Georgian Mathematical Journal 1(94), No. 5, 469-478

PASSAGE OF THE LIMIT THROUGH THE DOUBLE DENJOY INTEGRAL

N. CHELIDZE

Abstract. The conditions are given for passage of the limit through the double Denjoy integral defined by V.G. Chelidze.

As is well-known, there exist various definitions of the double Denjoy integral (see [1,2,5]). Conditions for passage of limits through these integrals have not yet been studied. The object of this paper is to investigate the conditions for passage of the limit through the double Denjoy integral defined by V.G. Chelidze (see [7]).

Here we shall use the well-known terms (see, for example, [8]). We recall only a few definitions.

Definition 1 (see [8], pp. 127-128). A function f : R_{0} → ℝ with R_{0} = [a, b] × [c, d] is said to be Denjoy integrable (D-integrable) on R_{0} or, briefly, f ∈ D(R_{0}) if there exists a generalized absolutely continuous function F (ACG-function) on R_{0}, briefly, F ∈ ACG(R_{0}) (see [8], pp. 99-100), with an approximate derivative ([8], p. 103) equal to f a.e. The function F is called a D-primitive of f and Δ(f; R_{0}) = F (b, d) − F (a, d) − F (b, c) + F (a, c) is called the D-integral of f on R_{0} which is written as

Δ(F, R_{0}) = (D) ∫∫_{R_{0}} f(x, y)dxdy.

A function f : R_{0} → ℝ is called D-integrable on a measurable ^{1} subset E ⊂ R_{0} (briefly f ∈ D(E)), if f_{E}(x, y) ∈ D(R_{0}) where

f_{E}(x, y) = \begin{cases}f(x, y) & (x, y) ∈ E, \\ 0 & (x, y) ∉ E. \end{cases}

469

1991 Mathematics Subject Classification. 28A35.

1Measurability is meant in the Lebesgue sense and |A| will stand for the Lebesgue measure of the set A.

1072-947X/94/0900-0469$07.00/0 \begin{matrix}c \\ ◯ \end{matrix} 1994 Plenum Publishing Corporation

Definition 2. A compact interval I is said to be of type 2 with respect to E if at least one pair of the opposite vertices of I lies in E.

Lemma 1. Let E_{1} be a compact subset of [a, b] whose contiguous intervals are \{r_{i}\}_{i}. If f ∈ D(E), f ∈ D(R̄_{i}), i = \overline{1, ∞}, where E = E_{1} × [c, d], R_{i} = r_{i} × [c, d], and

∑_{i=1}^{∞} O(D; f, R̄_{i}) < ∞,  (1)

where

O[D; f, R̄_{i}] = sup_{ρ⊂R̄_{i}} {|(D) ∫∫_{ρ} f(t, τ)dt dτ|}

and ρ stands for any measurable subset of R̄_{i}. Then f ∈ D(R_{0}) and for each subsegment R ⊂ R_{0} we have

(D) ∫∫_{R} f(t, τ)dt dτ = (D) ∫∫_{R∩E} f(t, τ)dt dτ + ∑_{i} ∫∫_{R∩R̄_{i}} f(t, τ)dt dτ.

Proof. Let us consider the function

F (x, y) = ∑_{i}(D) ∫∫_{R̄_{i}∩I(x,y)} f(t, τ)dt dτ,

where I(x, y) = [a, x; c, y].

We shall show that F is continuous on R_{0}, i.e., F is continuous at any point (x_{0}, y_{0}) of R_{0}; for ε > 0 there must exist a δ(ε) > 0 such that the inequality |F (x_{0} + k, y_{0} + l) − F (x_{0}, y_{0})| < ε. must take place for all k and l, |k| < δ(ε), |l| < δ(ε). Then

|F (x_{0} + k, y_{0} + l)−F (x_{0}, y_{0})|=| ∑_{i}(D) ∫∫_{R̄_{i}∩I(x_{0}+k,y_{0}+l)} f(t, τ)dt dτ −

− ∑_{i}(D) ∫∫_{R̄_{i}∩I(x_{0},y_{0})} f(t, τ)dt dτ| ≤ ∑_{i} |(D) ∫∫_{R̄_{i}∩P_{x_{0},y_{0}}} f(t, τ)dt dτ|,

where P_{x_{0},y_{0}} = [a, x_{0}; y_{0}, y_{0} + l] ∪ [x_{0}, x_{0} + k; c, y_{0} + l].

To estimate the last sum let us consider two subsystems of intervals from the system \{R_{i}\}_{i=\overline{1,∞}}: I′ = \{R′_{i}\} and I′′ = \{R′′_{i}\} such that system I′ satisfies the conditions:

R̄′_{i} ∩ P_{x_{0},y_{0}} ≠ ∅ and i > N

and system I′′ the conditions R̄′′_{i} ∩ P_{x_{0},y_{0}} ≠ ∅ and i ≤ N.

For the first system for all ε > 0 there is an N_{0} = N_{0}(ε) such that if N > N_{0}, then by the condition (1) we get the estimate

∑_{i>N} |(D) ∫∫_{R̄′_{i}∩P_{x_{0},y_{0}}} f(t, τ)dt dτ| < ∑_{i>N} O[D; f, R̄′_{i}] < \frac{ε}{2}.

Now consider system I′′. Since i ≤ N, this system is finite. Let N^{∗} be the number of members of this system. Then for ε > 0 there is a δ(ε) > 0 such that |P_{x_{0},y_{0}}| < δ and |P_{x_{0},y_{0}} ∩ R̄′′_{i}| < |P_{x_{0},y_{0}}| < δ. Hence it follows

(D) ∫∫_{R̄′′_{i}∩P_{x_{0},y_{0}}} f(t, τ)dt dτ < \frac{ε}{2N^{∗}} .

Now

∑_{i≤N} |(D) ∫∫_{R̄′′_{i}∩P_{x_{0},y_{0}}} f(t, τ)dt dτ| ≤ \frac{ε}{2N^{∗}} · N^{∗} = \frac{ε}{2}.

Finally, we obtain

|F (x_{0} + k, y_{0} + l) − F (x_{0}, y_{0})| ≤ ∑_{i} |(D) ∫∫_{R̄_{i}∩P_{x_{0},y_{0}}} f(t, τ)dt dτ| =

= ∑_{i>N} |(D) ∫∫_{R̄′_{i}∩P_{x_{0},y_{0}}}f(t, τ)dt dτ| + ∑_{i≤N} |(D) ∫∫_{R̄′′_{i}∩P_{x_{0},y_{0}}}f(t, τ)dt dτ| ≤ ε.

Thus F is continuous on R_{0}.

Now for the fixed i let us consider any finite system of pairwise disjoint compact intervals of type 2 with respect to R_{i}, say, {S_{k}}_{k=1}^{n}. If ∑_{k=1}^{n} |S_{k}| < δ, then, by the conditions of the lemma,

∑_{k} |Δ(F; S_{k})| = ∑_{k} | ∫∫_{S_{k}} f(t, τ)dt dτ| < ε.

Therefore F ∈ ACG (R̄_{i})_{i=\overline{1,∞}}. Now we shall show that F is absolutely continuous on E (or, briefly, is AC, written as F ∈ AC(E)) (see Definition 3 or [8], p. 97). To this end we define the function G as follows:

G(x, y) = \begin{cases}0, & when (x, y) ∈ E, \\ \frac{1}{|R̄_{i}|}(D) ∫∫R̄_{i} f(t, τ)dt dτ, & when (x, y) ∈ R_{i}, i = \overline{1, ∞}. \end{cases}

We show that G is Lebesgue integrable on R_{0}.

∫∫_{R_{0}} |G(x, y)|dx dy = ∑_{i} ∫∫R̄_{i} |G(x, y)|dx dy =

= ∑_{i} ∫∫_{R_{i}} (\frac{1}{|R̄_{i}|}(D)| ∫∫R̄_{i} f(t, τ)dt dτ|)dx dy =

= ∑_{i} |(D) ∫∫R̄_{i} f(t, τ)dt dτ| ∫∫R̄_{i} \frac{dx dy}{|R̄_{i}|} = ∑_{i} |(D) ∫∫R̄_{i} f(t, τ)dt dτ| < C.

Put

Ψ(x, y) = ∫_{a}^{x} ∫_{c}^{y} G(t, τ)dt dτ.

Since G is summable, for any ε > a there is a δ > 0 such that |e| < δ implies

| ∫∫_{e} G(t, τ)dt dτ| < ε  (2)

for each measurable subset e ⊂ R_{0}.

If we consider any finite system of pairwise disjoint compact intervals of type 2 with respect to E, say, {I_{k}}_{k=1}^{m}, and if ∑_{k=1}^{m} |I_{k}| < δ, then

∑_{k=1}^{m} | ∫∫_{I_{k}} G(t, τ)dt dτ| < ε.

We shall now show that Ψ is AC on E. Assume I_{k} = [α_{k}, β_{k}; γ_{k}, δ_{k}]. Then

∑_{k=1}^{m} |ΔΨ(I_{k})| = ∑_{k=1}^{m} | ∫∫_{I_{k}} G(t, τ)dt dτ| < ε.

It is not difficult to check that F (x, y) = Ψ(x, y) when (x, y) ∈ E. Since F is ACG on each R_{i} and F ∈ AC(E), it follows that F is ACG on R_{0}.

Moreover, DF (x, y) = DΨ(x, y) = G(x, y) = 0 almost everywhere on E and DF (x, y) = f(x, y) almost everywhere on R_{0}\E = ⋃_{i} R_{i}.

Thus F is a D-primitive of φ_{1} defined as follows:

φ_{1}(x, y) = \begin{cases}f(x, y) & when (x, y) ∈ ⋃_{i} R_{i}, \\ 0 & when (x, y) ∈ E. \end{cases}

On the other hand, we consider

φ_{2}(x, y) = \begin{cases}0 & when (x, y) ∈ ⋃_{i} R_{i}, \\ f(x, y) & when (x, y) ∈ E. \end{cases}

Since f ∈ D(E), we have φ_{2} ∈ D(R_{0}) and f ∈ D(R_{0}), f(x, y) = φ_{1}(x, y) + φ_{2}(x, y).

Finally, we obtain the equality

(D) ∫∫_{R} f(t, τ)dt dτ = (D) ∫∫_{R} φ_{1}(t, τ)dt dτ + (D) ∫∫_{R} φ_{2}(t, τ)dt dτ =

= (D) ∫∫_{R∩E} φ_{1}(t, τ)dt dτ + (D) ∫∫_{R∩(∪_{i}R̄_{i})} φ_{1}(t, τ)dt dτ +

+(D) ∫∫_{R∩E} φ_{2}(t, τ)dt dτ + (D) ∫∫_{R∩(∪_{i}R̄_{i})} φ_{2}(t, τ)dt dτ =

= (D) ∫∫_{R∩E} f(t, τ)dt dτ + (D) ∫∫_{R∩(∪_{i}R̄_{i})} f(t, τ)dt dτ =

= (D) ∫∫_{R∩E} f(t, τ)dt dτ + ∑_{i}(D) ∫∫_{R∩R̄_{i}} f(t, τ)dt dτ,

where R is a subsegment of R_{0}.

Lemma 2. Let F_{n} be continuous on R_{0}, n ∈ N, and lim_{n→∞} F_{n} = F, where F is also continuous. Then lim_{n→∞} O[F_{n}; R_{0}] ≥ O(F; R_{0}).

Proof. Since F is continuous, there exist points (x_{1}, y_{1}), (x_{2}, y_{2}) on R_{0} such that F (x_{2}, y_{2}) − F (x_{1}, y_{1}) = O(F; R_{0}). On the other hand, for any ε > 0 there is an N_{0} = N_{0}(ε) > 0 such that F_{n}(x_{i}, y_{i}) − ε ≤ F (x_{i}, y_{i}) ≤ F_{n}(x_{i}, y_{i}) + ε, i = 1, 2, for all n > N_{0}.

It follows that F (x_{2}, y_{2}) − F (x_{1}, y_{1}) ≤ F_{n}(x_{2}, y_{2}) − F_{n}(x_{1}, y_{1}) + 2ε and so O(F; R_{0}) ≤ O(F_{n}; R_{0}) + 2ε from which the result is immediate.

Definition 3. Let F_{n} : R_{0} → R be continuous, n ∈ N. Then {F_{n}; n ∈ N} is said to be a sequence of uniformly A_{x}-functions on the set P ⊂ [a, b] if P = ⋃_{k} P_{k} and F_{n} is uniformly AC on each H_{k} = P_{k} × [c, d] (see [9]), i.e., for all ε > 0 there is a δ = δ(ε) > 0 such that for any finite system of pairwise disjoint compact intervals of type 2 with respect to H_{k}, say, {R_{i}}, i = \overline{1, m}, the inequality ∑_{i=1}^{m} |R_{i}| < δ implies ∑_{i=1}^{m} |Δ(F_{n}; R_{i})| < ε for all n ∈ N.

Lemma 3. If f_{n} ∈ D(R_{0}), n ∈ N, E ⊂ [a, b] is a closed set and the sequence

F_{n}(x, y) = ∫_{a}^{x} ∫_{c}^{y} f_{n}(t, τ)dt dτ, n ∈ N,

is uniformly AC on E′ = E ×[γ, δ], then for all ε > 0 there is a δ = δ(ε) > 0 such that for all measurable sets, e, e ⊂ E and |e| < δ, we have

| ∫∫_{e′} f_{n}(t, τ)dt dτ| < ε, n = \overline{1, ∞}, where e′ = e × [γ, δ].

Proof. By the conditions of the lemma, if ε > 0, then there is a δ > 0 such that for any finite system \{R_{k}\}, k = \overline{1, m}, of disjoint compact intervals of type 2 with respect to E′ the inequality ∑_{k=1}^{m} |R_{k}| < δ implies

∑_{k=1}^{m} |F_{n}(R_{k})| = ∑_{k=1}^{m} | ∫∫_{R_{k}} f_{n}(t, τ)dt dτ| < ε  (3)

for all n ∈ N.

Let now e be a measurable subset of E with |e| < δ. Then there are open sets G_{m}, m ∈ N, such that G_{m} ⊃ G_{m+1} ⊃ e, m ∈ N, and lim_{m→∞} |G_{m}| = |e|. It can further be assumed that G_{m} = ∪_{k=1}^{∞}r_{k}^{(m)}, r_{k}^{(m)} ∩ r_{e}^{(m)} = ∅, k ≠ l, for each m, where r_{k}^{(m)} are open intervals. We set e_{m} = E ∩ G_{m} = ∪_{k}(E ∩ r_{k}^{(m)}). Denote by ρ_{k,j}^{(m)}, j ∈ N, contiguous intervals of E ∩ r_{k}^{(m)} on

r_{k}^{(m)}. Then G_{m}\e_{m} = ∪_{k} ∪_{j} ρ_{k,j}^{(m)}.

The endpoints of each ρ_{k,j}^{(m)} lie on E. Moreover, for δ = δ(ε) > 0 there is an N_{0}(δ) = N_{0} > 0 such that if m > N_{0}, then |G_{m}| < δ.

If m > N_{0}, from (3) we obtain

\begin{matrix}∑_{k} | ∫∫_{r_{k}^{(m)}×[γ,δ]} f_{n}(t, τ)dt dτ| < ε, \\ ∑_{k} ∑_{j} | ∫∫_{ρ_{k,j}^{(m)}×[γ,δ]} f_{n}(t, τ)dt dτ| < ε \end{matrix}  (4)

for all n ∈ N.

Now, by Lemma 1, for m > N_{0} we have

∫∫_{r_{k}^{(m)}×[γ,δ]} f_{n}(t, τ)dt dτ =

= ∫∫_{(r_{k}^{(m)}∩E)×[γ,δ]} f_{n}(t, τ)dt dτ + ∑_{j} ∫∫_{ρ_{k,j}^{(m)}×[γ,δ]} f_{n}(t, τ)dt dτ, n = \overline{1, ∞}.

Performing the summation of the last expression over k, from (4) we obtain

| ∫∫_{e_{m}×[γ,δ]} f_{n}(t, τ)dt dτ| < 2ε, n = \overline{1, ∞},

and so

| ∫∫_{e′} f_{n}(t, τ)dt dτ| < 2ε, \overline{1, ∞}. ■

Note that the following lemma of P. Romanovskii is very essential for our investigation.

Lemma 4. Let 𝔉 be a nonempty system of open subintervals of the open interval ]a, b[, having the following four properties:

1) If ]α, β[∈ 𝔉 and ]β, γ[∈ 𝔉, then ]α, γ[∈ 𝔉;

2) If ]α, β[∈ 𝔉, then every subinterval of ]α, β[ also belongs to 𝔉;

3) If every proper open subinterval of ]α, β[ belongs to 𝔉, then ]α, β[∈ 𝔉;

4) If all contiguous intervals on [a, b] of a nonempty perfect set E from ]a, b[ belong to 𝔉, then 𝔉 contains an interval ]α, β[ such that ]α, β[∩E ≠ ∅.

Then ]a, b[∈ 𝔉.

Theorem. If f_{n} ∈ D(R_{0}), lim_{n→∞} f_{n} = f almost everywhere on R_{0} and the sequence

F_{n}(x, y) = ∫_{a}^{x} ∫_{c}^{y} f_{n}(t, τ)dt dτ, n = \overline{1, ∞},

is uniformly continuous on R_{0} and uniformly A_{x} on [a, b], then f ∈ D(R_{0}) and

lim_{n→∞} ∫∫_{R_{0}} f_{n}(t, τ)dt dτ = ∫∫_{R_{0}} f(t, τ)dt dτ.

Proof. The subinterval ]α, β[⊂]a, b[ will be said to lie in the family of subintervals 𝔉 if for ]α, β[ and all all compact subintervals we have

lim_{n→∞} ∫_{α}^{β} ∫_{γ}^{δ} f_{n}(t, τ)dt dτ = ∫_{α}^{β} ∫_{γ}^{δ} f(t, τ)dt dτ,  (*)

where [γ, δ] ⊂ [c, d].

The theorem will be proved if we show that 𝔉 satisfies the conditions of Romanovskii’s lemma. The condition of the theorem implies that there exists a portion P such that {F_{n}}_{n≥1} is uniformly AC on P ×[c, d]. Then due to Vitali’s theorem on passage of limits through integrals and by virtue of some properties of the Denjoy integral there exists an interval [α^{∗}, β^{∗}, γ^{∗}, δ^{∗}] for which the equation (*) is fulfilled. Thus 𝔉 ≠ ∅.

Now assume that ]α, β[∈ 𝔉 and ]β, η[∈ 𝔉. Then ]α, η[∈ 𝔉. Indeed,

lim_{n→∞} ∫_{α}^{η} ∫_{γ}^{δ} f_{n}(t, τ)dt dτ = lim_{n→∞} ( ∫_{α}^{β} ∫_{γ}^{δ} f_{n}(t, τ)dt dτ + ∫_{β}^{η} ∫_{γ}^{δ} f_{n}(t, τ)dt dτ) =

= ∫_{α}^{β} ∫_{γ}^{δ} f(t, τ)dt dτ + ∫_{β}^{η} ∫_{γ}^{δ} f(t, τ)dt dτ = ∫_{α}^{η} ∫_{γ}^{δ} f(t, τ)dt dτ,

i.e., ]α, η[∈ 𝔉. Condition 2) follows from the definition of the family 𝔉. Now let us consider condition 3). It is assumed that every open proper

subinterval of ]α, β[ is contained in 𝔉. Then we must show that ]α, β[∈ 𝔉. Consider ]λ_{m}, μ_{m}[∈ 𝔉, ]λ_{m}, μ_{m}[⊂]α, β[, λ_{m} ↓ α, μ_{m} ↑ β, m → ∞. Since {F_{n}}_{n≥1} is uniformly continuous on R_{0}, for all ε > 0 there is an

N_{0} = N_{0}(ε) such that if m > N_{0}, then

∫_{λ_{m}}^{μ_{m}}∫_{γ}^{δ} f_{n}(t, τ)dt dτ − ε ≤ ∫_{α}^{β} ∫_{γ}^{δ} f_{n}(t, τ)dt dτ ≤ ∫_{λ_{m}}^{μ_{m}}∫_{γ}^{δ} f_{n}(t, τ)dt dτ + ε,

where [γ, δ] ⊂ [c, d] is arbitrary.

Since ]λ_{m}, μ_{m}[∈ 𝔉, we have

∫_{λ_{m}}^{μ_{m}}∫_{γ}^{δ} f(t, τ)dt dτ − ε ≤ \underline{lim}_{n→∞} ∫_{α}^{β} ∫_{γ}^{δ} f_{n}(t, τ)dt dτ ≤

≤ \overline{lim}_{n→∞} ∫_{α}^{β} ∫_{γ}^{δ} f_{n}(t, τ)dt dτ ≤ ∫_{λ_{m}}^{μ_{m}}∫_{γ}^{δ} f(t, τ)dt dτ + ε.  (5)

Hence, ε > 0 being arbitrary, (5) implies that

lim_{n→∞} ∫_{λ_{m}}^{μ_{m}}∫_{γ}^{δ} f(t, τ)dt dτ  (6)

exists and

lim_{n→∞} ∫_{α}^{β} ∫_{γ}^{δ} f_{n}(t, τ)dt dτ = ∫_{α}^{β} ∫_{γ}^{δ} f(t, τ)dt dτ.

Thus ]α, β[∈ 𝔉 and condition 3) is satisfied.

Finally, we shall check condition 4). It is assumed that E ⊂ [a, b] is a closed set. Since the sequence {F_{n}}_{n≥1} is uniformly A_{x} on [a, b], there exists a portion P = E ∩ [α, β] such that {F_{n}}_{n≥1} is uniformly AC on P × [c, d].

Let \{I_{k} =]α_{k}, β_{k}[, k = \overline{1, ∞}\} be the contiguous intervals of P on ]α, β[ which are the members of 𝔉. By Lemma 1, where I_{k} × [γ, δ] are the contiguous intervals of P × [γ, δ], k = \overline{1, ∞}, we have

∫_{α}^{β}∫_{γ}^{δ}f_{n}(t, τ)dt dτ =∫∫_{P ×[γ,δ]}f_{n}(t, τ)dt dτ +∑_{k} ∫∫_{I_{k}×[γ,δ]}f_{n}(t, τ)dt dτ. (7)

By Lemma 3 and Vitali’s theorem on passage of limits through integrals

lim_{n→∞} ∫∫_{P′} f_{n}(t, τ)dt dτ = ∫∫_{P′} f(t, τ)dt dτ,  (8)

where P′ = P × [γ, δ].

Since {F_{n}}_{n≥1} is uniformly AC on P′ for all ε > 0, there is an N_{0} = N_{0}(ε) such that if k > N_{0}, then

∑_{k>N_{0}} O(F_{n}; I_{k} × [γ, δ]) < ε.  (9)

Since I_{k} ∈ 𝔉, (9) implies

∑_{k>N_{0}} O(F; I_{k} × [γ, δ]) < ε, ∑_{k=1}^{∞} O(F; I_{k} × [γ, δ]) < +∞.  (10)

Thus f ∈ D([α, β; γ, δ]) and from (7), (8), (10) we obtain

lim_{n→∞} ∫_{α}^{β} ∫_{γ}^{δ} f_{n}(t, τ)dt dτ = ∫_{α}^{β} ∫_{γ}^{δ} f(t, τ)dt dτ,

i.e., ]α, β[∈ 𝔉 and ]α, β[∩E ≠ ∅. Therefore 𝔉 satisfies all conditions of Romanovskii’s lemma and ]a, b[∈ 𝔉.

References

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(Received 28.08.1992, revized version 25.06.1993)

Author’s address: Faculty of Mechanics and Mathematics I. Javakhishvili Tbilisi State University 2, University st., Tbilisi 380043 Republic of Georgia