GEORGIAN MATHEMATICAL JOURNAL: Vol. 6, No. 2, 1999, 149-168

ON THE INTEGRABILITY OF STRONG MAXIMAL FUNCTIONS CORRESPONDING TO DIFFERENT

FRAMES

G. ONIANI

Abstract. For the frame θ in ℝ^{n}, let B_{2}(θ)(x) (x ∈ ℝ^{n}) be a family of all n-dimensional rectangles containing x and having edges parallel to the straight lines of θ, and let M_{B_{2}(θ)} be a maximal operator corresponding to B_{2}(θ). The main result of the paper is the following

Theorem. For any function f ∈ L(1 + ln^{+} L)(ℝ^{n}) (n ≥ 2) there exists a measure preserving and invertible mapping ω : ℝ^{n} → ℝ^{n} such that

1. {x : ω(x) ≠ x} ⊂ supp f;

2. sup_{θ∈θ(ℝ^{n})} ∫_{{M_{B2(θ)}(f∘ω)>1}} M_{B_{2}(θ)}(f ∘ ω) < ∞.

This theorem gives a general solution of M. de Guzmán’s problem that was previously studied by various authors.

1. Definitions and the Notation

Let B be a mapping defined on ℝ^{n} such that, for every x ∈ ℝ^{n}, B(x) is a family of open bounded sets in ℝ^{n} containing x. The maximal operator M_{B} corresponding to B is defined as follows: for f ∈ L_{loc}(ℝ^{n}) and x ∈ ℝ^{n}

M_{B}(f)(x) = sup_{R∈B(x)} \frac{1}{|R|} ∫_{R} |f| if B(x) ≠ ∅,

and

M_{B}(f)(x) = 0 if B(x) = ∅.

A frame in ℝ^{n} will be called a set whose elements are n pairwise orthogonal straight lines passing through the origin O. Frames will be denoted by θ, θ = {θ^{1}, …, θ^{n}}. Under θ_{0} will be meant a frame {Ox^{1}, …, Ox^{n}}, where Ox^{1}, …, Ox^{n} are the coordinate axes of ℝ^{n}. A set of all frames in ℝ^{n} will be denoted by θ(ℝ^{n}).

1991 Mathematics Subject Classification. 28A15. Key words and phrases. Strong maximal operator, rectangle, frame.

149

1072-947X/99/0300-0149$15.00/0 \begin{matrix}c \\ ◯ \end{matrix} 1999 Plenum Publishing Corporation

A set congruent to a set of the form I_{1} × ⋯ × I_{n}, where I_{1}, …, I_{n} are intervals of positive length on the straight line, will be called an ndimensional rectangle or simply a rectangle in ℝ^{n}.

The frame θ = {θ^{1}, …, θ^{n}} for which the sides of the rectangle I are parallel to the corresponding straight lines θ^{j} (j = 1, …, n) will be called the frame of I which will be denoted by θ(I).

For a nonempty set E ⊂ θ(ℝ^{n}) we shall denote by B_{2}(E)(x) (x ∈ ℝ^{n}) a family of all rectangles I in ℝ^{n} with the properties x ∈ I, θ(I) ∈ E. Instead of B_{2}({θ}) we shall write B_{2}(θ) when E = {θ}, and B_{2} when θ = θ_{0}.

Since M_{B_{2}} is said to be a strong maximal operator, it is natural to call M_{B_{2}(θ)} the strong maximal operator corresponding to the frame θ.

By B_{1}(x) (x ∈ ℝ^{n}) we denote a family of all cubic intervals in ℝ^{n} containing x (for n = 1 a one-dimensional interval is understood here as a square interval).

The support {x ∈ ℝ^{n} : f(x) ≠ 0} of the function f : ℝ^{n} → ℝ will be denoted by supp f.

2. Formulation of the Question and the Main Result

The class L(1 + ln^{+} L)(ℝ^{n}) was characterized by Guzmán and Welland ([1, 2], Ch. II, §6) by means of the maximal operator M_{B_{1}}. In particular, they have shown that for f ∈ L(ℝ^{n}) the following conditions are equivalent:

1. f ∈ L(1 + ln^{+} L)(ℝ^{n}),
2. ∫_{{M_{B_{1}} (f)>1}} M_{B_{1}}(f) < ∞.

From the strong maximal Jessen–Marcinkiewicz–Zygmund’s theorem it follows that if

f ∈ L(1 + ln^{+} L)^{n}(ℝ^{n}),  (2.1)

then

∫_{{M_{B_{2}} (f)>1}} M_{B_{2}}(f) < ∞.  (2.2)

Guzmán (see [2], Ch. II, §6) posed the question whether it was possible to characterize the class L(1 + ln^{+} L)^{2}(ℝ^{2}) by the operator M_{B_{2}} as it was done for the class L(1 + ln^{+} L)(ℝ^{n}) using the operator M_{B_{1}}, i.e., whether conditions (2.1) and (2.2) are equivalent for f ∈ L(ℝ^{2}). Gogoladze [4, 5] and Bagby [6] answered this question in the negative.

It can be easily verified that much more than (2.2) is fulfilled for f ∈ L(1 + ln^{+} L)^{n}(ℝ^{n}), in particular,

sup_{θ∈θ(ℝ^{n})}  ∫_{{M_{B_{2}(θ)}(f)>1}} M_{B_{2}(θ)}(f) < ∞.  (2.3)

A question arises if it is possible to characterize the class L(1+ln^{+}L)^{n}(ℝ^{n}) by condition (2.3), i.e., if conditions (2.1) and (2.3) are equivalent for f ∈ L(ℝ^{n}) (n ≥ 2).

This question was answered in the negative for n = 2 in [7]. The answer remains negative for an arbitrary n > 2 as well. In particular, the following theorem is valid.

Theorem 1. For any function f ∈ L(1+ln^{+} L)(ℝ^{n}) (n ≥ 2) there exists a measure preserving and invertible mapping ω : ℝ^{n} → ℝ^{n} such that

1. {x : ω(x) ≠ x} ⊂ supp f,
2. sup_{θ∈θ(ℝ^{n})}  ∫_{{M_{B_{2}(θ)}(f∘ω)>1}} M_{B_{2}(θ)}(f ∘ ω) < ∞.

Note that we had to use many new arguments to proceed from the case to n = 2 to the case of arbitrary (n ≥ 2).

Theorem 1 was first formulated by us in a less general for in [8].

3. Auxiliary Statements

Throughout the discussion preceding Lemma 4 we shall consider the spaces ℝ^{n} with n ≥ 2.

We shall call a strip in ℝ^{n} an open set bounded by two different parallel hyperplanes, i.e., a set of the form

{x ∈ ℝ^{n} : a < α_{1}x^{2} + ⋯ + α_{n}x^{n} < b},

where a, b (a > b) and α_{1}, …, α_{n} (α_{1}^{2} + ⋯ + α_{n}^{2} > 0) are some real numbers, and x^{k} (k = 1, …, n) here and everywhere below denotes the k-th coordinate of the point x ∈ ℝ^{n}. The strip width will be called the distance between the hyperplanes that bound the strip, i.e., the number b − a will be called the strip width.

In the sequel it will always be assumed that χ_{A} is the characteristic function of the set A.

Lemma 1. For every x ∈ ℝ^{n} let B(x) be a family of open bounded and convex sets in ℝ^{n}, containing x, and let S be a strip in ℝ^{n} of width δ. Then

M_{B}(χ_{S})(x) < \frac{2^{n}δ}{dist(x, S)} when dist(x, S) ≥ δ.

Proof. Let dist(x, S) ≥ δ and R ∈ B(x), R ∩ S ≠ ∅.

Among the hyperplanes bounding S we denote by Γ the hyperplane which is the closest to x. It is obvious that R ∩ Γ ≠ ∅. For every y ∈ R ∩ Γ let Δ_{y} be a segment connecting x and y. It is assumed that K = ⋃_{y∈R∩Γ} Δ_{y}. Since

R is convex, we have

K ⊂ R.  (3.1)

Let H be the homothety centered at x and with the coefficient

α = \frac{dist(x, S) + δ}{dist(x, S)} .

Let us show that

R ∩ S ⊂ H(K)\K.  (3.2)

Indeed, assume that z ∈ R ∩ S and denote by y the point at which the segment connecting x and z intersects with Γ. Since x, z ∈ R, by virtue of the convexity of R we have y ∈ R. Therefore y ∈ R ∩ Γ. By the definitions of the set K and homothety H we easily obtain z ∈ H(Δ_{y}) ⊂ H(K). (R ∩ S) ∩ K = ∅. Therefore z ∉ K. Thus z ∈ H(K)\K. Thus, since z ∈ R ∩ S is arbitrary, we have proved (3.2).

Using (3.1), (3.2), the definition of H and obvious inequality α^{n} − 1 < \frac{2^{n}δ}{dist(x,S)} we can write

\frac{1}{|R|} ∫_{R} χ_{S} = \frac{|R ∩ S|}{|R|} ≤ \frac{|H(K)\K|}{|K|} = \frac{(α^{n} − 1)|K|}{|K|} < \frac{2^{n}δ}{dist(x, S)},

which, obviously, proves the lemma.

For the rectangle I in ℝ^{n} having pairwise orthogonal edges of lengths δ_{1}, δ_{2}, …, δ_{n}, where δ_{1} ≤ δ_{2} ≤ ⋯ ≤ δ_{n}, we introduce the notation:

(1) r(I) is a number δ_{2}/δ_{1};

(2) when r(I) > 1, for h ≥ 1, J(I, h) is an open rectangle with the following properties: J(I, h) has the same center and frame as I; the length of the edges of J(I, h) parallel to the edges of I of the length δ_{1} is equal to (2^{n+1}h + 1)δ_{1}, while the length of the edges of J(I, h) parallel to the edges of I of length δ_{j} (j = 2, …, n) is equal to 3δ_{j};

(3) for r(I) > 1, ℓ_{I} is a straight line passing through O and parallel to the edges of I of length δ_{1}.

For the straight line ℓ in ℝ^{n} and 0 < ε < π/4 we assume

E(ℓ, ε) = {θ ∈ θ(ℝ^{n}) : ∠(ℓ, θ^{j}) < π/2 − ε, j = 1, …, n},

where ∠(·, ·) is the angle lying between the two straight lines.

Lemma 2. Let I be a rectangle in ℝ^{n}, h > 1, 0 < ε < π/4, r(I) > \frac{nh}{sin ε}, and E = E(ℓ_{I}, ε). Then

{M_{B_{2}(E)}(hχ_{I}) > 1} ⊂ J(I, h),

and therefore

|{M_{B_{2}(E)}(hχ_{I}) > 1}| ≤ 9^{n}h|I|.

Proof. Without loss of generality we assume that

I = (−δ_{1}/2, δ_{1}/2) × · × (−δ_{n}/2, δ_{n}/2),

where δ_{1} < δ_{1} ≤ ⋯ ≤ δ_{n}. We write

S_{1} = \{x ∈ ℝ^{n} : |x^{1}| < (2^{n}h + \frac{1}{2})δ_{1}\},
S_{j} = {x ∈ ℝ^{n} : |x^{j}| < 3δ_{j}/2} (j = 2, …, n).

As is easily seen, J(I, h) is the intersection of the strips S_{1}, …, S_{n}.

Let S = {x ∈ ℝ^{n} : |x^{1}| < δ_{1}/2} and x ∈ S_{1}. Obviously, dist(x, S) ≥ 2^{n}hδ_{1}. Now by lemma 1 we write

M_{B_{2}(E)}(hχ_{I})(x) = hM_{B_{2}(E)}(χ_{I})(x) ≤ hM_{B_{2}(E)}(χ_{S})(x) < \frac{h2^{n}δ_{1}}{dist(x, S)} ≤ 1.

Hence we conclude that

{M_{B_{2}(E)}(hχ_{I}) > 1} ⊂ S_{1}.  (3.3)

Consider arbitrary 2 ≤ j ≤ n. Let x ∉ S_{j}, J ∈ B_{2}(E)(x), and J ∩ I ≠ ∅. Obviously, dist(x, I) ≥ δ_{j}, and we have

dist(x, I) ≤ diam I < t_{1} + t_{2} + ⋯ + t_{n},

where t_{1}, t_{2}, …, t_{n} are lengths of orthogonal edges of J. Therefore there exists a side of J with the length greater than δ_{j}/n. We can represent J as a union of pairwise nonintersecting intervals equal and parallel to abovementioned edge: J = ⋃_{α∈T} Δ_{α}. Obviously,

|Δ_{α}|_{1} > δ_{j}/n (α ∈ T).  (3.4)

(Here and everywhere below, for the set A contained in some k-dimensional (k = 1, …, n − 1) affine subspace ℝ^{n}, we denote by |A|_{k} k-dimensional measure of A.)

Let us prove that

\frac{h|Δ_{α} ∩ I|_{1}}{|Δ_{α}|_{1}} ≤ 1 (α ∈ T).  (3.5)

Indeed, let ℓ be the straight line containing the segment Δ_{α}. It is easy to see that |ℓ ∩ S|_{1} = δ_{1}/ cos ∠(ℓ, Ox^{1}). J ∈ B_{2}(E)(x), Therefore ∠(ℓ, Ox^{1}) < π/2 − ε. Consequently, |ℓ ∩ S|_{1} ≤ \frac{δ_{1}}{cos(π/2−ε)} = \frac{δ_{1}}{sin ε}, which by virtue of (3.4) and the inequality δ_{j} > r(I)δ_{1} ≥ \frac{nhδ_{1}}{sin ε} implies

\frac{h|Δ_{α} ∩ I|_{1}}{|Δ_{α}|_{1}} ≤ \frac{h|ℓ ∩ S|_{1}}{δ_{j}/n} ≤ \frac{hδ_{1}}{sin ε} \frac{sin ε}{hδ_{1}} = 1.

It is not difficult to verify that

\frac{1}{|J|} ∫_{J} hχ_{I} = \frac{h|J ∩ I|}{|J|} ≤ sup_{α∈T} \frac{h|Δ_{α} ∩ I|_{1}}{|Δ_{α}|_{1}} .

Hence, by (3.5),

\frac{1}{|J|} ∫_{J} hχ_{I} ≤ 1,

which, taking into account the arbitrariness of J ∈ B_{2}(E)(x), L ∩ I ≠ ∅, allowsus to conclude that

M_{B_{2}(E)}(hχ_{I})(x) ≤ 1 (x ∉ S_{j}, 2 ≤ j ≤ n).

This and (3.3) imply

{M_{B_{2}(E)}(hχ_{I}) > 1} ⊂ ⋂_{j=1}^{n}S_{j} = J(I, h).

Lemma 3. If among the pairwise different straight lines ℓ_{1}, …, ℓ_{k} (k ≥ n) in ℝ^{n} which pass through the same point none of n lie in the same hyperplane, then there exists ε > 0 such that for every straight line ℓ in ℝ^{n} and every 1 ≤ k_{1} < k_{2} < ⋯ < k_{n} ≤ k

min_{1≤j≤n} ∠(ℓ, ℓ_{kj}) < \frac{π}{2} − ε .

Proof. Let x_{j} ∈ ℝ^{n}, ∥x_{j}∥ = 1 (∥ · ∥ is the norm in ℝ^{n}, j = 1, …, n) be the direction vector of the straight line ℓ_{j}. If we assume the contrary to the assertion of the lemma, then for every m ∈ ℕ there exist y_{m} ∈ ℝ^{n}, ∥y_{m}∥ = 1, and numbers 1 ≤ k_{1}(m) < k_{2}(m) < ⋯ k_{n}(m) ≤ k such that

arccos |(y_{m}, x_{k_{j}(m)})| > \frac{π}{2} − \frac{1}{m}

for j = 1, …, n, where (·, ·) is the scalar product in ℝ^{n}. Hence by the compactness of the unit sphere in ℝ^{n} and the continuity of the scalar product there exist y ∈ ℝ^{n}, ∥y∥ = 1, and 1 ≤ k_{1} < k_{2} < ⋯ < k_{n} ≤ k such that

(y, x_{k_{j}}) = 0

for j = 1, …, n. This implies that the points x_{k_{1}}, …, x_{k_{n}} belong to the hyperplane which is orthogonal to y. Thus the straight lines ℓ_{k_{1}}, …, ℓ_{k_{n}} lie in the same hyperplane which contradicts the condition of the lemma.

Lemma 4. Let f be a continuous function on ℝ^{n}, θ ∈ θ(ℝ^{n}), λ > 0, and an open set G contain {M_{B_{2}(θ)}(f) > λ}. If for the rectangle I in ℝ^{n} with θ(I) = θ, I\G ≠ ∅, then

∫_{I∩G} |f| ≤ λ|I ∩ G|.

Proof. We prove the lemma by induction with respect to n. For n = 1 the proof is obvious. Consider the passage from n − 1 to n.

Without loss of generality we assume that θ = θ_{0} and I is closed.

Introduce the notation:

\begin{matrix}Γ_{t} & = \{x ∈ ℝ^{n} : x^{1} = t\}, \\ I_{t} & = I ∩ Γ_{t}, G_{t} = G ∩ Γ_{t}, \end{matrix} (t ∈ ℝ)

\begin{aligned}J & = \{t ∈ ℝ^{n} : I_{t} ≠ ∅\}, \\ S_{1} & = \{t ∈ J : I_{t} ⊂ G_{t}\}, \\ S_{2} & = \{t ∈ J : I_{t}\G_{t} ≠ ∅\}. \end{aligned}

It is easy to see that S_{1} is open by the natural topology on the interval J. Therefore S_{1} divides into pairwise nonintersecting intervals {δ_{k}}_{k∈T ⊂ℕ}. Obviously, the n-dimensional rectangles Δ_{k} = ⋃_{t∈δ_{k}} I_{t} (k ∈ T) satisfy the

conditions

∂Δ_{k} ∩ ∂G ≠ ∅ (k ∈ T),  (3.6)

where ∂Δ_{k} and ∂G are the boundaries of Δ_{k} and G, respectively;

θ(Δ_{k}) = θ(I) = θ_{0} and Δ_{k} ⊂ I ∩ G (k ∈ T),  (3.7)
Δ_{k} ∩ Δ_{m} = ∅ (k ≠ m).  (3.8)

By the conditions of the lemma, M_{B_{2}}(f)(x) ≤ λ for x ∈ ∂G and therefore, with (3.6) and (3.7) taken into account, we have

∫_{Δ_{k}} |f| ≤ λ|Δ_{k}| (k ∈ T),

which on account to (3.8) implies

∫_{⋃_{k∈T} Δ_{k}} |f| ≤ λ| ⋃_{k∈T}Δ_{k}|.  (3.9)

Estimate now the integral of |f| on (I ∩G)\ ⋃_{k∈T} Δ_{k}. Let M be an (n−1)-

dimensional strong maximal operator. For each t ∈ ℝ consider the function g_{t}(y) = f(t, y) (y ∈ ℝ^{n−1}) and assume that

F (t, y) = M(g_{t})(y) (t ∈ ℝ, y ∈ ℝ^{n−1}).

For t ∈ S_{2} we have

{F (t, ·) > λ} ⊂ G_{t}.  (3.10)

Indeed, assume the contrary, i.e., there exist t_{0} ∈ S_{2}, y_{0} ∈ ℝ^{n−1} and an (n − 1)-dimensional interval R such that (t_{0}, y_{0}) ∉ G_{t_{0}}, R ∋ y_{0}, and

∫_{R} |g_{t_{0}}(y)| dy ≥ λ|R|_{n−1}.

Then by the continuity of f, for a sufficiently small one-dimensional interval Δ ∋ t_{0} we shall have

∫_{Δ×R} |f(t, y)| dt dy > λ|Δ × R|.

Hence M_{B_{2}}(f)(t_{0}, y_{0}) > λ. On the other hand, since (t_{0}, y_{0}) ∉ G_{t_{0}}, we have (t_{0}, y_{0}) ∉ G ⊃ {M_{B_{2}}(f) > λ}. The obtained contradiction proves (3.10).

By virtue of (3.10) and the induction assumption we easily obtain

∫_{I_{t}∩G_{t}} |f(t, y)| dy ≤ λ|I_{t} ∩ G_{t}|_{n−1}

for t ∈ S_{2}.

Thus we can immediately write

∫_{I∩G\ ⋃_{k∈T} Δ_{k}} |f| = ∫_{S_{2}} [ ∫_{I_{t}∩G_{t}} |f(t, y)| dy] dt ≤

≤ ∫_{S_{2}} λ|I_{t} ∩ G_{t}| dt = λ|(I ∩ G)\ ⋃_{k∈T}Δ_{k}|.

whence by (3.7) and (3.9) we conclude that Lemma 4 is valid.

Denote by \overline{L}(ℝ^{n}) a class of all functions f ∈ L(ℝ^{n}) for each of which there exists, for ε > 0, a continuous function g ∈ L(ℝ^{n}) on ℝ^{n} such that |g(x)| ≤ |f(x)| almost everywhere on ℝ^{n}, and ∥f − g∥_{1} < ε.

Lemma 5. Let f ∈ \overline{L}(ℝ^{n}), θ ∈ θ(ℝ^{n}), λ > 0, and then open set G contain {M_{B_{2}(θ)}(f) > λ}. If for the rectangle I in ℝ^{n} with θ(I) = θ, I\G ≠ ∅, then

∫_{I∩G} |f| ≤ λ|I ∩ G|.

Proof. f ∈ \overline{L}(ℝ^{n}). Therefore for arbitrarily given ε > 0 there exists a continuous function g ∈ L(ℝ^{n}) on ℝ^{n} such that |g(x)| ≤ |f(x)| almost everywhere on ℝ^{n}, and ∥f − g∥_{1} < ε. It is obvious that

{M_{B_{2}(θ)}(g) > λ} ⊂ {M_{B_{2}(θ)}(f) > λ} ⊂ G.

Now by Lemma 4

∫_{I∩G} |g| ≤ λ|I ∩ G|,

and therefore

∫_{I∩G} |f| − ε ≤ λ|I ∩ G|,

whence by the arbitrariness of ε > 0 we conclude that Lemma 5 is valid.

Lemma 6. Let f_{k} ∈ \overline{L}(ℝ^{n}), f_{k} ≥ 0 (k ∈ ℕ), E ⊂ θ(ℝ^{n}), E ≠ ∅, λ > 0, and let for k, m ∈ ℕ and k ≠ m the following conditions be fulfilled:

supp f_{k} ∩ supp f_{m} = ∅,
supp f_{k} ∩ {M_{B_{2}(E)}(f_{m}) > λ} = ∅,
{M_{B_{2}(E)}(f_{k}) > λ} ∩ {M_{B_{2}(E)}(f_{m}) > λ} = ∅.

Then

{M_{B_{2}(E)}( ∑_{k=1}^{m} f_{k}) > λ} = ⋃_{k=1}^{∞}{M_{B_{2}(E)}(f_{k}) > λ}.

Proof. Denote G_{k} = {M_{B_{2}(E)}(f_{k}) > λ}, k ∈ ℕ. For each k ∈ ℕ

f_{k}(x) ≤ λ almost everywhere on ℝ^{n}\G_{k}.  (3.11)

Indeed, otherwise, since the differential bases B_{2}(θ), θ ∈ θ(ℝ^{n}), are dense (see, for e.g., [2], Ch.II, §3), for arbitrary θ ∈ E and A_{j} = (ℝ^{n}\G_{k}) ∩ {f_{k} > λ + 1/j} (j ∈ ℕ) we shall have

lim_{I∈B_{2}(θ)(x), diam I→0} \frac{|I ∩ A_{j}|}{|I|} for almost all x ∈ A_{j}.

Hence M_{B_{2}(E)}(f_{k}) ≥ M_{B_{2}(θ)}(f_{k}) > λ for almost all x ∈ (ℝ^{n}\G_{k}) ∩ {f_{k} > λ}, which contradicts the definition of G_{k}.

158 G. ONIANI

By (3.11) and the condition of the lemma we write

∑_{k=1}^{∞} f_{k}(x) ≤ λ for almost all x ∉ ⋃_{k=1}^{∞}G_{k}.

Hence, by the conditions of the lemma and by Lemma 5, we find that for

every x ∉ ⋃_{k=1}^{∞} G_{k} and I ∈ B_{2}(E)(x)

∫_{I} ∑_{k=1}^{∞} f_{k} ≤ ∑_{k=1}^{∞} ∫_{I∩G_{k}} f_{k} + ∫_{I\ ⋃_{k=1}^{∞} G_{k}} ∑_{k=1}^{∞} f_{k} ≤

≤ ∑_{k=1}^{∞} λ|I ∩ G_{k}| + λ|I\ ⋃_{k=1}^{∞}G_{k}| = λ|I|.

Therefore

M_{B_{2}(E)}( ∑_{k=1}^{∞} f_{k})(x) ≤ λ for x ∉ ⋃_{k=1}^{∞}G_{k}.

The next assertion belongs to Jessen, Marcinkiewicz, and Zygmund and is referred to as the strong maximal theorem (see [3] or [2], Ch. II, §3).

Theorem. If f ∈ L(1 + ln^{+} L)^{n−1}(ℝ^{n}), then

|\{M_{B_{2}}(f) > λ\}| ≤ c_{1} ∫_{ℝ^{n}} \frac{|f|}{λ} (1 + ln^{+} \frac{|f|}{λ})^{n−1} (λ > 0),

where c_{1} is the constant depending only on n.

The foollowing lemma is a simple improvement of this result.

Lemma 7. If f ∈ L(1 + ln^{+} L)^{n−1}(ℝ^{n}), then for every θ ∈ θ(ℝ^{n})

|\{M_{B_{2}(θ)}(f) > λ\}| ≤ c_{2} ∫_{\{|f|>λ/2\}} \frac{|f|}{λ} (1 + ln \frac{2|f|}{λ})^{n−1} (λ > 0),

where the constant c_{2} depends only on n. Proof. For arbitrary fixed λ>0 assume f_{∗} =fχ_{{|f|≤λ/2}} and f^{∗} =fχ_{{|f|>λ/2}}. f = f_{∗} + f^{∗}. Therefore M_{B_{2}}(f) ≤ M_{B_{2}}(f_{∗}) + M_{B_{2}}(f^{∗}). Hence

{M_{B_{2}} > λ} ⊂ {M_{B_{2}}(f_{∗}) > λ/2} ∪ {M_{B_{2}}(f^{∗}) > λ/2}.

But {M_{B_{2}}(f_{∗}) > λ/2} = ∅ and therefore by the strong maximal theorem

|\{M_{B_{2}}(f) > λ\}|≤|\{M_{B_{2}}(f^{∗}) > λ/2\}|≤c_{1} ∫_{ℝ^{n}} \frac{2|f^{∗}|}{λ} (1 + ln^{+} \frac{2|f^{∗}|}{λ})^{n−1} ≤

≤ 2c_{1} ∫_{\{|f|>λ/2\}} \frac{|f|}{λ} (1 + ln \frac{2|f|}{λ})^{n−1}.  (3.12)

Let γ_{θ}, θ ∈ θ(ℝ^{n}), be a rotation such that θ = {γ_{θ}(Ox^{1}), …, γ_{θ}(Ox^{n})}. In view of the fact that the rotation is a measure preserving mapping, we readily obtain

M_{B_{2}(θ)}(f)(x) = M_{B_{2}}(f ∘ γ_{θ})(γ_{θ}^{−1}(x)) (x ∈ ℝ^{n})  (3.13)

Therefore

|{M_{B_{2}(θ)}(f) > λ}| = |{M_{B_{2}}(f ∘ γ_{θ}) > λ}| (λ > 0).

By this and (3.12) we conclude that the lemma is valid.

Lemma 8. If f ∈ L(1 + ln^{+} L)^{n}(ℝ^{n}), then for every θ ∈ θ(ℝ^{n})

∫_{\{M_{B_{2}(θ)}(f)>λ\}} M_{B_{2}(θ)}(f) ≤ c_{3} ∫_{ℝ^{n}} |f|(1 + ln^{+} \frac{|f|}{λ})^{n} (λ > 0),

where the constant c_{3} depends only on n.

Proof. Let f ∈ L(1 + ln^{+} L)^{n}(ℝ^{n}) and λ > 0. We have

∫_{{M_{B_{2}(θ)}(f)>λ}} M_{B_{2}}(f) = − ∫_{λ}^{∞}t dF (t) = [ − tF (t)]_{λ}^{∞} + ∫_{λ}^{∞}F (t) dt,

where F (t) = |{M_{B_{2}}(f) > t}| (t > 0). By Lemma 7

tF (t) ≤ c_{2} ∫_{\{|f|>t/2\}} |f|(1 + ln \frac{2|f|}{t})^{n−1} (t > 0).  (3.14)

Hence

∫_{{M_{B_{2}(θ)}(f)>λ}} M_{B_{2}}(f) = λF (λ) + ∫_{λ}^{∞}F (t) dt.  (3.15)

Lemma 7 yields

∫_{λ}^{∞}F (t) dt ≤ c_{2} ∫_{λ}^{∞} ∫_{\{|f(x)|>t/2\}} \frac{|f(x)|}{t}(1 + ln \frac{2|f(x)|}{t})^{n−1}dx dt =

= c_{2} ∫_{\{|f(x)|>λ/2\}} ∫_{λ}^{2|f(x)|}\frac{|f(x)|}{t}(1 + ln \frac{2|f(x)|}{t})^{n−1}dt dx ≤

≤ c_{2} ∫_{\{|f(x)|>λ/2\}} ∫_{λ}^{2|f(x)|}\frac{|f(x)|}{t}(1 + ln \frac{2|f(x)|}{λ})^{n−1}dt dx ≤

≤ c_{2} ∫_{\{|f(x)|>λ/2\}} |f(x)|(1 + ln \frac{2|f(x)|}{λ})^{n}dx,

whence with regard for (3.14) and (3.15) we obtain

∫_{\{M_{B_{2}} (f)>λ\}} M_{B_{2}}(f) ≤ 2c_{2} ∫_{ℝ^{n}} |f|(1 + ln^{+} \frac{2|f|}{λ})^{n} (λ > 0)  (3.16)

for f ∈ L(1 + ln^{+} L)^{n}(ℝ^{n}).

(3.13) readily implies

∫_{{M_{B_{2}(θ)}(f)>λ}} M_{B_{2}(θ)}(f) =  ∫_{{M_{B_{2}} (f∘γ_{θ})>λ}} M_{B_{2}}(f ∘ γ_{θ}) (λ > 0) (3.17)

for f ∈ L(1 + ln^{+} L)^{n}(ℝ^{n}) and θ ∈ θ(ℝ^{n}).

Since the rotation is the measure preserving mapping, by (3.16) and (3.17) we immediately conclude that Lemma 8 is valid.

Lemma 9. Let f ∈ L(1 + ln^{+} L)(ℝ^{n}), g : ℝ^{n} → ℝ be a measurable function, and a, b > 0 and λ ≥ 0. If

|\{|g| > t\}| ≤ \frac{a}{t} ∫_{\{|f|>bt\}} |f| (t ≥ λ),  (3.18)

then

∫_{\{|g|>λ\}} |g| ≤ a ∫_{ℝ^{n}} |f|(1 + ln^{+} \frac{|f|}{bλ}).

Proof. We have

∫_{{|g|>λ}} |g| = − ∫_{λ}^{∞}t dF (t) = [ − tF (t)]_{λ}^{∞} + ∫_{λ}^{∞}F (t) dt,

where F (t) = |{|g| > t}| (t ≥ 0). By (3.18)

tF (t) ≤ a ∫_{{|f|>bt}}|f| (t ≥ λ).  (3.19)

Hence

∫_{{|g|>λ}} |g| = λF (λ) + ∫_{λ}^{∞}F (t) dt.  (3.20)

By (3.18)

∫_{λ}^{∞}F (t) dt ≤ a ∫_{λ}^{∞} \frac{1}{t} ∫_{\{|f(x)|>bt\}} |f(x)| dx dt =

= a ∫_{\{|f(x)|>bλ\}} |f(x)| ∫_{λ}^{|f(x)|/b}\frac{dt}{t} dx = a ∫_{\{|f(x)|>bλ\}} |f(x)| ln \frac{|f(x)|}{bλ} dx.

Hence with (3.19) and (3.20) taken into account, we conclude that Lemma 9 is valid.

Lemma 10. Let f_{1} and f_{2} be the nonnegative measurable functions defined on ℝ^{n}. Then

∫_{{f_{1}+f_{2}>2λ}} (f_{1} + f_{2}) ≤ (1 + λ)( ∫_{{f_{1}>λ}} f_{1} + ∫_{{f_{2}>λ}} f_{2}) (λ ≥ 0).

Proof. The validity of the lemma follows from the following relations easy to verify:

(1)  ∫_{{f_{1}+f_{2}>2λ}} (f_{1} + f_{2}) ≤  ∫_{{f_{1}>λ}∪{f_{2}>λ}} (f_{1} + f_{2});

(2) ∫_{\{f_{1}>λ\}∪\{f_{2}>λ\}} f_{j} ≤ ∫_{\{f_{j}>λ\}}f_{j} + λ|\{f_{i} > λ\}|, where j, i ∈ \overline{1, 2} and j ≠ i.

The set E ⊂ ℝ^{n} is called elementary if it is a union of a finite number of n-dimensional intervals.

Lemma 11. Let A be a subset of ℝ^{n} of positive measure. Then for each

δ_{k} > 0 (k ∈ ℕ) with ∑_{k=1}^{∞} δ_{k} < |A| and ε_{k} > 0 (k ∈ ℕ) there exist pairwise

nonintersecting elementary sets G_{k} (k ∈ ℕ) such that

|G_{k}| = δ_{k} and |G_{k}\A| < ε_{k}.

Proof. Let us construct the sequence {G_{k}} with the needed properties. For this we shall need the following simple facts:

(1) For each measurable set E and number δ with 0 ≤ δ ≤ |E| there exists a measurable set E′ ⊂ E with |E′| = δ;

(2) For each open set E ⊂ ℝ^{n} and number δ with 0 < δ < |E| there exists an elementary set E′ ⊂ E with |E′| = δ.

By virtue of (1), there exists E ⊂ A with |E| = δ_{1}. Let an open set Q be such that Q ⊃ E, |Q| > |E| = δ_{1} and |Q\E| < ε_{1}. According to (2), there exists an elementary set G_{1} ⊂ Q with |G_{1}| = δ_{1}. Obviously, |G_{1}\A| ≤ |Q\E| < ε_{1}.

Suppose the pairwise nonintersecting elementary sets G_{1}, …, G_{k} with the properties

|G_{j}| = δ_{j} and |G_{j}\A| < ε_{j} (j ∈ \overline{1, k})

have already been constructed. Then

|A\ ⋃_{j=1}^{k}(\overline{G}_{j} ∩ A)| ≥ |A| − ∑_{j=1}^{k} δ_{j} > δ_{k+1},

where \overline{G}_{j} is the closure of G_{j}. Therefore by (1), there exists

E ⊂ A\ ⋃_{j=1}^{k}(\overline{G}_{j} ∩ A)

with |E| = δ_{k+1}. We can easily obtain an open set Q ⊃ E with the properties

Q ∩ ⋃_{j=1}^{k}\overline{G}_{j} = ∅, |Q| > |E| = δ_{k+1}, |Q\E| < ε_{k+1}.

By (2), we can choose an elementary set G_{k+1} ⊂ Q such that |G_{k+1}| = δ_{k+1}. By virtue of the properties of Q we have

|G_{k+1}\A| ≤ |Q\E| < ε_{k+1},

G_{k+1} ∩ ⋃_{j=1}^{k}G_{j} = ∅,

which obviously proves Lemma 11.

We shall need the following simple lemma (see [2], Ch. III, §1).

Lemma 12. Let G be an open bounded set in ℝ^{n}, and K be a compact set in ℝ^{n} with |K| > 0. Then there exists a sequence {K_{k}} of pairwise nonintersecting sets, homothetic to K, contained in G and such that |G\ ⋃_{k} K_{k}| = 0.

We shall also need the following well-known fact from the measure theory (see, e.g., [9], Ch. “Uniform Approximation”).

Lemma 13. For every measurable sets A_{1}, A_{2} ⊂ ℝ^{n}, |A_{1}| = |A_{2}|, there exists a measure preserving and invertible mapping ω : A_{1} → A_{2}.

4. Proof of Theorem 1

Without loss of generality we assume that f ≥0 and f ∉(1+ln^{+} L)^{n}(ℝ^{n}). Denote

G = supp f, A_{k} = {k − 1 ≤ f < k} (k ∈ ℕ),
k_{0} = min {k ≥ 2 : ∑_{m=k}^{∞} 9^{n}m|A_{m}| < |G|},
N = {k ≥ k_{0} : |A_{k}| > 0}.

Choose natural numbers m_{k} ≥ n (k ∈ N) such that

∑_{k∈N} \frac{k(ln k)^{n}|A_{k}|}{m_{k}} < 1 .  (4.1)

For k ∈ N, let ℓ_{k,1}, …, ℓ_{k,m_{k}} be the straight lines passing through the origin with none of n lying in the same hyperplane. Then by Lemma 2 there exists ε_{k} > 0 such that

min_{1≤j≤n} ∠(ℓ, ℓ_{k,ν_{j}}) < \frac{π}{2} − ε_{k}  (4.2)

for every 1 ≤ ν_{1} < ν_{2} < ⋯ < ν_{n} ≤ m_{k} and for every straight line ℓ.

For every k ∈ N and m ∈ \overline{1, m_{k}} let us consider the rectangle I_{k,m} with the properties:

r(I_{k,m}) ≥ \frac{4kn}{sin ε_{k}} , |I_{k,m}| = \frac{|A_{k}|}{m_{k}}, ℓ_{I_{k,m}} = ℓ_{k,m}.  (4.3)

Denote J_{k,m} = J(I_{k,m}, 4k), E_{k,m} = E(ℓ_{k,m}, ε_{k}) (k ∈ N, m ∈ \overline{1, m_{k}}). By Lemma 2

{M_{B_{2}(E_{k,m})}(4kχ_{I_{k,m}}) > 1} ⊂ J_{k,m}.

From the definition of k_{0} and J_{k,m} and from (4.3), we conclude by virtue of Lemma 11 that there exist pairwise nonintersecting open sets Q_{k,m} such that

|Q_{k,m}| = |J_{k,m}| and |Q_{k,m}\G| < \frac{1}{2^{k}m_{k}} .

For each k ∈ N and m ∈ \overline{1, m_{k}} we complete Q_{k,m} with pairwise nonintersecting rectangles {J_{k,m,q}} which are homothetic to the rectangle J_{k,m} (see Lemma 12), i.e.,

J_{k,m,q} = H_{k,m,q}(J_{k,m}), where H_{k,m,q} is the homothety (q ∈ ℕ), J_{k,m,q} ⊂ Q_{k,m} (q ∈ ℕ),

J_{k,m,q} ∩ J_{k,m,q′} = ∅ (q ≠ q′),

|Q_{k,m}\ ⋃_{q∈ℕ}J_{k,m,q}| = 0.

Let I_{k,m,q} = H_{k,m,q}(I_{k,m}) (k ∈ N, m ∈ \overline{1, m_{k}}, q ∈ ℕ). Because of the homothety properties we can easily see that

J_{k,m,q} = J(I_{k,m,q}, 4k),  (4.4)
{M_{B_{2}(E_{k,m})}(4kχ_{I_{k,m,q}}) > 1} ⊂ J_{k,m,q}  (4.5)

for k ∈ N, m ∈ \overline{1, m_{k}}, q ∈ ℕ, and

∑_{q∈ℕ} |I_{k,m,q}| = |I_{k,m}| = \frac{|A_{k}|}{m_{k}}  (4.6)

for k ∈ N, m ∈ \overline{1, m_{k}}.

Denote

g_{k,m} = sup \{kχ_{I_{k,m,q}} : q ∈ ℕ\} (k ∈ N, m ∈ \overline{1, m_{k}}),
g = sup \{g_{k,m} : k ∈ N, m ∈ \overline{1, m_{k}}\},

and prove that

sup_{θ∈θ(ℝ^{n})}  ∫_{{M_{B_{2}(θ)}(g)>1/2}} M_{B_{2}(θ)}(g) < ∞.  (4.7)

The following estimate is valid:

card S_{θ,k} < n^{2} (θ ∈ θ(ℝ^{n}), k ∈ N),  (4.8)

where S_{θ,k} = \{m ∈ \overline{1, m_{k}} : θ ∉ E_{k,m}\}. Indeed, let us assume the contrary, i.e., that card S_{θ,k} ≥ n^{2} for some θ ∈ θ(ℝ^{n}) and k ∈ N. Then there exist 1 ≤ ν_{1} < ⋯ < ν_{n^{2}} ≤ m_{k} such that θ ∈ E_{k,ν_{j}} (j ∈ \overline{1, n^{2}}), i.e., max_{1≤i≤n} ∠(θ^{i}, ℓ_{k,ν_{j}}) ≥ \frac{π}{2} − ε_{k} (j ∈ \overline{1, n^{2}}). Hence there exist a straight line

θ^{i} ∈ θ and indices ν′_{1}, …, ν′_{n} ∈ \{ν_{1}, …, ν_{n^{2}}\} such that ∠(θ^{i}, ℓ_{k,ν′_{j}}) ≥ \frac{π}{2} − ε_{k} (j ∈ \overline{1, n}), which contradicts (4.2). Therefore (4.8) is proved.

Let us consider an arbitrary frame θ. Suppose

g_{θ} = \begin{cases}sup\{g_{k,m} : k ∈ N, m ∈ S_{θ,k}\} & if ⋃_{k∈N} S_{θ,k} ≠ ∅, \\ 0 & if ⋃_{k∈N} S_{θ,k} = ∅. \end{cases}

By Lemma 8, (4.1), (4.3), (4.6) and (4.8) we have

∫_{{M_{I(θ)}(g_{θ})>1/4}} M_{I(θ)}(g_{θ}) ≤ c_{3} ∫_{ℝ^{n}} g_{θ}(1 + ln^{+} 4g_{θ})^{n} <

< c_{3} ∑_{k∈N} n^{2}k(1 + ln 4k)^{n} \frac{|A_{k}|}{m_{k}} < 5^{n}n^{2}c_{3} ∑_{k∈N} \frac{k(ln k)^{n}|A_{k}|}{m_{k}} < 5^{n}n^{2}c_{3}. (4.9)

Denote

T = \{(k, m, q) : k ∈ N, m ∈ \overline{1, m_{k}} \S_{θ,k}, q ∈ ℕ\},

J_{k,m,q}(λ) = J(I_{k,m,q}, k/λ) for (k, m, q) ∈ T and 1/4 ≤ λ < k.

Obviously,

r(I_{k,m,q}) > \frac{4kn}{sin ε_{k}} ≥ \frac{kn}{λ sin ε_{k}}

for (k, m, q) ∈ T and 1/4 ≤ λ < k, whence on account of (4.4), (4.5) and Lemma 2

{M_{B_{2}(E_{k,m})}(kχ_{I_{k,m,q}}) > λ} =

= \{M_{B_{2}(E_{k,m})}(\frac{k}{λ} χ_{I_{k,m,q}} ) > 1\} ⊂ J_{k,m,q}(λ) ⊂ J_{k,m,q}.

Consequently, since θ ∈ E_{k,m}, we have

{M_{B_{2}(θ)}(kχ_{I_{k,m,q}}) > λ} ⊂ J_{k,m,q}(λ) ⊂ J_{k,m,q}.  (4.10)

On the other hand, it is clear that

{M_{B_{2}(θ)}(kχ_{I_{k,m,q}}) > λ} = ∅  (4.11)

for (k, m, q) ∈ T and λ ≥ k.

It is easy to see that the functions kχ_{I_{k,m,q}} belong to the class \overline{L}(ℝ^{n}) and therefore, keeping in mind that the rectangles J_{k,m,q} are pairwise nonintersecting and using (4.10), (4.11) and Lemma 6 we have

{M_{B_{2}(θ)}(g − g_{θ}) > λ} ⊂ ⋃_{(k,m,q)∈T, k>λ}J_{k,m,q}(λ) for λ ≥ 1/4.

The above inequality, (4.3), (4.6) and the definition of g imply that

|{M_{B_{2}(θ)}(g − g_{θ}) > λ}| ≤ ∑_{(k,m,q)∈T, k>λ} |J_{k,m,q}(λ)| ≤

≤ ∑_{k∈N, k>λ} ∑_{m=1}^{m_{k}} ∑_{q=1}^{∞} 9^{n} \frac{k}{λ} |I_{k,m,q}| = \frac{9^{n}}{λ} ∑_{k∈N, k>λ} k|A_{k}| ≤ \frac{2 · 9^{n}}{λ} ∫_{\{f>λ/2\}} f

for λ ≥ 1/4.

Consequently, by Lemma 9 we obtain

∫ ; {M_{B_{2}(θ)}(g−g_{θ})>1/4} M_{B_{2}(θ)}(g − g_{θ}) ≤ 2 · 9^{n} ∫_{ℝ^{n}} f(1 + ln^{+} 8f).  (4.12)

From (4.9), (4.12) and Lemma 10 we find that

∫ ; {M_{B_{2}(θ)}(g)>1/2} M_{B_{2}(θ)}(g) < 2 · 5^{n}n^{2}c_{3} + 4 · 9^{n} ∫_{ℝ^{n}} f(1 + ln^{+} 8f),

whence by virtue of the arbitrariness of θ we conclude that (4.7) is valid.

Denote

P_{k} = ⋃_{m=1}^{m_{k}} ⋃_{q=1}^{∞}(I_{k,m,q} ∩ G) (k ∈ N).

By our choice of sets Q_{k,m} we easily see that

0 ≤ |A_{k}| − |P_{k}| < \frac{1}{2^{k}} (k ∈ N).  (4.13)

Let A′_{k} ⊂ A_{k} (k ∈ N) be some measurable set with |A′_{k}| = |P_{k}|. By Lemma 13 there exists a measure preserving and invertible mapping ω : ℝ^{n} → ℝ^{n} such that

ω(P_{k}) = A′_{k} (k ∈ N), ω(G\ ⋃_{k∈N}P_{k}) = G\ ⋃_{k∈N}A′_{k},

ω(x) = x (x ∈ ℝ^{n}\G).  (4.14)

Suppose

φ_{1} = (f ∘ ω)χ_{⋃_{k∈N} P_{k}} and φ_{2} = (f ∘ ω)χ_{ℝ^{n}\ ⋃_{k∈N} P_{k}} .

Obviously, f ∘ ω = φ_{1} + φ_{2}. We have

∫_{ℝ^{n}} φ_{2}(1 + ln^{+} φ_{2})^{n} = ∫_{ℝ^{n}\ ⋃_{k∈N} A′_{k}} f(1 + ln^{+} f)^{n} =

= ∫_{{0≤f<k_{0}−1}} f(1 + ln^{+} f)^{n} + ∑_{k∈N} ∫_{A_{k}\A′_{k}} f(1 + ln^{+} f)^{n} = α_{1} + α_{2}.

It can be seen that α_{1} < ∞, and by (4.13)

α_{2} ≤ ∑_{k∈N} \frac{k(1 + ln k)^{n}}{2^{k}} < ∞.

Thus φ_{2} ∈ L(1 + ln^{+} L)^{n}(ℝ^{n}). Therefore, by the obvious inequality φ_{1} ≤ g, (4.7) and Lemmas 8 and 10, we conclude that

sup_{θ∈θ(ℝ^{n})} ∫_{{M_{B_{2}(θ)}(f∘ω)>1}} M_{B_{2}(θ)}(f ∘ ω) < ∞,

which together with (4.14) completes the proof of Theorem 1.

5. Remarks

(1) By the equality M_{B_{2}(θ)}(αf) = αM_{B_{2}(θ)}(f) (α > 0), we can easily verify that Theorem 1 remains valid if instead of {M_{B_{2}(θ)}(f ∘ ω) > 1} we shall take the integrals on {M_{B_{2}(θ)}(f ∘ω) > λ}, where λ > 0 is an arbitrarily fixed number.

(2) Theorem 1 immediately yields the following improvement:

Theorem 2. For every function f ∈ L(1 + ln^{+} L)(ℝ^{n}) (n ≥ 2) and measurable sets G_{1}, G_{2} ∈ ℝ^{n} such that fχ_{ℝ^{n}\G_{1}} ∈ L(1 + ln^{+} L)^{n}(ℝ^{n}) and |G_{1}| = |G_{2}| there exists a measure preserving and invertible mapping ω : ℝ^{n} → ℝ^{n} such that

1) ω(G_{1}) = G_{2} and {x : ω(x) ≠ x} ⊂ G_{1} ∪ G_{2},
2) sup_{θ∈θ(ℝ^{n})}  ∫_{{M_{B_{2}(θ)}(f∘ω)>1}} M_{B_{2}(θ)}(f ∘ ω) < ∞.

Proof. Let ω_{1} : ℝ^{n} → ℝ^{n} be a measure preserving and invertible mapping such that (see Lemma 12) ω_{1}(G_{1}) = G_{2} and {x : ω(x) ≠ x} ⊂ G_{1} ∪ G_{2}. Consider the function g = (f ∘ ω_{1})χ_{G_{2}} . Then supp g ⊂ G_{2}, and by virtue of Theorem 1 (see Remark (1)) there exists a measure preserving and invertible mapping ω_{2} : ℝ^{n} → ℝ^{n} such that

{x : ω_{2}(x) ≠ x} ⊂ supp g ⊂ G_{2} and sup_{θ∈θ(ℝ^{n})} ∫_{{M_{B_{2}(θ)}(g∘ω_{2})>1/2}} M_{B_{2}(θ)}(g∘ω_{2})<∞.

Obviously, (f ∘ ω_{1})χ_{ℝ^{n}\G_{2}} ∈ L(1 + ln^{+} L)^{n}(ℝ^{n}). Therefore, by Lemmas 8 and 10, one can take ω_{2} ∘ ω_{1} as ω.

(3) For arbitrary ε > 0, a mapping ω “correcting” the function f ∈ L(1 + ln^{+} L)(ℝ^{n}) can be chosen so that

|{f ∘ ω ≠ f}| < ε.

For this it is enough in Theorem 2 to take G_{1} and G_{2} with measures less than ε/2.

(4) When G_{1} = {|f| > 1}, and G_{2} is a cubic interval, Theorem 2 has been proved for n = 2 in [7] and announced for n ≥ 2 in [8].

References

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3. B. Jessen, J. Marcinkiewicz, and A. Zygmund, Note on the differentiability of multiple integrals. Fund. Math. 25(1935), 217–234.

4. L. D. Gogoladze, On maximal Hardy–Littlewood functions. (Russian) Soobshch. Akad. Nauk Gruz. SSR 3(1983), 257–259.

5. L. D. Gogoladze, On the summation of multiple trigonometric series and conjugate functions. (Russian) Dissertation for the Doctor of Science Degree, Tbilisi, 1984.

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(Received 20.06.1997)

Author’s address: A. Tsereteli Kutaisi State University 55, Queen Tamar St., Kutaisi 384000 Georgia