ(B(x, ϱ)) = 0 for some ϱ > 0, then lim_{t→0} t^{n−q}u(x, t) = 0; therefore our hypotheses are unaffected if we replace μ by μ_{B(x,1)} , and since our conclusion is also unaffected by such a change, we may assume that μ(R^{n}) < ∞. We shall also take x = 0.

The definitions of H_{q} and v give

(6)  (H_{q} ∗ v)(r) = qr^{−q} ∫_{R^{n}} dμ(y) ∫_{0}^{r} P (y, s)s^{n−1} ds.

If M_{q}(r) = r^{−q}μ(B(0, r)) for all r ∈]0, ∞[, then M_{q} is bounded. The function k_{q}, defined for all t ∈]0, ∞[ by

k_{q}(t) = 2s_{n+1}^{−1}qt^{n−q+1}(1 + t^{2})^{−(n+1)/2},

satisfies

P (y, ∥y∥t) = q^{−1}t^{q−n}∥y∥^{−n}k_{q}(t)

whenever y ≠ 0. Since M_{q} is bounded and q > 0, we have μ({0}) = 0. Therefore, if we replace s by ∥y∥t in (6), we obtain

(H_{q} ∗ v)(r) = (M_{q} ∗ k_{q})(r),

so that the bounded function M_{q} satisfies lim_{r→0}(M_{q} ∗ k_{q})(r) = λ. Furthermore, the Fourier transform of k_{q} is given by

\begin{aligned}k̂_{q}(s) & = 2s_{n+1}^{−1}q ∫_{0}^{∞}(1 + t^{2})^{−(n+1)/2}t^{n−q−is} dt \\ & = s_{n+1}^{−1}qΓ(\frac{q + is}{2})Γ(\frac{n + 1 − q − is}{2}) / Γ(\frac{n + 1}{2}), \end{aligned}

since q ∈]0, n + 1[. Hence k̂_{q} has no zeros, and k̂_{q}(0) = c_{n,q}^{−1} . It now follows from Wiener’s tauberian theorem that

lim_{r→0}(M_{q} ∗ f)(r) = c_{n,q}λ

for every f such that

∫_{0}^{∞} f(r)r^{−1} dr = 1.

The desired conclusion, that lim_{r→0} M_{q}(r) = c_{n,q}λ, now follows.

3. A relative Fatou theorem

Here we give a precise form of the relative Fatou theorem for approach along lines normal to the boundary. This is needed for the proofs of most of the subsequent results. The existence of parabolic limits of quotients of Gauss–Weierstrass integrals, and of nontangential limits of quotients of Poisson integrals, is proved in [6, p. 292 and p. 31, respectively]. However, we give a simpler argument for the present situation. The case K = W, ν = m_{n}, is given in [15].

Theorem 2. Let u = Kμ and v = Kν, where μ is signed and ν is positive, let

f(x) = lim_{t→0} \frac{u(x, t)}{v(x, t)}

whenever the limit exists, let Z^{+} = {x ∈ R^{n} : f(x) = ∞}, and let Z^{−} = {x ∈ R^{n} : f(x) = −∞}. Then f is defined and finite ν -a.e. on R^{n}, and there are positive ν -singular measures σ^{+} and σ^{−}, concentrated on Z^{+} and Z^{−} respectively, such that

(7)  dμ = f dν + dσ^{+} − dσ^{−}.

Proof. The inequalities (1) hold for all x ∈ X(ν), and hence for ν -almost all x, so that f(x) exists and is equal to

(8)  lim_{r→0} \frac{μ(B(x, r))}{ν(B(x, r))}

for ν -almost all x at which the limit (8) exists. Therefore f is defined and finite ν -a.e., by [3, Theorem 2]. Furthermore, [4, Theorem 6] shows that the limit (8) is the Radon–Nikodým derivative of μ with respect to ν, so that (7) holds with σ the ν -singular part of μ.

It remains to prove that σ^{+} and σ^{−} are concentrated on Z^{+} and Z^{−} respectively. Since ν and μ − σ^{+} are both σ^{+} -singular, it follows from Lemma 1 that v(x, t) = o(Kσ^{+}(x, t)) and K(μ − σ^{+})(x, t) = o(Kσ^{+}(x, t)) as t → 0, for σ^{+} -almost all x. Hence

\frac{u(x, t)}{v(x, t)} = \frac{K(μ − σ^{+})(x, t) + Kσ^{+}(x, t)}{v(x, t)} → ∞

as t → 0 for σ^{+} -almost all x, so that σ^{+} is concentrated on Z^{+} . Similarly σ^{−} is concentrated on Z^{−}.

4. Boundary singularities and rectifiable sets

The main result in this section is Theorem 4, which establishes a relationship between the sets of points x where lim_{t→0} t^{δ(n−q)}Kμ(x, t) exists and the rectifiable subsets of R^{n}. In its proof, we require a result that sharpens parts of both [13, Theorem 6] and the unstated version of [1, Theorem 4] which relates to normal limits.

Theorem 3. Let u = Kμ for some positive measure μ, and let q ∈ [0, n]. Then the set

Y = {x ∈ R^{n} : lim sup_{t→0} t^{δ(n−q)}u(x, t) > 0}

is a Borel set which is σ -finite with respect to m_{q}.

Proof. It follows from (3) that

(9)  Y ⊆ {x ∈ R^{n} : lim sup_{r→0} r^{−q}μ(B(x, r)) > 0}.

Furthermore, for any x ∈ R^{n},

\begin{aligned}t^{−qδ}μ(B(x, t^{δ})) & ≤ κt^{δ(n−q)} ∫_{B(x,t^{δ})} K(x − y, t) dμ(y) \\ & ≤ κt^{δ(n−q)}u(x, t), \end{aligned}

where κ = 2^{n}π^{n/2}e^{1/4} if K = W, and κ = 2^{(n−1)/2}s_{n+1} if K = P. Therefore

(10)  lim sup_{r→0} r^{−q}μ(B(x, r)) ≤ κ lim sup_{t→0} t^{δ(n−q)}u(x, t),

so that equality holds in (9). Hence Y is a Borel set. For each j ∈ N, let μ_{j} = μ_{B(0,j+1)} . Then Y = ⋃_{k,j=1}^{∞} Y_{k,j}, where

Y_{k,j} = {x ∈ B(0, j) : lim sup_{r→0} r^{−q}μ_{j}(B(x, r)) > k^{−1}}.

If q > 0, then [8, p. 181 (3)] shows that m_{q}(Y_{k,j}) ≤ kμ_{j}(R^{n}) < ∞; and if q = 0 then obviously m_{q}(Y_{k,j}) < ∞. Hence Y is σ -finite with respect to m_{q}.

Following Federer [8, p. 251], we say that a Borel set Z is countably (m_{q}, q) rectifiable if it can be written in the form

⋃_{j=1}^{∞} f_{j}(B_{j}) ∪ Y,

where m_{q}(Y) = 0 and each f_{j} is a Lipschitz function from a bounded subset B_{j} of R^{q} into R^{n}.

Theorem 4. Let u = Kμ, let q ∈ [0, n] and let Z be a Borel subset of R^{n} such that m_{q}(Z) > 0.

(i) If q is an integer, μ is signed, and Z is a countably (m_{q}, q) rectifiable set with σ -finite m_{q} -measure, then

lim_{t→0} t^{δ(n−q)}u(x, t) = 2^{q}c_{n,q}^{−1}f(x)

for m_{q} -almost all x ∈ Z, where c_{n,q} is given by (4) and f is the Radon– Nikodým derivative of μ_{Z} with respect to m_{qZ}.

(ii) Conversely, if μ is positive and

(11)  0 < lim_{t→0} t^{δ(n−q)}u(x, t) < ∞

for m_{q} -almost all x ∈ Z, then q is an integer and Z is a countably (m_{q}, q) rectifiable set with σ -finite m_{q} -measure.

Proof. (i) If q = 0 then Z is countable, and for every x ∈ Z it follows from (3) that

f(x) = μ({x}) = lim_{r→0} μ(B(x, r)) = c_{n,0} lim_{r→0} t^{δn}u(x, t).

Now suppose that q ∈]0, n]. (The case q = n is well-known, but is included for completeness.) We can assume that m_{q}(Z) < ∞. If ν = m_{qZ}, then lim_{r→0} r^{−q}ν(B(x, r)) = 2^{q} for ν -almost all x, by a theorem of Federer [8, p. 256] if q < n. It follows from (3) that

lim_{t→0} t^{δ(n−q)}Kν(x, t) = 2^{q}c_{n,q}^{−1}

for the same values of x. By Theorem 2, the function

f(x) = lim_{t→0} \frac{u(x, t)}{Kν(x, t)}

is defined for ν -almost all x, and is the Radon–Nikodým derivative of μ_{Z} with respect to ν. The result follows.

(ii) There is nothing to prove if q = n. If q = 0, then (11) holds for all x ∈ Z, and it follows from (3) that

0 < lim_{r→0} μ(B(x, r)) = μ({x})

for all x ∈ Z, so that Z is countable.

Now suppose that q ∈]0, n[, and put ν = m_{qZ} again. By Theorem 3, the set Z is σ -finite with respect to m_{q}, and so we may suppose that 0 < m_{q}(Z) < ∞. By (11) and either [13, Theorem 4] or Theorem 1 above,

0 < lim_{r→0} r^{−q}μ(B(x, r)) < ∞

for ν -almost all x. By [3, Theorem 2],

lim_{r→0} \frac{μ(B(x, r))}{ν(B(x, r))}

exists and is finite for ν -almost all x, and so it follows that

(12)  lim_{r→0} r^{−q}ν(B(x, r))

exists and is strictly positive for ν -almost all x. By [7, Corollary 2.5],

lim sup_{r→0} r^{−q}ν(B(x, r)) ≤ 2^{q},

so that the limit (12) is finite, for ν -almost all x. Therefore, by a theorem of Marstrand [9], q is an integer. It follows that Z is countably (m_{q}, q) rectifiable, by a theorem of Preiss [10, p. 613].

5. Positivity of sets for signed measures

We begin by establishing two conditions which ensure that a set is null for a positive measure.

Theorem 5. Let u = Kμ for some positive measure μ, let q ∈ [0, n], and let Z be a Borel subset of R^{n}.

(i) If m_{q}(Z) = 0 and

lim sup_{t→0} t^{δ(n−q)}u(x, t) < ∞

for μ-almost all x ∈ Z, then μ(Z) = 0. (ii) If Z is σ -finite with respect to m_{q} and

lim_{t→0} t^{δ(n−q)}u(x, t) = 0

for μ-almost all x ∈ Z, then μ(Z) = 0.

Proof. If q = 0 then (i) is trivial, and in case (ii) Z is countable and (10) shows that μ({x}) = 0 for all x ∈ Z, so that μ(Z) = 0.

Suppose that q ∈]0, n]. In case (i), it follows from (10) that μ_{Z} is concentrated on ⋃_{k=1}^{∞} S_{k}, where

S_{k} = {x ∈ Z : lim sup_{r→0} r^{−q}μ_{Z}(B(x, r)) < k}.

Furthermore, by [8, p. 181 (1)], we have μ_{Z}(S_{k}) ≤ km_{q}(S_{k}) ≤ km_{q}(Z) = 0, and therefore μ_{Z} is null. In case (ii), we may suppose that m_{q}(Z) < ∞. By (10), for every γ > 0 the measure μ_{Z} is concentrated on

S = {x ∈ Z : lim sup_{r→0} r^{−q}μ_{Z}(B(x, r)) < γ},

and as before μ_{Z}(S) ≤ γm_{q}(Z) < ∞. Hence μ_{Z} is again null.

Remark. In the case q = n, Theorem 5(ii) was proved in [14] for K = W and in [5] for K = P.

We are now in a position to prove three theorems on the positivity of sets for signed measures. In the case K = W, the first generalizes [15, Theorem 7] in several directions, and also [14, Theorem 6]. In the case K = P, the analogue of the latter result is [2, Theorem 2], which is complicated by the fact that behaviour at infinity has to be taken into account; that aside, the conditions therein which ensure the positivity of the appropriate boundary measure are relaxed and made applicable to an arbitrary Borel set.

Theorem 6. Let u = Kμ and v = Kν, where μ is signed and ν is positive, let q ∈ [0, n] and let Z be a Borel subset of R^{n}. Let Y be a ν -null Borel subset of Z, and suppose that

lim sup_{t→0} \frac{u(x, t)}{v(x, t)} > −∞

for all x ∈ Z \ Y , and that

(13)  lim sup_{t→0} \frac{u(x, t)}{v(x, t)} ≥ 0

for ν -almost all x ∈ Z \ Y .

(i) If m_{q}(Y) = 0 and

(14)  lim inf_{t→0} t^{δ(n−q)}u(x, t) > −∞

for μ-almost all x ∈ Y , then μ_{Z} is positive. (ii) If Y is σ -finite with respect to m_{q} and

(15)  lim inf_{t→0} t^{δ(n−q)}u(x, t) ≥ 0

for μ-almost all x ∈ Y , then μ_{Z} is positive.

Proof. By Theorem 2, there are positive ν -singular and mutually singular measures σ^{+} and σ^{−} on Z such that

dμ_{Z} = f dν_{Z} + dσ^{+} − dσ^{−},

where f(x) is the (upper) limit in (13) and σ^{−} is concentrated on Y . By (13), f is positive. The measures μ+σ^{−} and σ^{−} are mutually singular, so that by Lemma 1 K(μ + σ^{−})(x, t) = o(Kσ^{−}(x, t)) as t → 0 for σ^{−} -almost all x. Therefore

(16)  lim_{t→0} \frac{u(x, t)}{Kσ^{−}(x, t)} = −1

for σ^{−} -almost all x.

(i) It follows from (14) and (16) that

lim sup_{t→0} t^{δ(n−q)}Kσ^{−}(x, t) < ∞

for σ^{−} -almost all x, so that σ^{−}(Y) = 0 by Theorem 5(i). Hence σ^{−} is null and μ_{Z} is positive.

(ii) It follows from (15) and (16) that

lim_{t→0} t^{δ(n−q)}Kσ^{−}(x, t) = 0

for σ^{−} -almost all x, so that Theorem 5(ii) can be used to show that μ_{Z} is positive.

If Z is σ -finite with respect to m_{q}, for some q ∈ [0, n], we can derive from Theorem 6 two criteria for μ_{Z} to be positive, neither of which explicitly involves an auxiliary function v. The necessity of the conditions is shown in our next result.

Lemma 2. Let u = Kμ for some signed measure μ, let q ∈ [0, n], and let Z be a Borel subset of R^{n} which is σ -finite with respect to m_{q}. If μ_{Z} is positive, then

(17)  lim inf_{t→0} t^{δ(n−q)}u(x, t) ≥ 0

for μ-almost all and m_{q} -almost all x ∈ Z.

Proof. If v = Kμ_{Z}, then Lemma 1 implies that

K(μ − μ_{Z})(x, t) = o(v(x, t))

as t → 0 for μ_{Z} -almost all x. Therefore u(x, t) ∼ v(x, t) for the same values of x, so that (17) holds for μ_{Z} -almost all x because the corresponding inequality for v is obvious.

For the second part, we may suppose that m_{q}(Z) < ∞. Let ω = m_{qZ} and w = Kω. By Theorem 2,

f(x) = lim_{t→0} \frac{u(x, t)}{w(x, t)}

is defined and finite for ω -almost all x, and there is an ω -singular measure σ on Z such that dμ_{Z} = f dω + dσ. The function f is positive because μ_{Z} is, and since

lim inf_{t→0} t^{δ(n−q)}w(x, t) ≤ c_{n,q}^{−1} lim sup_{r→0} r^{−q}ω(B(x, r)) < ∞

for ω -almost all x by (3) and [7, Corollary 2.5], it follows that

lim inf_{t→0} t^{δ(n−q)}u(x, t) = f(x) lim inf_{t→0} t^{δ(n−q)}w(x, t) ≥ 0

for ω -almost all x.

We shall prove in Theorem 8 that, if (17) holds for μ-almost all x ∈ Z then μ_{Z} is positive. The alternative hypothesis that (17) holds m_{q} -a.e. is not generally sufficient. If q = 0, Z is countable, and (17) holds for (m_{0} -almost) all x ∈ Z, then (3) shows that

μ({x}) = lim_{r→0} μ(B(x, r)) = c_{n,0} lim_{t→0} t^{δn}u(x, t) ≥ 0

for all x ∈ Z, so that μ_{Z} is positive. However, if q ∈]0, n] then −t^{δ(n−q)}K(0, t) = −c_{n,0}^{−1}t^{−δq} → −∞ and −t^{δ(n−q)}K(x, t) → 0 as t → 0 if x ≠ 0, so that (17) can hold m_{q} -a.e. on Z without μ_{Z} being positive. Hence an extra condition is required in this case, which is labelled (19) in the theorem below. Observe that if q = 0 then condition (18) below is assumed to hold for all x ∈ Z, so that (19) is vacuous.

Theorem 7. Let u = Kμ for some signed measure μ, let q ∈ [0, n] and let Z be a Borel subset of R^{n} which is σ -finite with respect to m_{q}. Then μ_{Z} is positive if and only if both

(18)  lim inf_{t→0} t^{δ(n−q)}u(x, t) ≥ 0

for m_{q} -almost all x ∈ Z, and

(19)  lim inf_{t→0} t^{δ(n−q)}u(x, t) > −∞

for μ-almost all x ∈ Z.

Proof. Suppose that (18) and (19) hold as described. We may suppose that m_{q}(Z) < ∞. If ν = m_{qZ} then

lim sup_{r→0} r^{−q}ν(B(x, r)) ≥ 1

for ν -almost all x, by [7, Corollary 2.5] if q ∈]0, n[. It therefore follows from (10) that

lim sup_{t→0} t^{δ(n−q)}Kν(x, t) ≥ κ^{−1}

for the same values of x. This combines with (18) to yield

lim sup_{t→0} \frac{u(x, t)}{Kν(x, t)} ≥ 0

for ν -almost all x, so that it follows from (19) and Theorem 6(i) (with Y the set of x ∈ Z for which the last inequality fails) that μ_{Z} is positive.

The converse is given in Lemma 2.

In Theorem 7, if μ is negative and m_{q}(Z) = 0, then the result reduces to Theorem 5(i) and its converse. However, if we suppose only that μ is negative, then we do not obtain Theorem 5(ii), but the following variant thereof.

Corollary. Let u = Kμ for some positive measure μ and let Z be a Borel subset of R^{n} which is σ -finite with respect to m_{q}. Then μ(Z) = 0 if and only if both

lim_{t→0} t^{δ(n−q)}u(x, t) = 0

for m_{q} -almost all x ∈ Z, and

lim sup_{t→0} t^{δ(n−q)}u(x, t) < ∞

for μ-almost all x ∈ Z.

Theorem 8 below reduces to Theorem 5(ii) and its converse when μ is negative. In its proof, we use the following terminology from [12]. For any q ∈]0, n] and x ∈ R^{n} the convex upper q -derivate of a positive measure ν at x is the supremum of the set of extended-real numbers l such that there exists a sequence of convex sets {C_{j}}, with strictly positive diameters d(C_{j}), such that x ∈ C_{j} for all j, d(C_{j}) → 0 as j → ∞, and

lim_{j→∞} \frac{ν(C_{j})}{d(C_{j})^{q}} = l.

It is denoted by \overline{D}_{q}ν(x).

Theorem 8. Let u = Kμ for some signed measure μ, let q ∈ [0, n], and let Z be a Borel subset of R^{n} which is σ -finite with respect to m_{q}. Then μ_{Z} is positive if and only if

(20)  lim inf_{t→0} t^{δ(n−q)}u(x, t) ≥ 0

for μ-almost all x ∈ Z.

Proof. If μ_{Z} is positive, then Lemma 2 shows that (20) holds for μ-almost all z ∈ Z.

For the converse, we may suppose that m_{q}(Z) < ∞ and that (20) holds for all x ∈ Z. If q ∈]0, n] then, by [12, Theorem 3], there is a finite, positive measure ν such that \overline{D}_{q}ν(x) ≥ 1 for all x ∈ Z. If x ∈ Z and C is a convex superset of {x} with diameter d(C) > 0, then

\frac{ν(C)}{d(C)^{q}} ≤ \frac{ν(B(x, d(C)))}{d(C)^{q}}

and hence

(21)  1 ≤ \overline{D}_{q}ν(x) ≤ lim sup_{r→0} r^{−q}ν(B(x, r)).

If q = 0 we take ν = m_{0Z}, and the corresponding inequality is then trivial. It now follows from (10) that

(22)  κ^{−1} ≤ lim sup_{t→0} t^{δ(n−q)}Kν(x, t)

for all x ∈ Z. Furthermore, (20) and (22) together imply that

lim sup_{t→0} \frac{u(x, t)}{Kν(x, t)} ≥ 0

for all x ∈ Z, and so it follows from Theorem 6 that μ_{Z} is positive.

6. Absolute continuity with respect to Hausdorff measures

Here we establish the following characterization of those Borel sets Z of σ - finite m_{q} -measure for which μ_{Z} is absolutely continuous with respect to m_{q}.

Theorem 9. Let u = Kμ for some signed measure μ, let q ∈ [0, n], and let Z be a Borel subset of R^{n} which is σ -finite with respect to m_{q}. If

(23)  lim sup_{t→0} t^{δ(n−q)}|u(x, t)| < ∞

for μ-almost all x ∈ Z, then μ_{Z} is absolutely continuous with respect to m_{q}. Conversely, if μ_{Z} is absolutely continuous with respect to m_{q}, then (23) holds for m_{q} -almost all x ∈ Z.

Proof. Suppose that (23) holds μ-a.e. on Z. Since it suffices to prove the result locally, we may suppose that |μ|(R^{n}) < ∞. We may also assume that m_{q}(Z) < ∞.

Let ν = m_{qZ}. By Theorem 2, if

f(x) = lim_{t→0} \frac{u(x, t)}{Kν(x, t)},

then f is defined and finite ν -a.e., and there is a ν -singular measure τ such that dμ_{Z} = f dν + dτ. The measures μ + τ^{−} and τ^{−} are mutually singular, so that

K(μ + τ^{−})(x, t) = o(Kτ^{−}(x, t))

as t → 0 for τ^{−} almost all x, by Lemma 1. Therefore

lim_{t→0} \frac{u(x, t)}{Kτ^{−}(x, t)} = −1

for the same values of x, and so it follows from (23) that

lim sup_{t→0} t^{δ(n−q)}Kτ^{−}(x, t) < ∞

for τ^{−} -almost all x. Since τ^{−} is ν -singular, it is concentrated on a subset Y of Z such that m_{q}(Y) = 0, and it therefore follows from Theorem 5(i) that τ^{−}(Y) = 0, so that τ^{−} is null. Similarly τ^{+} is null, so that dμ_{Z} = f dν.

For the converse, we need consider only the case where μ is positive and 0 < m_{q}(Z) < ∞.

Let ν = m_{qZ} again, and f be the Radon–Nikodým derivative of μ with respect to ν. Then

f(x) = lim_{r→0} \frac{μ(B(x, r))}{ν(B(x, r))} < ∞

for ν -almost all x, by [4, Theorem 6] and [3, Theorem 2]. Furthermore,

lim sup_{r→0} r^{−q}ν(B(x, r)) ≤ 2^{q}

for ν -almost all x, by [7, Corollary 2.5] if q ∈]0, n[. In view of (3), it follows that

c_{n,q} lim sup_{t→0} t^{δ(n−q)}u(x, t) ≤ lim sup_{r→0} r^{−q}μ(B(x, r))

= f(x) lim sup_{r→0} r^{−q}ν(B(x, r)) < ∞

for ν -almost all x.

Theorem 9 gives rise to the following apparently weaker test for absolute continuity.

Corollary. Let q ∈]0, n], let {q_{i}} be a sequence in [0, q[, let u = Kμ for some signed measure μ, let Z be a Borel subset of R^{n} which is σ -finite with respect to m_{q}, and let {Z_{i}} be a sequence of disjoint Borel subsets of Z such that m_{q_{i}}(Z_{i}) = 0 for all i. If

lim sup_{t→0} t^{δ(n−q)}|u(x, t)| < ∞

for μ-almost all x ∈ Z \ ⋃_{i=1}^{∞} Z_{i}, and for every i

lim sup_{t→0} t^{δ(n−q_{i})}|u(x, t)| < ∞

for μ-almost all x ∈ Z_{i}, then μ_{Z} is absolutely continuous with respect to m_{q}.

Proof. Let Y = Z \ ⋃_{i=1}^{∞} Z_{i}. By Theorem 9, μ_{Y} is absolutely continuous with respect to m_{q}; and each μ_{Z_{i}} is absolutely continuous with respect to the corresponding m_{q_{i}}, and hence is null.

Remarks. If q is an integer and Z is a countably (m_{q}, q) rectifiable set, we know from Theorem 4 that the Radon–Nikodým derivative f of μ_{Z} with respect to m_{qZ} is given by

f(x) = 2^{−q}c_{n,q} lim_{t→0} t^{δ(n−q)}u(x, t).

In general, such precise information is not available. However, if μ is positive we can obtain estimates for f in terms of the upper limits in (23), as follows. We know from [4, Theorem 6] and [3, Theorem 2] that if ν = m_{qZ} then

f(x) = lim_{r→0} \frac{μ(B(x, r))}{ν(B(x, r))} < ∞,

The study of Gauss–Weierstrass and Poisson integrals 131

and from [7, Corollary 2.5] that

1 ≤ lim sup_{r→0} r^{−q}ν(B(x, r)) ≤ 2^{q},

for ν -almost all x. It therefore follows from (3) and (10) that

2^{−q}c_{n,q} lim sup_{t→0} t^{δ(n−q)}u(x, t) ≤ 2^{−q} lim sup_{r→0} r^{−q}μ(B(x, r))

= 2^{−q}f(x) lim sup_{r→0} r^{−q}ν(B(x, r)) ≤ f(x)

and

κ lim sup_{t→0} t^{δ(n−q)}u(x, t) ≥ lim sup_{r→0} r^{−q}μ(B(x, r))

= f(x) lim sup_{r→0} r^{−q}ν(B(x, r)) ≥ f(x)

for ν -almost all x.

132 Neil A. Watson

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Received 22 December 1992