GEORGIAN MATHEMATICAL JOURNAL: Vol. 4, No. 1, 1997, 81-100

ON DIFFERENTIAL BASES FORMED OF INTERVALS

G. ONIANI AND T. ZEREKIDZE

In memory of young mathematician A. Berekashvili

Abstract. Translation invariant subbases of the differential basis B_{2} (formed of all intervals), which differentiates the same class of all nonnegative functions as B_{2} does, are described. A possibility for extending the results obtained to bases of more general type is discussed.

1. Definitions and Notation

A mapping B defined on ℝ^{n} is said to be a differential basis in ℝ^{n} if, for every x ∈ ℝ^{n}, B(x) is a family of open bounded sets containing the point x such that there exists a sequence {R_{k}} ⊂ B(x), diam R_{k} → 0 (k → ∞).

For f ∈ L_{loc}(ℝ^{n}) the numbers

\overline{D}_{B}( ∫ f, x) = \overline{lim}_{diam R→0,R∈B(x)} \frac{1}{|R|} ∫_{R} f

and

\underline{D}_{B}( ∫ f, x) = \underline{lim}_{diam R→0,R∈B(x)} \frac{1}{|R|} ∫_{R} f

are said to be respectively the upper and the lower derivative of the integral of f at the point x. If the upper and the lower derivative coincide, then their common value is called the derivative of the integral of f at the point x, and we denote it by D_{B}( ∫ f, x). They say that the basis B differentiates the integral of f if D_{B}( ∫ f, x) = f(x) for almost all x. The set of those functions f ∈ L_{loc}(ℝ^{n}), f ≥ 0, whose integrals are differentiable with respect to the basis B will be denoted by F_{B}^{+}. Under M_{B} we mean the maximal operator

M_{B}(f)(x) = sup_{R∈B(x)} \frac{1}{|R|} ∫_{R} |f| (f ∈ L_{loc}(ℝ^{n}), x ∈ ℝ^{n}),

81 corresponding to the basis B. It will be assumed here that \overline{B} = ⋃_{x∈ℝ^{n}}B(x).

1991 Mathematics Subject Classification. 28A15.

Key words and phrases. Differential basis, translation invariant bases, interval, locally regularity.

1072-947X/97/0100-0081$12.50/0 \begin{matrix}c \\ ◯ \end{matrix} 1997 Plenum Publishing Corporation

B is said to be a subbasis of B′ (writing B ⊂ B′), if B(x) ⊂ B′(x) (x ∈ ℝ^{n}). The basis B is said to be translation invariant or the T I-basis, if B(x) = {x + R : R ∈ B(O)} (x ∈ ℝ^{n}) (here O is the origin in ℝ^{n}). Let us have the bases B and B′. We shall say that the family \overline{B} is locally regular with respect to the family \overline{B′} (writing \overline{B} ∈ LR(\overline{B′})), if there exist δ > 0 and c > 0 such that for any R ∈ \overline{B}, diam R < δ, there is R′ ∈ \overline{B}′ such that R ⊂ R′ and |R′| < c|R|.

We shall agree that I^{n} = [0, 1]^{n} and f ∈ L(I^{n}), if f ∈ L(ℝ^{n}) and supp f ⊂ I^{n}.

2. T I-Bases Formed of Intervals

Let B_{2} be the basis in ℝ^{2} for which B_{2}(x) (x ∈ ℝ^{2}) consists of all twodimensional intervals containing the point x.

The theorem below characterizes B ⊂ B_{2}, T I-bases for which F_{B}^{+} = F_{B_{2}}^{+}.

Theorem 1. Let B ⊂ B_{2} be a T I-basis. Then the following conditions are equivalent:

(a) F_{B}^{+} = F_{B_{2}}^{+};

(b) for every f ∈ L(ℝ^{2}), f ≥ 0, a.e. on ℝ^{2}

\overline{D}_{B}( ∫ f, x) = \overline{D}_{B_{2}}( ∫ f, x) and \underline{D}_{B}( ∫ f, x) = \underline{D}_{B_{2}}( ∫ f, x);

(c) \overline{B}_{2} is locally regular with respect to \overline{B}.

The implication (b) ⇒ (a) is evident. Therefore to prove Theorem 1 it suffices to show that (a) ⇒ (c) and (c) ⇒ (b).

The implication (a) ⇒ (c) follows from the following assertion.

Theorem 2. Let B ⊂ B_{2} be a T I-basis. If \overline{B}_{2} is not locally regular with respect to \overline{B}, then there exists a function f ∈ L(I^{2}), f ≥ 0, such that

\overline{D}_{B_{2}}( ∫ f, x) = ∞ a.e. on I^{2},
D_{B}( ∫ f, x) = f(x) a.e. on I^{2}.

Before proving Theorem 2 we shall give several lemmas.

For the interval I we denote by αI (α > 0) the interval H(I), where H is the homothety with the coefficient α whose center is the center of I.

Lemma 1. Let I ∈ \overline{B}_{2} and h > 1. Then \{M_{B_{2}}(hχ_{I}) > 1\} ⊂ (2h + 1)I and |{M_{B_{2}}(hχ_{I}) > 1}| ≥ h(ln h)|I|.

The validity of this lemma can be shown by a direct checking.

Projections of I onto the ox^{1}- and ox^{2}-axes will be denoted by pr_{1} I and pr_{2} I, respectively. For I ∈ \overline{B}_{2} and h > 0 we have

R_{1}(I, h) = (2h + 1) pr_{1} I × 3 pr_{2} I,
R_{2}(I, h) = 3 pr_{1} I × (2h + 1) pr_{2} I,
R(I, h) = R_{1}(I, h) ∪ R_{2}(I, h).

It is clear that |R(I, h)| ≤ 18h|I|.

Lemma 2. Let B ⊂ B_{2} be a T I-basis; h > 1. Suppose that for I ∈ \overline{B}_{2} there is no J ∈ \overline{B}, J ⊃ I, such that |J| ≤ h|I|. Then \{M_{B}(hχ_{I}) > 1\} ⊂ R(I, h).

Proof. Let J ∈ \overline{B} and (1/|J|) ∫_{J} hχ_{I} > 1. From the condition of the lemma we can easily find that either

|pr_{1}J|_{1} < | pr_{1} I|_{1} or | pr_{2} J|_{1} < | pr_{2} I|_{1},  (1)

where | · |_{1} is the Lebesgue measure on ℝ.

By Lemma 1, J ⊂ (2h + 1)I, which by virtue of (1) implies that either J ⊂ R_{1}(I, h) or J ⊂ R_{2}(I, h), so that we have J ⊂ R(I, h). It follows from the latter inclusion that {M_{B}(hχ_{I}) > 1} ⊂ R(I, h).

It can be easily seen that the following two lemmas are valid.

Lemma 3. Let I, J ∈ \overline{B}_{2}; h > 1. If either | pr_{1} J|_{1} ≥ h| pr_{1} J|_{1} or | pr_{2} J|_{1} ≥ h| pr_{2} J|_{1}, then h|J ∩ I| ≤ |I ∩ R(I, h)|.

Lemma 4. Let I, J ∈ \overline{B}_{2}. If J ∩ I ≠ ∅ and J\R(I, h) ≠ ∅, then either | pr_{1} J|_{1} ≥ | pr_{1} I|_{1}, or | pr_{2} J|_{1} ≥ | pr_{2} I|_{1}.

The following lemma is also valid.

Lemma 5. Let B ⊂ B_{2}; h > 1, and let I_{1}, …, I_{k} ∈ \overline{B}_{2} be the equal intervals. If

\{M_{B}(hχ_{Im}) > 1\} ⊂ R(I_{m}, h) (m = \overline{1, k}),
R(I_{m}, h) ∩ R(I_{m′}, h) = ∅ (m ≠ m′),

then

{M_{B}( ∑_{m=1}^{k} hχ_{Im} ) > 1} ⊂ ⋃^{k} ; m=1R(I_{m}, h).  (2)

Proof. Inequality (2) is equivalent to the following inequality: if

x ∉ ⋃^{k} ; m=1R(I_{m}, h) and I ∈ B(x), then

\frac{1}{|I|} ∫_{I} ∑_{m=1}^{k} hχ_{Im} ≤ 1.  (3)

Let x ∉ ⋃^{k} ; m=1R(I_{m}, h) and I ∈ B(x). Then we may have three cases. Let

us consider each of them separately.

(i) I intersects none of the intervals I_{m} (m = \overline{1, k}); in this case the validity of (3) is evident.

(ii) I intersects only one interval I_{m} (m = \overline{1, k}).

Let I_{n} be the interval for which I ∩ I_{n} ≠ ∅. x ∉ R(I_{m}, h), and therefore by the condition of the lemma,

\frac{1}{|I|} ∫_{I} hχ_{In} ≤ M_{B}(hχ_{In})(x) ≤ 1,

from which, taking into account the equality I ∩ I_{m} = ∅ (m ≠ n), we obtain (3).

(iii) I intersects more than one interval I_{m} (m = \overline{1, k}).

It follows from the equality R(I_{m}, h) ∩ R(I_{m′}, h) = ∅ (m ≠ m′) that

(2h + 1)I_{m} ∩ I_{m′} = ∅  (4)

Denote P = {m ∈ [1, k] : I ∩ I_{m} ≠ ∅}. By (4) we have

I ∩ I_{m} ≠ ∅, I\(2h + 1)I_{m} ≠ ∅ (m ∈ P),

which gives either

| pr_{1} I|_{1} ≥ h| pr_{1} I_{m}|_{1} or | pr_{2} I|_{1} ≥ h| pr_{2} I_{m}|_{1} (m ∈ P).

Now according to Lemma 3 we can conclude that

h|I ∩ I_{m}| ≤ |I ∩ R(I_{m}, h)| (m ∈ P).

From the obtained inequality, taking into consideration the pairwise nonintersection of R(I_{m}, h) (m = \overline{1, k}), we write

∫_{I} ∑_{m=1}^{k} hχ_{Im} = ∑_{m∈P} ∫_{I} hχ_{Im} = ∑_{m∈P} h|I ∩ I_{m}| ≤ ∑_{m∈P} |I ∩ R(I_{m}, h)| ≤ |I|.

Thus inequality (3) and Lemma 5 are proved.

Lemma 6. Let B ⊂ B_{2}, h > 1, and for every m ∈ [1, k] let {I_{m,q}}_{q=1}^{q_{m}} ⊂ \overline{B}_{2} be a family of equal intervals. If

\begin{matrix}\{M_{B}(hχ_{Im,q}) > 1\} ⊂ R_{m,q} \\ (R_{m,q} = R(I_{m,q}, h); m = \overline{1, k}) q = \overline{1, q_{m}}), \end{matrix}  (5)
R_{m,q} ∩ R_{m′,q′} = ∅ (m, q) ≠ (m, q′),  (6)
\begin{matrix}| pr_{1} I_{m,1}|_{1} ≥ h| pr_{1} I_{m+1,1}|_{1}, \\ | pr_{2} I_{m,1}|_{1} ≥ h| pr_{2} I_{m+1,1}|_{1}, \end{matrix} (m = \overline{1, k − 1})  (7)

then

{M_{B}( ∑_{m=1}^{k} ∑_{q=1}^{q_{m}} hχ_{Im,q} ) > 2} ⊂ ⋃^{k} ; m=1 ⋃^{q_{m}} ; q=1R_{m,q}.

Proof. The inclusion we have to prove is equivalent to the following inequal-

ity: if x ∉ ⋃^{k} ; m=1 ⋃^{q_{m}} ; q=1R_{m,q} and I ∈ B(x), then

\frac{1}{|I|} ∫_{I} ∑_{m=1}^{k} ∑_{q=1}^{q_{m}} hχ_{Im,q} ≤ 2.  (8)

Assume x ∉ ⋃^{k} ; m=1 ⋃^{q_{m}} ; q=1R_{m,q} and I ∈ B(x). It is clear that (8) is fulfilled for

I ∩ I_{m,q} = ∅ (m = \overline{1, k}, q = \overline{1, q_{m}}). Denote otherwise n = min\{m ∈ [1, k] : ∃I_{m,q}(q = \overline{1, q_{m}}), I ∩ I_{m,q} ≠ ∅\} and consider first the case with 1 < n < k.

By virtue of Lemma 5 (see (5) and (6)), we write

M_{B}( ∑_{q=1}^{q_{n}} hχ_{In,q} )(x) ≤ 1,

which implies

∫_{I} ∑_{q=1}^{q_{n}} hχ_{In,q} = ∑_{q=1}^{q_{n}} h|I ∩ I_{n,q}| ≤ |I|.  (9)

For some q ∈ [1, q_{n}] I ∩ I_{n,q} ≠ ∅, it is clear that I\R_{n,q} ≠ ∅. Therefore, according to Lemma 4, we have either

| pr_{1} I|_{1} ≥ | pr_{1} I_{n,q}|_{1} or | pr_{2} I|_{1} ≥ | pr_{2} I_{n,q}|_{1}.

Taking into account that the intervals \{I_{m,q}\}_{q=1}^{q_{m}} (m = \overline{1, k}) are equal and using (7), we can write that either

| pr_{1} I|_{1} ≥ h| pr_{1} I_{m,q}|_{1} (n < m ≤ k, q = \overline{1, q_{m}})

or

| pr_{2} I|_{1} ≥ h| pr_{2} I_{m,q}|_{1} (n < m ≤ k, q = \overline{1, q_{m}}),

whence by Lemma 3 we get

h|I ∩ I_{m,q}| ≤ |I ∩ R_{m,q}| (n < m ≤ k; q = \overline{1, q_{m}}).  (10)

Clearly, I ∩ I_{m,q} = ∅ (n ≤ m < n; q = \overline{1, q_{m}}), so that by (6), (9), (10) and 1 ≤ m < n we have

∫_{I} ∑_{m=1}^{k} ∑_{q=1}^{q_{m}} hχ_{Im,q} = ∑_{m=1}^{k} ∑_{q=1}^{q_{m}} h|I ∩ I_{m,q}| = ∑_{m=1}^{n−1} ∑_{q=1}^{q_{m}} h|I ∩ I_{m,q}| +

+ ∑_{q=1}^{q_{n}} h|I ∩ I_{n,q}| + ∑_{m=n+1}^{k} ∑_{q=1}^{q_{m}} h|I ∩ I_{m,q}| =

= A_{1} + A_{2} + A_{3} ≤ 0 + |I| + ∑_{m=n+1}^{k} ∑_{q=1}^{q_{m}} |I ∩ R_{m,q}| ≤ |I| + |I| = 2|I|.

Inequality (8) can be proved in a simpler way for n = 1 or n = k, since in these cases we do not have the terms A_{1} and A_{3}.

For the basis B let us define the operator

M_{B}^{∗}(f)(x) = sup_{R∈B(x)} \frac{1}{|R|}| ∫_{R} f| (f ∈ L_{loc}(ℝ^{n}), x ∈ ℝ^{n}).

Lemma 7. Let the basis B differentiate the integrals of the functions f_{k} ∈ L(ℝ^{n}) (k ∈ ℕ), ∑_{k=1}^{∞} ∥f_{k}∥_{1} < ∞. If

∑_{k=1}^{∞} |{M_{B}^{∗}(f_{k}) > λ_{k}}|_{e} < ∞,

where λ_{k} > 0 (k ∈ ℕ), ∑_{k=1}^{∞} λ_{k} < ∞ and | · |_{e} is an outer measure, then B differentiates the integral of the function ∑_{k=1}^{∞} f_{k}.

Note that since M_{B}^{∗}(f) ≤ M_{B}(f) (f ∈ L(ℝ^{n})), the conclusion of Lemma 7 will be the more so valid when the inequality ∑_{k=1}^{∞} |{M_{B}(f_{k}) > λ_{k}}|_{e} < ∞ is fulfilled.

Proof. Let us note that:

(i) ∑_{k=1}^{∞} ∥f_{k}∥_{1} < ∞; therefore the set A_{0} = {x ∈ ℝ^{n} : ∑_{k=1}^{∞} |f_{k}(x)| < ∞} is of measure 0;

(ii) B differentiates ∫ f_{k} (k ∈ ℕ), i.e., the sets A_{k} ={x∈ℝ^{n}:D_{B}(∫ f, x)= f(x)} (k ∈ ℕ) have a complete measure on ℝ^{n}.

(iii) ∑_{k=1}^{∞} |\{M_{B}^{∗}(f_{k}) > λ_{k}\}|_{e} < ∞; therefore the set \overline{lim}_{k→∞}\{M_{B}^{∗}(f_{k}) > λ_{k}} is of measure 0.

It follows from (i)–(iii) that the set A = ∞ ; ⋃_{k=1}A_{k}\(\overline{lim}_{k→∞}\{M_{B}^{∗}(f_{k}) >

λ_{k}} ⋃ A_{0}) is of complete measure. Let us show that for every x ∈ A we

have D_{B}( ∫ ∑_{k=1}^{∞} f_{k}, x) = ∑_{k=1}^{∞} f_{k}(x), which will prove the lemma. Let x ∈ A and ε > 0. From the inclusion x ∈ A we can conclude that:

1) there is k_{0} ∈ ℕ (k_{0} = k(x, ε)) such that x ∉ ∞ ; ⋂_{k=k_{0}}{M_{B}^{∗}(f_{k}) > λ_{k}},

∑_{k=k_{0}}^{∞} λ_{k} < ε/3 and ∑_{k=k_{0}}^{∞} |f_{k}(x)| < ε/3; 2) there is δ > 0 such that for every R ∈ B(x), diam R < δ

|\frac{1}{|R|} ∫_{R} ∑_{k=1}^{k_{0}−1} f_{k} − ∑_{k=1}^{k_{0}−1} f_{k}(x)| < ε/3.

According to 1) and 2) we can write that for every R ∈ B(x), diam R < δ,

|\frac{1}{|R|} ∫_{R} ∑_{k=1}^{∞} f_{k} − ∑_{k=1}^{∞} f_{k}(x)| ≤ |\frac{1}{|R|} ∫_{R} ∑_{k=1}^{k_{0}−1} f_{k} − ∑_{k=1}^{k_{0}−1} f_{k}(x)| +

+ ∑_{k=k_{0}}^{∞} \frac{1}{|R|}| ∫_{R} f_{k}| + ∑_{k=k_{0}}^{∞} |f_{k}(x)| < ε/3 + ∑_{k=k_{0}}^{∞} M_{B}^{∗}(f_{k})(x) +

+ε/3 ≤ 2ε/3 + ∑_{k=k_{0}}^{∞} λ_{k} < ε.

Since ε > 0 is arbitrary, we conclude that the equality D_{B}( ∫ ∑_{k=1}^{∞} f_{k}, x) = ∑_{k=1}^{∞} f_{k}(x) is valid.

Proof of Theorem 2. Let k ≥ 3. Choose h_{k} such that \frac{1}{k} ln \frac{h_{k}}{k} ≥ 18 · 2^{2k}. Let α_{k} = \frac{1}{k} ln \frac{h_{k}}{k} /2^{2(k+2)}h_{k}. Obviously, 0 < α_{k} < 1. For I ∈ \overline{B}_{2} and h > 0 denote Q^{0}(I, h) = (2h + 1)I. Let 2^{−m} <

| pr_{1} Q^{0}(I, h)|_{1} ≤ 2^{−m+1} and 2^{−m′} < | pr_{2} Q^{0}(I, h)|_{1} ≤ 2^{−m′+1}, where m, m′ ∈ ℕ. Denote by Q(I, h) the interval concentric with Q^{0}(I, h), and | pr_{1} Q(I, h)|_{1} = 2^{−m+1}, | pr_{2} Q(I, h)|_{1} = 2^{−m′+1}.

For the basis B denote by M_{B}^{(r)} (r > 0) the operator

M_{B}^{(r)}(f)(x) = sup_{R∈B(x),diam R<r} \frac{1}{|R|} ∫_{R} |f| (f ∈ L(ℝ^{n}), x ∈ ℝ^{n}).

\overline{B}_{2} ∉ LR(\overline{B}), since there is I ∈ \overline{B}_{2} such that:

1) there is no J ∈ \overline{B}, J ⊃ I for which |J| < 2^{k}h_{k}|I|;

2) diam Q(I, 2^{k}h_{k}) < 1/k.

Divide I^{2} into intervals equal to Q(I, 2^{k}h_{k}) and denote them by Q_{1,q} (1 ≤ q ≤ q_{1}). Take I_{1,q} (1 ≤ q ≤ q_{1}) equal to T_{q}(I), where T_{q} is a shift translating Q(I, 2^{k}h_{k}) to Q_{1,q}.

Let the families {I_{1,q}}_{q=1}^{q_{1}} … {I_{m,q}}_{q=1}^{q_{m}} consisting of equal intervals be already constructed. Consider the sets

A_{m}^{1} = ⋃^{m} ; j=1 ⋃^{q_{j}} ; q=1{M_{B_{2}}^{(1/k)}(h_{k}, χ_{I_{j,q}}) > k},

A_{m}^{2} = ⋃^{m} ; j=1 ⋃^{q_{j}} ; q=1R_{j,q} (R_{j,q} = R(I_{j,q}, 2^{k}h_{k})).

If |A_{m}^{1}| > 1 − \frac{1}{k}, then we stop the construction. If |A_{m}^{1}| ≤ 1 − \frac{1}{k}, then

we shall construct the family {I_{m+1,q}}_{q=1}^{q_{m+1}} as follows:

Consider the set A_{m} = I^{2} = I^{2}\(A_{m}^{1} ∪ A_{m}^{2}) which, obviously, can be represented as G_{1} ∪G_{2}, where G_{1} is open and G_{2} consists of a finite number of smooth closed lines. It is clear that there is δ ∈ (0, 1/k) such that if we divide I^{2} into equal intervals {J_{j}} with diameters less than δ, then

|A_{m} ∩ ⋃_{J_{j}⊂A_{m}}J_{j}| ≥ (1 − \frac{α_{k}}{4} )|A_{m}|.

\overline{B}_{2} ∉ LR(\overline{B}), since there is I ∈ \overline{B}_{2} such that:

1) there is no J ∈ \overline{B}, J ⊃ I for which |J| ≤ 2^{k}h_{k}|I|;

2) diam Q(I, 2^{k}h_{k}) < δ;

3) | pr_{1} I_{m,1}|_{1} ≥ 2^{k}h_{k}| pr_{1} I|_{1} and | pr_{2} I_{m,1}|_{1} ≥ 2^{k}h_{k}| pr_{2} I|_{1}. Divide I^{2} into intervals equal to Q(I, 2^{k}h_{k}). Denote the intervals included

in A_{m} by Q_{m+1,q} (1 ≤ q ≤ q_{m+1}). Take I_{m+1,q} (1 ≤ q ≤ q_{m+1}) equal to T_{q}(I), where T_{q} is a shift translating Q(I, 2^{k}h_{k}) to Q_{m+1,q}.

By our construction we obtain

|A_{m} ∩ ⋃^{q_{m+1}} ; q=1Q_{m+1,q}| ≥ (1 − \frac{α_{k}}{4} )|A_{m}|.  (11)

By Lemma 1,- for q = \overline{1, q_{m+1}}

\{M_{B_{2}}(h_{k}χ_{I_{m+1,q}}) > k\} ⊂ (2\frac{h_{k}}{k} − 1)I_{m+1,q} ⊂ Q_{m+1,q},  (12)
|\{M_{B_{2}}(h_{k}χ_{I_{m+1,q}}) > k\}| > \frac{h_{k}}{k} ln \frac{h_{k}}{k} |I_{m+1,q}|.  (13)

Since diam Q_{m+1,q} < 1/k (q = \overline{1, q_{m+1}}), because of (12) we have

\{M_{B_{2}}(h_{k}χ_{I_{m+1,q}}) > k\} = \{M_{B_{2}}^{(1/k)}(h_{k}χ_{I_{m+1,q}}) > k\} (q = \overline{1, q_{m+1}}). (14)

ON DIFFERENTIAL BASES FORMED OF INTERVALS 89

It can be easily seen that (see (13), (14))

|\{M_{B_{2}}^{(1/k)}(h_{k}χ_{I_{m+1,q}}) > k\}| ≥ α_{k}|Q_{m+1,q}| (q = \overline{1, q_{m+1}}),

which by (11) readily implies

| ⋃^{q_{m+1}} ; q=1{M_{B_{2}}^{(1/k)}(h_{k}χ_{I_{m+1,q}}) > k}| ≥ α_{k}| ⋃^{q_{m+1}} ; q=1Q_{m+1,q}| ≥

≥ α_{k}(1 − \frac{α_{k}}{4} )|A_{m}| > \frac{α_{k}}{2} |A_{m}|.  (15)

According to the construction and Lemma 2 we have for q = \overline{1, q_{m+1}}

{M_{B}(h_{k}χ_{I_{m+1,q}}) > 1/2^{k}} ⊂ R_{m+1,q} (R_{m+1,q} = R(I_{m+1,q}, 2^{k}h_{k})). On account of our choice of h_{k}, because of (13) and (14) we write

|\{M_{B_{2}}^{(1/k)}(h_{k}χ_{I_{m+1,q}}) > k\}| > 2^{k}|R_{m+1,q}| (q = \overline{1, q_{m+1}}),

whence

| ⋃^{q_{m+1}} ; q=1{M_{B_{2}}^{(1/k)}(h_{k}χ_{I_{m+1,q}}) > k}| > 2^{k}| ⋃^{q_{m+1}} ; q=1R_{m+1,q}|.  (16)

Let us show that for sufficiently large mk the construction ceases, i.e., we shall have the inequality

| ⋃^{m_{k}} ; m=1 ⋃^{q_{m}} ; q=1\{M_{B_{2}}^{(1/k)}(h_{k}χ_{I_{m+1,q}}) > k\}| > 1 − \frac{1}{k} .  (17)

Assume the contrary, i.e., |A_{m}^{1}| ≤ 1 − \frac{1}{k} (m ∈ ℕ). Introduce the notation

\begin{matrix}A_{m}^{3} = ⋃^{q_{m}} ; q=1\{M_{B_{2}}^{(1/k)}(h_{k}χ_{Im,q}) > k\}, \\ A_{m}^{4} = ⋃^{q_{m}} ; q=1R_{m,q}. \end{matrix} (m ∈ ℕ)

Clearly, (15) and (16) will be valid for all m ∈ ℕ, and therefore we write

|A_{m+1}^{3}| > \frac{α_{k}}{2} |A_{m}| = \frac{α_{k}}{2} (1 − ∑_{j=1}^{m} |A_{j}^{3} ∪ A_{j}^{4}|) ≥

≥ \frac{α_{k}}{2} (1 − ∑_{j=1}^{m} (1 + \frac{1}{2^{k}} )|A_{j}^{3}|) ≥ \frac{α_{k}}{2} (1 − \frac{1}{2^{k}} − ∑_{j=1}^{m} |A_{j}^{3}|),

which by the equality |A_{m}^{1}| = ∑_{j=1}^{m} |A_{m}^{3}| implies that |A_{m}^{1}| > 1 − \frac{1}{2^{k−1}} >

1 − \frac{1}{k} for sufficiently large m. From this contradiction we conclude that (17) is valid.

Consider the function f_{k} = ∑_{m=1}^{m_{k}} ∑_{q=1}^{q_{m}} h_{k}χ_{Im,q}. Clearly, f_{k} ∈ L(I^{2}) and f_{k} ≥ 0. From (17) we get

|\{M_{B_{2}}^{(1/k)}(f_{k}) > k\}| > 1 − \frac{1}{k} .  (18)

From the construction we can easily see that 2^{k}h_{k} and the families {I_{1,q}}_{q=1}^{q_{1}}, …, {I_{m_{k},q}}_{q=1}^{q_{m_{k}}} satisfy all the conditions of Lemma 6. Therefore by Lemma 6 we write

{M_{B}(f_{k}) > 1/2^{k−1}} = {M_{B}(2^{k}f_{k}) > 2} =
= {M_{B}( ∑_{m=1}^{m_{k}} ∑_{q=1}^{q_{m}} 2^{k}h_{k}χ_{Im,q} ) > 2} ⊂ ⋃^{m_{k}} ; m=1 ⋃^{q_{m}} ; q=1R_{m,q}.  (19)

Obviously, (16) is fulfilled for all m ∈ [1, m_{k} − 1], so that taking the construction into account, we have

2^{k}| ⋃^{m_{k}} ; m=1 ⋃^{q_{m}} ; q=1R_{m,q}| < | ⋃^{m_{k}} ; m=1 ⋃^{q_{m}} ; q=1{M_{B_{2}}^{(1/k)}(h_{k}χ_{Im,q}) > k}| ≤ 1.  (20)

Due to (19) and (20) we write

|\{M_{B}(f_{k}) > 1/2^{k−1}\}| < \frac{1}{2^{k}} .  (21)

It can be easily seen that

\frac{h_{k}}{k} ln \frac{h_{k}}{k} | ⋃^{m_{k}} ; m=1 ⋃^{q_{m}} ; q=1I_{m,q}| < | ⋃^{m_{k}} ; m=1 ⋃^{q_{m}} ; q=1\{M_{B_{2}}^{(1/k)}(h_{k}χ_{Im,q}) > k\}| ≤ 1,

Hence, due to our choice of h_{k}, we obtain

∥f_{k}∥_{1} = h_{k}| ⋃^{m_{k}} ; m=1 ⋃^{q_{m}} ; q=1I_{m,q}| < 1/\frac{1}{k} ln \frac{h_{k}}{k} < \frac{1}{2^{k}} .  (22)

Consider the function f = ∑_{k=3}^{∞} f_{k}. Clearly, f ∈ L(I^{2}) (see (22)) and f ≥ 0.

It is obvious that if x ∈ \overline{lim_{k→∞}}\{M_{B_{2}}^{(1/k)}(f_{k}) > k\}, then \overline{D}_{B_{2}}(∫ ∑_{k=3}^{∞} f_{k}, x)=

∞. Because of (18) \overline{lim_{k→∞}}\{M_{B_{2}}^{(1/k)}(f_{k}) > k\} has a complete measure on I^{2}.

Therefore the latter equality holds almost everywhere on I^{2}.

It is clear that B differentiates ∫ f_{k} (k ∈ ℕ). Moreover, owing to (21) we get ∑_{k=3}^{∞} |\{M_{B}(f_{k}) > 1/2^{k−1}\}| < ∑_{k=3}^{∞} \frac{1}{2^{k}} < ∞, which by Lemma 7 implies that B differentiates ∫ f.

After making some technical changes in the proof of Theorem 2 we can obtain the following generalization.

Theorem 3. Let B ⊂ B_{2} be a T I-basis. If \overline{B}_{2} is not locally regular with respect to \overline{B} then for any function f ∈ L\L ln^{+} L(I^{2}), f ≥ 0, there is a Lebesgue measure-preserving and invertible mapping ω : ℝ^{2} → ℝ^{2}, {x : ω(x) ≠ x} ⊂ I^{2} such that

\overline{D}_{B_{2}}( ∫ f ∘ ω, x) = ∞ a.e. on I^{2},
D_{B}( ∫ f ∘ ω, x) = (f ∘ ω)(x) a.e. on I^{2}.

To prove the implication (c) ⇒ (b) let us show the validity of

Lemma 8. Let B ⊂ B_{2} be a T I-basis. If \overline{B}_{2} is locally regular with respect to \overline{B}, then for every f ∈ L(ℝ^{2})

|\{M_{B_{2}}^{(r)}(f) > λ\}| ≤ 25|\{M_{B}^{(cr)}(f) > \frac{λ}{4c}\}| (λ > 0, 0 < r < δ),

where δ and c are the constants from the definition of local regularity of \overline{B}_{2} with respect to \overline{B}.

Proof. Note first that if I and I′ (I ⊂ I′) are the one-dimensional intervals and I^{∗} is either the left or the right half of the interval I, then for the point y ∈ I′ the set E = {z ∈ ℝ : (I′ + z − y) ⊃ I^{∗}} possesses the following properties:

1) E is an interval;

2) E ∩ I′ ≠ ∅;

3) |E|_{1} = |I′|_{1} − \frac{1}{2}|I|_{1} ≥ \frac{1}{2}|I′|_{1}.

It follows from these properties that

4) 5E ⊃ I′ is likewise valid.

Let f ∈ L(ℝ^{2}). Denote by P the set of all two-dimensional intervals

E ∈ \overline{B}_{2} such that E ⊂ \{M_{B}^{(cr)}(f) > \frac{λ}{4c}\} and show the inclusion

{M_{B_{2}}^{(r)}(f) > λ} ⊂ ⋃_{E∈P}5E (λ > 0, 0 < r < δ).  (23)

Let x ∈ {M_{B_{2}}^{(r)}(f) > λ} (λ > 0, 0 < r < δ). Then there is I ∈ B_{2}(x) such that diam I < r and

\frac{1}{|I|} ∫_{I} |f| > λ.  (24)

Since \overline{B}_{2} ∈ LR(\overline{B}), there is an interval I′ from \overline{B} such that I ⊂ I′ and |I′| ≤ c|I|. Then it is obvious that diam I′ ≤ c diam I < cr. Let y ∈ I′ be the point for which I′ ∈ B(y). Divide the interval I into four equal intervals. Because of (24) the integral mean will be greater than λ at least on one of these intervals. Denote this interval by I^{∗}. Thus

\frac{1}{|I^{∗}|} ∫_{I^{∗}} |f| > λ.  (25)

Consider the set E = {z ∈ ℝ^{2} : (I′ +z −y) ⊃ I^{∗}}. By virtue of the above discussion we can easily note that

1) E ∈ \overline{B}_{2};

2) E ∩ I′ ≠ ∅;

3) |E| ≥ \frac{1}{4}|I′|;

4) 5E ⊃ I′.

Since x ∈ I ⊂ I′, then to prove (23) it suffices to show that E ∈ P. Let z ∈ E. Then (I′ + z − y) ⊃ I^{∗}. Since B is the T I-basis and I′ ∈ B(y), we have (I′ + z − y) ∈ B(z). Moreover, owing to (25), we have

\frac{1}{|I′ + z − y|} ∫_{I′+z−y} |f| ≥ \frac{1}{|I′|} ∫_{I^{∗}} |f| ≥ \frac{1}{c|I|} ∫_{I^{∗}} |f| = \frac{1}{c4|I^{∗}|} ∫_{I^{∗}} |f| > \frac{λ}{4c},

from which by the inequality diam(I′ + z − y) < cr, we conclude that

z ∈ \{M_{B}^{(cr)}(f) > \frac{λ}{4c}\}. Hence E ∈ P. Thus inclusion (23) is proved. Now, using (23) and Lemma 1 from [2], we will have

|\{M_{B_{2}}^{(r)}(f) > λ\}| ≤ | ⋃_{E∈P}5E| ≤ 25| ⋃_{E∈P}E| ≤ 25|\{M_{B}^{(cr)}(f) > \frac{λ}{4c}\}|

for λ > 0, 0 < r < δ.

The basis B is said to be regular if there is a constant c < ∞ such that for every R ∈ \overline{B} there exists a cubic interval Q with the following properties: Q ⊃ R, |Q| ≤ c|R|.

According to the well-known Lebesgue theorem on differentiability of integrals, the regular basis B differentiates L, i.e., D_{B}( ∫ f, x) = f(x) a.e. on ℝ^{n} for every f ∈ L(ℝ^{n}).

We shall say that the basis B possesses property (E) if for every f ∈ L(ℝ^{n}), \underline{D}_{B}( ∫ f, x) ≤ f(x) ≤ \overline{D}_{B}( ∫ f, x) a.e. on ℝ^{n}.

Obviously, the basis B, containing a regular subbasis, possesses the property (E).

It directly follows from the relation \overline{B}_{2} ∈ LR(\overline{B}) that B contains a regular subbasis, and hence possesses property (E). Now, to prove the implication (b) ⇒ (c), it suffices to consider the following assertion.

Lemma 9. Let B be a subbasis of B_{2} with property (E). Suppose that there exist positive constants c_{1}, c_{2}, c_{3} and δ such that for every f ∈ L(ℝ^{2}),

|\{M_{B_{2}}^{(r)}(f) > λ\}| ≤ c_{1}|\{M_{B}^{(c_{2}r)}(f) > \frac{λ}{c_{3}} \}| (λ > 0, 0 < r < δ).

Then for every f ∈ L(ℝ^{2}), f ≥ 0,

\underline{D}_{B}( ∫ f, x) = \underline{D}_{B_{2}}( ∫ f, x) and \overline{D}_{B}( ∫ f, x) = \overline{D}_{B_{2}}( ∫ f, x),

a.e. on ℝ^{2}.

The proof of Lemma 9 is based on the well-known Besicovitch theorem on possible values of upper and lower derivative numbers (see [1], Ch.IV, §3) and is carried out analogously to the proof of Theorem 1 from [2].

3. On a Possibility of Extending Theorem 1 to More General Bases

It should be noted that beyond the scope of T I-bases Theorem 1 becomes invalid. Moreover, even for bases which are similar enough to T I-bases the local regularity of \overline{B}_{2} with respect to \overline{B} is insufficient for the equality F_{B}^{+} = F_{B_{2}}^{+} to be fulfilled.

A basis B will be called a T I^{∗}-basis if for every x ∈ ℝ^{n}, R ∈ B(x) and y ∈ ℝ^{n} there is a translation T such that T (R) ∈ B(y).

The following theorem is valid.

Theorem 4. There is a T I^{∗}-basis B ⊂ B_{2} with the properties:

1) \overline{B}_{2} is locally regular with respect to \overline{B}, and B(O) = B_{2}(O);

2) there is a function f ∈ L(I^{2}), f ≥ 0 such that

\overline{D}_{B_{2}}( ∫ f, x) = ∞ a.e. on I^{2},
D_{B}( ∫ f, x) = f(x) a.e. on I^{2}.

Proof. Consider the sequences α_{k} ↑ ∞, α_{k} > 0 and h_{k} ↑ ∞, h_{k} > 10 with the properties

∑_{k=1}^{∞} \frac{α_{k}}{h_{k}} < ∞, ∑_{k=1}^{∞} \frac{ln h_{k}}{h_{k}} = ∞.  (26)

Let q_{k} = 2^{m_{k}}, where m_{k} ∈ ℕ (k ∈ ℕ) and m_{k} ↑ ∞. For every k, we divide I^{2} into q_{k}^{2} equal square interval and denote them by I_{k,q}^{2}. The length of square intervals sides will be denoted by Δ_{k}. For every I_{k,q}^{2} let us consider the square interval I_{k,q} concentric with I_{k,q}^{2} having the side length δ_{k} = Δ_{k}/(2h_{k} + 1). We shall assume q_{k} ↑ ∞ so that δ_{k} > Δ_{k+1} (k ∈ ℕ).

Let f_{k} = sup\{α_{k}h_{k}χ_{I_{k,q}} : q = \overline{1, q_{k}^{2}}\} (k ∈ ℕ) and f = ∑_{k=1}^{∞} f_{k}. Clearly, f ≥ 0. It can be easily checked that

∥f_{k}∥_{1} = α_{k}h_{k}| q_{k}^{2} ; ⋃_{q=1}I_{k,q}| = α_{k}h_{k} ∑_{q=1}^{q_{k}^{2}} \frac{1}{(2h_{k} + 1)^{2}} |I_{k,q}^{2}| < \frac{α_{k}}{h_{k}} ,

which according to (26) implies

∥f∥_{1} ≤ ∑_{k=1}^{∞} ∥f_{k}∥_{1} < ∑_{k=1}^{∞} \frac{α_{k}}{h_{k}} < ∞,

i.e., f ∈ L(I^{2}).

By Lemma 1 for k ∈ ℕ, q = \overline{1, q_{k}^{2}} we have

{M_{B_{2}}(α_{k}h_{k}χ_{I_{k,q}}) > α_{k}} ⊂ (2h_{k} + 1)I_{k,q} = I_{k,q}^{2},  (27)

and

|{M_{B_{2}}(α_{k}h_{k}χ_{I_{k,q}}) > α_{k}}| ≥ h_{k} ln h_{k}|I_{k,q}|.

Thus we write

| ⋃^{q_{k}^{2}} ; q=1{M_{B_{2}}(α_{k}h_{k}χ_{I_{k,q}}) > α_{k}}| = ∑_{q=1}^{q_{k}^{2}} |{M_{B_{2}}(α_{k}h_{k}χ_{I_{k,q}}) > α_{k}}| ≥

≥ ∑_{q=1}^{q_{k}^{2}} h_{k} ln h_{k}|I_{k,q}| = ∑_{q=1}^{q_{k}^{2}} h_{k} ln h_{k} \frac{|I_{k,q}^{2}|}{(2h_{k} + 1)^{2}} > \frac{ln h_{k}}{9h_{k}}  (28)

for k ∈ ℕ.

For every k ∈ ℕ let us consider those intervals I_{k+1,q}^{2} (q = \overline{1, q_{k+1}^{2}}) which

are contained in ⋃^{q_{k}^{2}} ; q=1{M_{B_{2}}(α_{k}h_{k}χ_{I_{k,q}}) > α_{k}}. Denote their union by O_{k}. It

can be easily seen that if {q_{k}} tends rapidly enough to ∞, then

|O_{k}| ≥ \frac{1}{2}| ⋃^{q_{k}^{2}} ; q=1\{M_{B_{2}}(α_{k}h_{k}χ_{I_{k,q}}) > α_{k}\}| (k ∈ ℕ).  (29)

Using the property of dyadic intervals, we can see that {O_{k}} is the sequence of independent sets, i.e., for every n ≥ 2 and for arbitrary pairwise different natural numbers k_{1}, k_{2} …, k_{n},

| ⋂^{n} ; m=1O_{k_{m}} | = ∏_{m=1}^{n} |O_{k_{m}}|.

Owing to (26), (28), and (29), ∑_{k=1}^{∞} |O_{k}| > ∑_{k=1}^{∞} \frac{1}{2} \frac{ln h_{k}}{9h_{k}} = ∞. Using

now Borel–Kantelly’s lemma, we get

\overline{lim_{k→∞}}O_{k}, has a complete measure in I^{2}.  (30)

From (27) we have for k ∈ ℕ, q = \overline{1, q_{k}^{2}}

{M_{B_{2}}(α_{k}h_{k}χ_{I_{k,q}}) > α_{k}} = {M_{B_{2}}^{(Δ_{k})}(α_{k}h_{k}χ_{I_{k,q}}) > α_{k}},

which for k ∈ ℕ implies

O_{k} ⊂ ⋃^{q_{k}^{2}} ; q=1{M_{B_{2}}(α_{k}h_{k}χ_{I_{k,q}}) > α_{k}} =

= ⋃^{q_{k}^{2}} ; q=1{M_{B_{2}}^{(Δ_{k})}(α_{k}h_{k}χ_{I_{k,q}}) > α_{k}} ⊂ {M_{B_{2}}^{(Δ_{k})}(f_{k}) > α_{k}}.

Because of (30), from the latter inclusion we obtain

\overline{lim_{k→∞}}\{M_{B_{2}}^{(Δ_{k})}(f_{k}) > α_{k}\} has a complete measure in I^{2}.  (31)

Since α_{k} ↑ ∞, Δ_{k} ↓ 0 (k → ∞), we can easily see that if x ∈ \overline{lim_{k→∞}}\{M_{B_{2}}^{(Δ_{k})}(f_{k}) > α_{k}\}, then \overline{D}_{B_{2}}( ∫ f, x) = ∞. By virtue of (31) we

conclude that the latter equality holds a.e. on I^{2}.

Assume k to be fixed. For every I_{k,q} (q = \overline{1, q_{k}^{2}}) let us consider the intervals J_{k,q}^{1} (pr_{1} J_{k,q}^{1} = 3 pr_{1} I_{k,q}, pr_{2} J_{k,q}^{1} = (0, 1)) and J_{k,q}^{2} (pr_{1} J_{k,q}^{2} =

(0, 1), pr_{2} J_{k,q}^{2} = 3 pr_{2} I_{k,q}). Denote G_{k}^{i} = ⋃^{q_{k}^{2}} ; q=1J_{k,q}^{i} (i = \overline{1, 2}), G_{k} = G_{k}^{1} ∪ G_{k}^{2}.

It can be easily verified (see (26)) that ∑_{k=1}^{∞} |G_{k}| < ∞ which implies that \overline{lim_{k→∞}}G_{k} has zero measure.

Determine now an unknown basis B. If x ∈ (\overline{lim_{k→∞}}G_{k}) ∪ ℚ^{2} ∪ (ℝ^{2}\I^{2})

(ℚ is a set of rational numbers), then we put B(x) = B_{2}(x). For x ∈ I^{2}\(\overline{lim_{k→∞}}G_{k} ∪ ℚ^{2}) choose B(x) such that if d_{I} is the length of the lesser side of the interval I ∋ x, and d_{I} ∉ ⋃_{k≥k(x)}[δ_{k}, Δ_{k}/2] (for x ∉ \overline{lim_{k→∞}}G_{k}, k(x)

is the number with the property x ∉ G_{k} for k ≥ k(x)), then I ∈ B(x), while if d_{I} ∈ [δ_{k}, Δ_{k}/2] for some k ≥ k(x), then I ∈ B(x) iff either I ∩ G_{k}^{1} = ∅ or I ∩ G_{k}^{2} = ∅.

It can be easily verified that for every I ∈ \overline{B}_{2} and x ∈ ℝ^{2} there is a translation T, T (I) ∈ B(x); this implies that B is the T I^{∗}-basis. The local regularity of \overline{B}_{2} with respect to \overline{B} is also easily checked. By proving the inclusion f ∈ F_{B}^{+} the proof of the theorem will be completed.

Denote A_{k} = {x ∈ ℝ^{n} : D_{B}( ∫ f_{k}, x) = f_{k}(x)} (k ∈ ℕ). Clearly, A_{k} (k ∈

ℕ) has a complete measure on ℝ^{2}. Denote A = (I^{2} ∩ ⋂^{∞} ; k=1A_{k})\(\overline{lim_{k→∞}}G_{k} ∪

ℚ^{2}). Since f(x) = 0 for x ∉ I^{2}, B differentiates ∫ f on ℝ^{2}\I^{2}; A is the set

of a complete measure on I^{2}. Therefore to prove the inclusion f ∈ F_{B}^{+}, it suffices to show the differentiability of ∫ f with respect to B on the set A.

Let x ∈ A and ε > 0. Find δ > 0 for which |(1/|I|) ∫_{I} f − f(x)| < ε when I ∈ B(x), diam I < δ. Consider k′(x) ≥ k(x) for which

∑_{k≥k′(x)} 16∥f_{k}∥_{1} < ε/2.  (32)

It is clear that ∑_{k=1}^{∞} f_{k}(x) = ∑_{k=1}^{k′(x)−1} f_{k}(x), and B differentiates ∫ ∑_{k=1}^{k′(x)−1} f_{k} at the point x. This implies that there is δ ∈ (0, Δ_{k′(x)}/2) such that

|\frac{1}{|I|} ∫_{I} ∑_{k=1}^{k′(x)−1} f_{k} − ∑_{k=1}^{k′(x)−1} f_{k}(x)| < ε/2, for I ∈ B(x), diam I < δ. (33)

Consider an arbitrary interval I ∈ B(x), diam I < δ. For k_{I} ∈ ℕ let Δ_{K_{I}+1}/2 ≤ d_{I} ≤ Δ_{k_{I}} /2. Clearly, k_{I} ≥ k′(x). By virtue of the inequality Δ_{k+1} < δ_{k} (k ∈ ℕ) and the condition x ∉ G_{k} (k ≥ k(x)), we have I ∩ supp f_{k} = ∅ for k′(x) ≤ k ≤ k_{I} − 1, which gives

∫_{I} f_{k} = 0 for k′(x) ≤ k ≤ k_{I} − 1.  (34)

Analogously, ∫_{I} f_{k_{I}} = 0 if d_{I} < δ_{k_{I}}, while if δ_{k_{I}} ≤ d_{I} ≤ Δ_{k_{I}} /2, then ∫_{I} f_{k_{I}} = 0 because I ∩ supp f_{k} = ∅ by construction of the basis B. Thus

∫_{I} f_{k_{I}} = 0.  (35)

Let is show that

|I ∩ supp f_{k}| ≤ 16|I| | supp f_{k}| for k ≥ k_{I} + 1.  (36)

To this end it is sufficient to consider the following easily verifiable facts:

1) d_{I} ≥ Δ_{k}/2 for k ≥ k_{I} + 1;

2) let P_{k} be a set of vertices of the squares I_{k,q}^{2} (q = \overline{1, q_{k}^{2}}). If the vertices of the interval J ⊂ I^{2} belong to the set P_{k}, then |J ∩supp f_{k}| = |I|| supp f_{k}|;

3) for every interval J ⊂ I^{2}, d_{J} ≥ Δ_{k}/2 there is an interval J′ ⊃ J, |J′| ≤ 16|J| with vertices at the points of the set P_{k}.

By (36) we can write that for k ≥ k_{I} + 1,

∫_{I} f_{k} = α_{k}h_{k}|I ∩ supp f_{k}| ≤ 16α_{k}h_{k}|I| | supp f_{k}| = 16|I| ∥f_{k}∥_{1},

which, owing to (32) and the inequality k_{I} ≥ k′(x), implies

\frac{1}{|I|} ∫_{I} ∑_{k≥k_{I}+1} f_{k} = \frac{1}{|I|} ∑_{k≥k_{I}+1} ∫_{I} f_{k} ≤ \frac{1}{|I|} ∑_{k≥k_{I}+1} 16|I| ∥f_{k}∥_{1} < ε/2. (37)

By virtue of relations (33)–(35) and (37) (it should also be noted that f_{k}(x) = 0 for k ≥ k′(x)), we have

|\frac{1}{|I|} ∫_{I} ∑_{k=1}^{∞} f_{k} − ∑_{k=1}^{∞} f_{k}(x)| ≤ |\frac{1}{|I|} ∫_{I} ∑_{k=1}^{k′(x)−1} f_{k} − ∑_{k=1}^{k′(x)−1} f_{k}(x)| +

+\frac{1}{|I|} ∫_{I} ∑_{k=k′(x)}^{k_{I}} f_{k} + \frac{1}{|I|} ∫_{I} ∑_{k≥k_{I}+1} f_{k} + ∑_{k≥k′(x)} f_{k}(x) < \frac{ε}{2} + 0 + \frac{ε}{2} + 0 = ε.

Due to the arbitrariness of x ∈ A and ε > 0, we conclude that D_{B}( ∫ f, x) = f(x) for every x ∈ A.

4. Remarks

(1) Let us consider one application of the obtained results. Let B_{Z} be the basis in ℝ^{2} for which B_{Z}(x) (x ∈ ℝ^{2}) consists of all intervals I ∋ x, D_{I}^{2} ≤ d_{I} ≤ D_{I} ≤ 1, where D_{I} and d_{I} are the lengths, respectively, of the greater and of the lesser side of I. This basis was introduced by Zygmund and it was he who initiated (see [1], Ch. VI, §4) the study of the differential properties of B_{Z}.

Morion showed (see [1], Appendix IV) that for the integral classes B_{Z} behaves like B_{2}, i.e., B_{Z} does not differentiate a wider integral class than L ln^{+} L.

A question arises: does there exist in general a function whose integral does not differentiate B_{2} and differentiates B_{Z}? From the above proven theorems the answer is positive.

B_{Z} is the T I-basis. The fact that \overline{B}_{2} ∉ LR(\overline{B}_{z}) is easily verified. By Theorem 1 this implies that the strict inclusion F_{B_{2}}^{+} ⊂ F_{B_{Z}}^{+} holds. Moreover, Theorem 3 implies that by perturbing the values ω of any function f ∈ L\L ln^{+} L(I^{2}), f ≥ 0, one can get the function f ∘ ω with the following properties: \overline{D}_{B_{2}}( ∫ f ∘ ω, x) = ∞ a.e. on I^{2}, and B_{Z} differentiates ∫ f ∘ ω.

Thus, for the integral classes the bases B_{2} and B_{Z} behave similarly, while for individual nonnegative functions the basis B_{Z} behaves better than the basis B_{2}.

(2) The bases B and B′ are said to be positive equivalent (B ⇔^{+} B′) if for every f ∈ L(ℝ^{n}), f ≥ 0, \overline{D}_{B}( ∫ f, x) = \overline{D}_{B′}( ∫ f, x) and \underline{D}_{B}( ∫ f, x) = \underline{D}_{B′}( ∫ f, x) a.e. on ℝ^{n} (i.e., condition (b) in Theorem 1 means that the bases B and B′ are positive equivalent).

B is said to be a Busemann–Feller type basis (BF -basis) if for any R ∈ \overline{B} and x ∈ R we have R ∈ B(x).

We shall say that B exactly differentiates φ(L) (writing B ∈ D(φ(L)) if B differentiates φ(L) and does not differentiate a wider integral class than φ(L).

For integral classes the behavior of B ⊂ B_{2}, BF, T I-bases was studied by Stokoloc in [3], where he introduced the property (S) and proved that if B possesses the property (S), then B ∈ D(L ln^{+} L), and if B does not possess this property, then B ∈ D(L).

One can easily see that ignoring the BF property, this result remains valid. Thus, for the integral classes B ⊂ B_{2}, the T I-basis behaves like B_{2} or B_{1} (B_{1} is the basis formed of square intervals). The B_{Z}-basis illustrates that an analogous fact does not hold for nonnegative individual functions and, generally speaking, if we combine Stokoloc’s assertion and Theorem 1, then we shall have:

1) if \overline{B}_{2} ∈ LR(\overline{B}), then B ⇔^{+} B_{2};

2) if \overline{B}_{2} ∉ LR(\overline{B}) and B possesses the property (S), then B ∈ D(L ln^{+} L) and F_{B_{2}}^{+} ⊂ F_{B}^{+} (strictly);

3) if B does not possess the property (S), then B ∈ D(L).

(3) Let B ⊂ B_{2} be a T I-basis. Consider the intervals I ∈ \overline{B} of the type I = (0, x^{1}) × (0, x^{2}). Denote the set of points (x^{1}, x^{2}) by A_{B}. The set A_{B} indicates how rich the family \overline{B} is.

One can easily prove the following criterion of local regularity of \overline{B}_{2} with respect to \overline{B} : (\overline{B}_{2} ∈ LR(\overline{B})) ⇔ (∃m, k_{0} ∈ ℕ : A_{B} ∩ ([1/m^{k}, 1/m^{k−1}) × [1/m^{k′}, 1/m^{k′−1})) ≠ ∅ for k, k′ > k_{0}).

(4) The basis B is said to be invariant (HI-basis) with respect to homotheties if for every x ∈ ℝ^{n} and every homothety H centered at x we have B(x) = {H(R) : R ∈ B(x)}.

If \overline{B} is locally regular with respect to \overline{B}′ and, moreover, if δ is equal to ∞, then \overline{B} is called regular with respect to \overline{B}′.

For the basis B ⊂ B_{2} define the sets: R_{1,B} = \{r > 1 : ∃I ∈ \overline{B}, | pr_{1} I|_{1} = r| pr_{2} I|_{1}\}, R_{2,B} = \{r > 1 : ∃I ∈ \overline{B}, | pr_{2} I|_{1} = r| pr_{1} I|_{1}\}.

For the basis B ⊂ B_{2} which is simultaneously the T I- and HI basis one can easily verify that (\overline{B}_{2} ∈ LR(\overline{B})) ⇔ (\overline{B}_{2} is regular with respect to B) ≡↔ (∃m ∈ ℕ : R_{i,B} ∩ [m^{k}, m^{m+1}) ≠ ∅, k ∈ ℕ, i = \overline{1, 2}).

(5) Let B_{2} (B_{2} = B_{2}(ℝ^{n})) be the basis in ℝ^{n} for which B_{2}(x) (x ∈ ℝ^{n}) consists of all n-dimensional intervals containing the point x.

Theorems 1–4 are also valid for B ⊂ B_{2}(ℝ^{n}) (n ≥ 3), T I-bases. Proofs for the n-dimensional case are similar to those for the two-dimensional case.

(6) Let δ_{k}^{1} ↓ 0, …, δ_{k}^{n} ↓ 0 (k → ∞). Denote Δ_{k,m}^{i} = [(m−1)δ_{k}^{i}, mδ_{k}^{i}) (i = \overline{1, n}; k, m ∈ ℕ). Let B be the basis in ℝ^{n} for which B(x) = \{ ∏_{i=1}^{n} Δ_{k^{i},m^{i}}^{i} : ∏_{i=1}^{n} Δ_{k^{i},m^{i}}^{i} ∋ x; k^{i}, m^{i} ∈ ℕ). Such bases are sometimes called nets. We call them N-bases.

Let B be the N-basis in ℝ^{n} (n ≥ 2). Denote by B_{T} the least T I-basis containing B. For B the following analogue of Theorem 1 is valid: the following conditions are equivalent: (i) F_{B}^{+} = F_{B_{2}}^{+}; (ii) B ⇔^{+} B_{2}; (iii) \overline{B}_{2} is locally regular with respect to \overline{B}_{T}.

Obviously, (ii)⇒(i), the implication (i)⇒(iii) follows directly from Theorem 1. Thus it remains only to show that (iii)⇒(i). To this end, note that using Lemma 1 of [2] one can easily obtain the inequality

|\{M_{B_{T}}^{(r)}(f) > λ\}| ≤ 3^{n}|\{M_{B}^{(r)}(f) > \frac{λ}{2^{n}} \}| (f ∈ L(ℝ^{n}); λ, r > 0).

From this and Lemma 5 we get the upper bound of the distribution function of M_{B_{2}}^{(r)}(f) by means of the distribution function of M_{B}^{(cr)}(f). It easily follows from the relation \overline{B}_{2} ∈ LR(\overline{B}) that B contains a regular subbasis. Hence B possesses the property (E). Next, we obtain the implication (iii)⇒(ii) from Lemma 6.

Note that for δ_{k}^{i} = 1/2^{k} (i = \overline{1, n}; k ∈ ℕ) the relation B ⇔^{+} B_{2} was proved earlier in [2].

(7) The basis B constructed in Theorem 4 is not a BF -basis, which is not a casual fact. In particular, the following assertion is valid: let B ⊂ B_{2} (B_{2} = B_{2}(ℝ^{n})) be a BF -basis and let \overline{B}_{2} ∈ LR(\overline{B}). Then B ⇔^{+} B_{2}.

To prove the above assertion we have to consider the following facts which easily follow from the Busemann-Feller property of the basis B and from the relation \overline{B}_{2} ∈ LR(\overline{B}: (i) B contains a regular subbasis; (ii) for every f ∈ L(ℝ^{n}),

|\{M_{B_{2}}^{(r)}(f) > λ\}| ≤ |\{M_{B}^{(cr)}(f) > \frac{λ}{c}\}| (λ > 0, 0 < r < δ),

where c and δ are the constants from the definition of local regularity of \overline{B}_{2} with respect to \overline{B}.

From (i) we have that B possesses the property (E). Now, taking into account (ii) and using Lemma 6, we can conclude that the relation B ⇔^{+} B_{2} is valid.

Acknowledgement

The authors express their gratitude to the referee for his useful remarks.

References

1. M. de Guzmán, Differentiation of integrals in ℝ^{n}. Lecture Notes in Math. 481, Springer, 1975.

2. T. Sh. Zerekidze, Convergence of multiple Fourier–Haar series and strong differentiability of integrals. (Russian) Trudy Tbilis. Mat. Inst. A. Razmadze 76(1985), 80–99.

3. A. M. Stokoloc, On the differentiation of integrals of functions from Lφ(L). Studia Math. 88(1988), 103–120.

(Received 17.05.1995)

Authors’ address: Faculty of Mechanics and Mathematics I. Javakhishvili Tbilisi State University 2, University St., Tbilisi 380043 Republic of Georgia