On ordering GQ Directory UMM :Journals:Journal_of_mathematics:OTHER:

358 D. D. Ang – K. Schmitt – L. K. Vy Theorem 1 GQ, k · k is a separable Banach space, isomorphic to ˆ CQ, k · k ∞ . We next consider some basic properties of the integral just defined. Since the mapping f 7→ F f is linear, we immediately deduce Z Q sf + tg = s Z Q f + t Z Q g, ∀f, g ∈ GQ, s, t ∈ R. More properties are given below.

1.2 On ordering GQ

The ordering in this case is much more complicated than that in the one-dimensional case. As usual, for f, g ∈ GQ, we say that f ≥ g if and only if f − g is a positive measure on Q. For f ∈ D ′ Q, we know, by the Riesz Representation Theorem, that if f is a positive or bounded Radon measure on Q then it can be identified with a Borel measure in the set function sense µ = µ f , which is also positive or bounded. We can define in these cases F x, y = F f x, y = µ f a, x × c, y, x, y ∈ Q. We need the following lemma for our study of ordering on GQ. Lemma 3 i If f is a bounded measure on Q then F f ∈ L ∞ Q is the distributional primitive of f, i.e., ∂F f = f in D ′ Q. ii If in i, we assume furthermore that f ∈ GQ, then F f x, y = F fx, y, ∀x, y ∈ Q. 2 iii If f ∈ GQ is a positive measure then 2 holds. Proof. We sketch here only the main features, and leave the details to the reader. i We first define Zx, y, s, t = χ a,x×c,y s, t, x, y, s, t ∈ Q. We have F x, y = Z Q Zx, y, s, tdµs, t and F ∈ L ∞ Q. For g ∈ DQ, hF, ∂gi = Z Q Z Q Zx, y, s, t∂gx, ydµs, t dxdy = Z Q Z Q Zx, y, s, t∂gx, ydxdy dµs, t. But Z Q Zx, y, s, t∂gx, ydxdy = Z b s Z d t ∂ 12 gx, ydxdy = − Z b s ∂ 1 gx, tdx = gs, t, An analogue of the Denjoy-Perron-Henstock-Kurzweil integral 359 hence hF, ∂gi = Z Q gs, tdµs, t = hf, gi, ∀g ∈ DQ. ii Let f ∈ GQ, and let J = F − F f. Then J ∈ L ∞ Q and ∂J = 0 in D ′ Q. We first prove that lim x →a+ F x, y = lim y →c+ F x , y = 0, ∀x ∈ [a, b], ∀y ∈ [c, d]. In fact, we have the Hahn decomposition µ = µ + −µ − , where µ ± are positive bounded measures on Q. Let {x n } be any sequence decreasing to a, since ∩ n ≥1 [a, x n × c, y n ] = ∅, lim µ ± a, x n × c, y = 0. Hence F x n , y = µ + a, x n × c, y − µ − a, x n × c, y → 0. On the other hand, we have, as in Lemma 1, H ∈ D ′ a, b, K ∈ D ′ c, d such that J = 1 a,b ⊗ K + H ⊗ 1 c,d . Since F, F f are bounded, we have H ∈ L ∞ a, b, K ∈ L ∞ c, d. As in the last part of the proof of Lemma 1, this implies that J x, y = Hx + Ky, a.e. x, y ∈ Q. 3 By Fubini’s theorem, there exists B ⊂ [c, d], m 1 [c, d]\B = 0 m 1 is the one- dimensional Lebesgue measure such that 3 holds for all y ∈ B, a.e. x ∈ [a, b]. Let a ′ , b ′ ⊂ a, b. Since lim y →c J x, y = 0, ∀x ∈ [a, b] and |J x, y| ≤ |µ|G + kF fk ∞ , x, y ∈ Q, the dominated convergence theorem gives lim y →c Z b ′ a ′ J x, ydx = 0. This implies that b ′ − a ′ −1 Z b ′ a ′ Hxdx = lim y →c, y∈B Ky for all subintervals a ′ , b ′ ⊂ a, b. Applying Lebesgue’s theorem, one has Hx = lim y →c,y∈B Ky fora.e. x ∈ a, b, i.e., H=const a.e. on a, b. Similarly, K=const a.e. on c, d. Thus, J =const a.e. on Q. This constant must be 0 since lim y →c J x, y = 0. Hence F x, y = F fx, y a.e. on Q. Using the continuity of F f and the Hahn decom- position of µ, we see that this equality actually holds for all x, y ∈ Q. iii Let f ∈ GQ be a positive measure on Q. We choose sequences {a n }, {b n }, {c n }, {d n } such that a n b n , c n d n , [a n , b n ] × [c n , d n ] ⊂ Q, ∀n and a n ց a, b n ր b, c n ց c, d n ր d as n → ∞. Put Q n = a n , b n × c n , d n and f n = f| Q n , µ n = µ| Q n . µ n is a positive measure on Q n and µ n represents f n . Since Q n is compact, µ n is bounded. Applying ii, we have F f n x, y = µ n a n , x × c n , y = µa n , x × c n , y, 360 D. D. Ang – K. Schmitt – L. K. Vy for all x, y ∈ Q n . Let x, y ∈ Q. Choosing m large enough such that for x, y ∈ Q m , we have µa n , x × c n , y = F fx, y − F fa n , y − F fx, c n + F fa n , c n , ∀n ≥ m. On the other hand, we have lim µa n , x × c n , y = µ   [ n ≥m a n , x × c n , y   = µa, x × c, y. Letting n → ∞, we have F x, y = µa, x × c, y = F fx, y − F fa, y − F fx, c + F fa, c = F fx, y. This show that F = F f in Q. The extension of this equality to the boundary of Q is straightforward. We immediately obtain from Lemma 3 the order preserving property of the G- integral, i.e., Z Q f ≥ Z Q g, whenever f ≥ g, f, g ∈ GQ. In fact, since f − g ∈ GQ is a positive measure, Lemma 3 iii gives Z Q f − Z Q g = F f − gb, d = µ f −g Q ≥ 0. We also have other usual relations between the G-integral and the ordering. For example, we have the following result: Corollary 1 If f, g, h ∈ D ′ Q, f ≤ g ≤ h and if f and h are G-integrable, then g is also G-integrable. Proof. Without loss of generality, we can assume that f = 0, i.e., g, h are positive measures on Q and g ≤ h. Letting µ g , µ h be the Borel measures corresponding to g and h, we have 0 ≤ µ g ≤ µ h . Lemma 3 iii implies that µ h is a bounded measure on Q and F hx, y = µ h a, x × c, y, x, y ∈ Q. Then µ g is also bounded. Putting F x, y = µ g a, x × c, y, x, y ∈ Q, one has from Lemma 3 ii that ∂F = g. We check that F ∈ CQ. Let x, y ∈ Q. If F is not continuous at x, y then we can find a sequence {x n , y n } converging to x, y such that inf n |F x n , y n − F x, y| 0. Divide Q into four subrectangles: Q 1 = a, x] × c, y], Q 2 = x, b × c, y, Q 3 = [x, b × [y, d, Q 4 = a, x × y, d. Passing to a subsequence if necessary, we can assume that x n , y n ∈ Q i for some i ∈ {1, 2, 3, 4}. Assume i = 1. Passing once more to a subsequence, we can assume that x n ր x, y n ր y. Hence µ g a, x n × c, y n → µ g   [ n ≥1 a, x n × c, y n   = µ g a, x × c, y, An analogue of the Denjoy-Perron-Henstock-Kurzweil integral 361 contradicting the choice of x n , y n . Assume now i = 2. As above we can also assume that x n ց x, y n ր y. Simple calculations show that |F x, y − F x n , y n | ≤ µ g a, x × [y n , y + µ g [x n , x × c, y n ≤ µ h a, x × [y n , y + µ h [x n , x × c, y n = F hx, y − F hx, y n − F hx n , y + F hx n , y n . By the continuity of F h, the right hand side of this inequality tends to 0. We obtain again a contradiction. The cases i = 3, 4 are treated in the same way. We have proved that F is continuous at x, y ∈ Q. The continuity of F on ∂Q is established similarly.

1.3 Some further properties