Abstract: This study introduce and advance the characteristics of S-lattice measurable function and T-lattice measurable function. It has been proved that the integrations of these lattice measurable functions are made equal. Also it establishes the result that the two iterated integrals of lattice measurable functions are finite and equal. Finally, it confirms that for the lattice σ-finiteness, the Lebesgue lattice measure cannot be omitted and the condition that f is a lattice measurable with respect to the lattice σ-algebra.
INTRODUCTION
The concept of lattice measure was initiated by Gabor (1964). For the extension of this theme literature took some time. Afterward, the concepts of lattice sigma algebra and lattice measure on a lattice sigma algebra launched by Tanaka (2009). Recently the concepts of lattice measurable set, lattice measure space and lattice σ-finite measure were established by Kumar et al. (2011a, b).
The perception of measurable Borel lattices was introduced and studied by Kumar et al. (2011a). Further Radon-Nikodym theorem for signed lattice measure was expanded by Kumar et al. (2011a). A class of super lattice measurable sets was introduced by Pramada et al. (2011). Lebesgue decomposition and its uniqueness of a signed lattice measure were studied successfully by Kumar et al. (2012). A class of positive lattice measurable sets and positive lattice measurable functions was obtained by Pramada et al. (2012a). A characterization of complex integrable lattice functions and μ-free lattices was made by Pramada et al. (2012c). Further recently a characterization of boolean valued star and mega lattice functions was obtained by Pramada et al. (2012b).
This manuscript is aimed to the study of concept of product lattice measurable functions and their various characterizations. In particular these functions are observed by defined over topological spaces. Also it has been investigated the characteristics of S-lattice measurable function and T-lattice measurable functions. The concept of iterated integral of a product lattice measurable function has been defined in order to identify that the two iterated integrals of a product lattice measurable function are finite and equal. It is also aimed to get a condition that product lattice measurable function is lattice measurable is obtained and the condition that product lattice measurable function is lattice measurable cannot be dropped.
PRELIMINARIES
This section briefly reviews the well-known facts of Birkhoff (1967) lattice theory.
The system (L,
| • | (L1) commutative law: x∧y = y∧x and x∨y = y∨x |
| • | (L2) associative law: x∧(y∧z) = (x∧y)∧z and x∨(y∨z) = (x∨y)∨z |
| • | (L3) absorption law: x∨(y∧x) = x and x∧(y∨x) = x. Hereafter,
the lattice (L, |
| • | (L4) distributive law holds: x∨(y∧z) = (x∨y) |
A lattice L is called complete if, for any subset A of L, L contains the supremum ∨ A and the infimum ∧ A. If L is complete, then L itself includes the maximum and minimum elements which are often denoted by 1 and 0 or I and O, respectively.
A distributive lattice is called a Boolean lattice if for any element x in L, there exists a unique complement xc such that:
| x∨xc = 1(L5) | the law of excluded middle |
| x∧xc = 0(L6) | the law of non-contradiction |
Let L be a lattice and c: L→L be an operator. Then c is called a lattice complement in L if the following conditions are satisfied.
|
(L5) and (L6): |
∀x∈L, x∨xc = 1 and x∧xc = 0 |
| (L7) the law of contrapositive: | ∀x, y∈L, x<y implies xc>yc |
| (L8) the law of double negation: | ∀x∈L(xc)c = x |
Definition 1: If a lattice L satisfies the following conditions, then it is called a lattice s-algebra:
|
(1) |
for all h∈L, hc∈L |
| (2) | if hn ∈ L for n = 1, 2, 3 ....., then: |
We denote σ(L) = ß, as the lattice σ-algebra generated by L.
Example 1 (Halmos, 1974): 1: {Φ, X}(that is the empty set together with entire set) is a lattice σ-algebra. 2: P (X) power set of any nonempty set X is a lattice σ-algebra.
Example 2: Let X =
Here, L is a lattice and σ(L) = ß is a lattice σ-algebra generated
by L, where,
Example 3: Let X be any non-empty set, L = {All topologies on X}. Here, L is a complete lattice but not a σ-algebra.
Example 4 (Halmos, 1974): Let X =
Here L is lattice algebra but not lattice σ-algebra.
Definition 2: The entire set X together with a lattice σ-algebra
ß is said to be lattice measurable space, it is denoted by the ordered
pair (
Example 5: Let X =
Definition 3: If the mapping μ: ß→R∪{∞} satisfies the following properties, then μ is called a lattice measure on the lattice σ-algebra σ(L).
| (1) |
μ(φ) = μ(0) = 0 |
| (2) | For all h, g∈ß, such that μ(h), μ(g)≥0 and h≤g⇒μ(h)≤μ(g) |
| (3) | For all h, g∈ß, μ(h∨g)+μ(h∧g) = μ(h)+μ(g) |
| (4) | If hn⊂ß, n∈N such that h1≤h2≤ ... ≤hn≤ ...., then: |
Note 1: Let μ1 and μ2 be lattice measures defined on the same lattice σ-algebra ß. If one of them is finite, then the set function μ(E) = μ1 (E)-μ2 (E), Eεß is well defined and is countably additive on ß.
Example 6 (Royden, 1981): Let X be any set and ß = P(X) be the class of all sub sets of X. Define for any Aεß, μ(A) = +∞ if A is infinite = |A| if A is finite, where |A| is the number of elements in A.
Then μ is a countable additive set function defined on ß and hence μ is a lattice measure on ß.
Definition 4: A set A is said to be lattice measurable set or lattice measurable if A belongs to ß.
Example 7 (Kumar et al., 2011a): The interval (a, ∞) is a lattice measurable under usual ordering.
Example 8 (Kumar et al., 2011a, b):
The closed interval [0, 1]<
Let X =
Example 9 (Kumar et al., 2011a, b): Every Borel lattice is a lattice measurable.
Definition 5: The lattice measurable space (X, ß) together with a lattice measure μ is called a lattice measure space and it is denoted by (X, β, μ).
Example 10: Suppose
Example 11: Let
Definition 6: Let (X,
Example 12: The lattice Lebesgue measure on the closed interval [0, 1] is a lattice finite measure.
Example 13: When a coin is tossed, either head or tail comes when the coin falls. Let us assume that these are the only possibilities. Let X = {H, T}, H for head and T for tail. Let ß = {φ, {H}, {T}, X}. Define the mapping P: ß→[0, 1] by P (φ) = 0 P ({H}) = P ({T}) = ½, P (X) = 1. Then P is a lattice finite measure on the lattice measurable space (X, ß).
Definition 7: If μ is a lattice finite measure, then (X,
Example 14: Let ß be the class of all Lebesgue measurable sets
of [0, 1] and μ be a lattice Lebesgue measure on [0, 1]. Then ([0, 1],
Definition 8: Let (X,
Example 15: The lattice Lebesgue measure on (
and μ(-n,n) = 2n is finite for every n.
Definition 9: If μ be a lattice σ-finite measure, then (X,
Example 16: Let ß be the class of all Lebesgue measurable sets on:
and μ be a lattice measure on
| Fig. 1(a-c): | (a) 2 point lattice L, (b) 4 point lattice M and (c) LxM; the Cartesian product of lattices L and M |
Definition 10 (Gabor, 1964): Let X and Y be two lattices. Then their Cartesian product denoted by XxY is defined as XxY = {(x, y)/xεX, yεY}. It is called product lattice.
Example 17: Let L and M be two lattices shown in the Fig. 1.
Consider LxM in Fig. 1, w h ere, l = (x2, y4), d = (x2, y2), e = (x1, y4), f = (x2, y3), a = (x1, y2), b = (x2, y1), c = (x1, y3) and O = (x1, y1).
Definition 11: The lattice measure m defined on SxT is called the product of the lattice measures μ and λ and is denoted by μxλ.
Example 18: If μ is a lattice measure on R, then m = μxμ is a product lattice measure on RxR.
Definition 12: If A<X and B<Y, then AxB<XxY. Any lattice of the form AxB is called super lattice in XxY.
| Example 19: If A⊂B and C⊂D, then (AxC)⊂(BxD). | |
| Let (x, y) be any element of AxC. Then by definition of product lattice we have xεA, yεC. | |
| But it is given that A⊂B and C⊂D. | |
| Therefore xεB and yεD. | |
| That is (x, y) is an element of BxD. Hence, (AxC)⊂(BxD) is a super lattice in BxD. | |
Remark 1: Counting measure: Let X be a non-empty set. Let σ (L) = P (X).
Define μ: σ (L)→[0, ∞] by |E| = number of lattice measurable sets in E, if E is finite, ∞if E is infinite. Then μ is a lattice measure on P (X) called the lattice counting measure on X.
Definition 13 (Pramada et al., 2011): Let f be a complex lattice measurable function on X. Then |f| is a lattice measurable function from X→[0, ∞]. If:
then we say that f is a complex integrable lattice function with respect to μ. The set of all complex integrable lattice measurable functions with respect to μ on X is denoted by L1.
Definition 14 (Pramada et al., 2011): Let f = u+iv where u and v are real lattice measurable functions on X. Let fεL1. Then we define:
for every lattice measurable set E, where, u+ = max {u, 0}, u¯ = -min {u, 0} and v+ = max {v, 0}, v¯ = -min{v, 0}.
Definition 15 (Kumar et al., 2011a): If E is a lattice measurable set and then the characteristic function χE(x) is defined as if χE(x) = 1, if xεE = 0, if x∉E.
Remark 2: Let (X, S) (Y, T) be lattice measurable spaces.
Then S is a lattice σ-algebra in X and T is a lattice σ-algebra in Y.
Definition 16: If AεS and BεT, then the lattice of the form AxB is called super lattice measurable set where S, T are lattice σ-algebras on X and Y, respectively.
Example 19: Every member of SxT is a super lattice measurable set.
Definition 17: Let E<XxY where xεX, yεY. We define x-section lattice of E by Ex = {y/ (x, y)εE} and y-section lattice of Ey = {x/(x, y)εE}.
Note 2: Ex<Y and Ey<X.
Definition 18 (Kumar et al., 2011a): Let f be an extended real valued measurable function on the lattice of real numbers such that {xεL/f(x)>α} is lattice for each αεL. Then f is lattice measurable function.
Definition 19 (Kumar et al., 2011a): A function s on a lattice measurable space X whose range consists of only finitely many points in [0, ∞] is called a simple lattice measurable function.
Theorem 1 (Pramada et al., 2011): If EεSxT, then ExεT and EyεS for every xεX and yεY.
Theorem 2 (Rudin, 1987) and (Pramada et al., 2011): Let f: X→[0,∞] be a lattice measurable function. Then there exists simple lattice measurable functions sn on X such that:
| i: | 0≤s1≤s2≤…….≤f |
| ii: | sn(x) →f(x) as n →∞for every xεX |
Theorem 3 (Rudin, 1987) and (Pramada et al., 2011): Let {fn} be a sequence of lattice measurable functions on X such that 0≤f1(x)≤f2(x)......≤∞ for every xεX and fn(x)→f(x) as n→∞ for every xεX. Then f is lattice measurable and:
Note 3: Let E = [a, b]. Then:
CHARACTERIZATION OF LATTICE MEASURABLE FUNCTIONS ON PRODUCT LATTICES
Definition 20: A lattice measurable function f: XxY→Z where z is a topological space. For each xεX, we define fx: Y→Z by fx(y) = f(x, y). Then fx is called Y-lattice measurable function. For each yεY, we define fy : X→Z by fy(x) = f(x, y). Then fy is called X-lattice measurable function.
Theorem 4: Let f be an (SxT) lattice measurable function on XxY, The:
(1): |
For each xεX, fx is a T-lattice measurable function |
| (2): | For each yεY, fy is a S-lattice measurable function |
Proof: Let V be an open set in Z. Let Q = {(x, y) εXxY: f(x, y)εV}.
Since f is SxT lattice measurable, QεSxT.
Qx = {y: (x, y)εQ} = {y: f(x,y)εV}
= {y: fx (y)εV} |
By theorem 1, QxεT. Therefore fx is a T-lattice measurable function.
A similar argument shows that fy is an S-lattice measurable function.
Theorem 5: Let (X, S, μ) and (Y, T, λ) be a lattice σ-finite measure spaces. Let f be an (SxT)-lattice measurable function on product lattice XxY. Then the following conditions are hold good:
| 1: | If 0≤f≤∞ and if: |
then Φ is S-lattice measurable. Ψ is T-lattice measurable and:
| 2: | If f is complex and if: |
and
then fεL1(μxλ).
| 3: | If fεL1(μxλ) then fxεL1(λ) for almost all xεX, fyεL1(μ) for almost all yεY; the functions Φ and Ψ defined by: |
almost every where, are in L1(μ) and L1(λ), respectively and:
Proof: By theorem 4, we get fx is a T-lattice measurable function for each xεX and fy is an S-lattice measurable function for each yεY. Hence the definitions of Φ and Ψ make sense.
Part (a): Let QεSxT. Let f =χQ
Then:
Similarly:
Therefore, by theorem 4:
Hence, we get (a) for characteristic functions.
Let f be a non-negative (S x T) simple lattice measurable function.
Then:
Let:
Similarly:
Now:
Therefore:
That is:
That is:
Hence, (a) holds for all non-negative (SxT)-simple lattice measurable functions S.
Let f be any (S x T)-lattice measurable function. Then by theorem 2, there exist (S x T)-simple lattice measurable functions sn on X x Y such that 0≤s1≤s2≤……. ≤f and sn (x, y) →f(x, y) as n→∞ for every (x, y) εX x Y. Let Φn be associated with sn in the same way as Φ is associated to f. we have:
| (1) |
Now:
Therefore, if we apply theorem 3, on (Y, T, λ) then this shows that:
That is, Φn(x) increase to Φ(x) for every xεX as n→∞.
Again applying theorem 3, to the integrals of (1) we get:
By interchanging the role of x and y we get:
Therefore:
This completes proof of (a).
Part (b): Let f be complex. Then 0≤|f|≤∞.
Let:
Given:
Then by using (a) for |f|, we get:
therefore, fεL1(μxλ).
| Proof of (c): First we prove for real f∈L1 (μxλ) | |
| Let f be in L1 (μxλ) and let f be real. Then 0≤f+≤ ∞ and 0≤f¯≤∞. | |
| Let Φ1 and Φ2 correspond to f+ and f– , respectively as Φ corresponds to f. | |
| Now fεL1 (μxλ) and f+≤ |f| | |
| Since, (a) holds for f+, we get that: | |
| Therefore, Φ1εL1 (μ). | |
| Similarly, Φ2ε L1 (μ). | |
| Now fx = (f+)x-(f¯)x.
Also |
|
| Similarly (f–)xε L1(λ). | |
| Therefore fxεL1 (λ) for every x for which both Φ1 (x) and Φ2(x) are<∞. | |
| Since, Φ1, Φ2, εL1(μ) we get that Φ1(x) and Φ1(x) are <∞, almost every where. | |
| Hence, fxε L1 for almost all xεX. | |
| For such x, we have Φ (x) = Φ1 (x)-Φ2(x) | |
| Hence, Φε L1(μ) using (a) we get: | |
Therefore:
That is:
Similarly, we can prove that:
by using fy in place of fx and Ψ in the place of φ.
Suppose f is complex and fεL1(μxλ).
Let f = u+iv. Then u, vεL1 (μxλ) and u, v are real.
Then applying what we proved above to u, v we get:
where, Φu, Φv corresponds to u, v as Φ corresponds to f.
Thus:
That is:
This proves (c).
Hence the theorem.
Note 4:
can be written as:
The integrals at the ends are the so called iterated integrals of f.
The middle integral is often referred to as a double integral.
Result 1: The two iterated integrals are finite and equal.
Proof: From part (b) and part (c) we get the following useful result.
Let f is (SxT)-lattice measurable and let:
Then ΦεL1(μ). ΨεL1(λ) that is:
Therefore, the two iterated integrals are finite and equal.
Note 5: The order of integration may be reversed for (SxT)-lattice measurable functions f whenever f≥0 or when ever one of the iterated integrals of |f|is finite.
Result 2: For lattice σ-finiteness μ can not be omitted.
Proof: Let X = [0, 1] = Y, μ = Lebesgue measure on [0,1], λ = lattice counting measure on Y.
Let f(x, y) = 1 if x = y, f(x, y) = 0 if x≠y:
since, for a given y, f (x, y) = 1 when x = y and 0 at all other x. also Lebesgue lattice measure of a single point is 0.
That is:
Hence:
So:
Hence:
To show that f in (SxT)-lattice measurable.
(where, S is the class of all Lebesgue lattice measurable sets in [0, 1] and T consists of all subsets of [0, 1]).
Since f (x, y) = 1 if x = y, f (x, y) = 0 if x ≠ y, we see that f = χD where, D is the diagonal of the unit square.
Given a positive integer n:
Let Qn = (I1 x I1) V (I2 x I2) V………….. V ( In x In)
Where n = 1,
I1 = [0, 1], Q1 = I1 x I1 is the
unit square.
When:
| n | = | 2 |
| I1 | = | [0, 1/2], I2= [1/2, 1] |
| Q2 | = | [0, 1/2] x [0, 1/2] V [1/2, 1] x [1/2, 1] |
| Q3 | = | [0, 1/3] x [0, 1/3] V [1/3, 2/3] x [1/3, 2/3] V [2/3, 1] x [2/3, 1] etc. |
| Thus, Qn is finite union of super lattice measurable sets and D = ΛQn. | |
| Hence, DεSxT. | |
| Therefore, f = χD is SxT-lattice measurable. | |
| Since λ is the lattice counting measure, if: |
| a disjoint union such that λ (Yn) < ∞ for all n, then every Yn is a finite set. | |
| Hence, Y is countable, a contradiction since Y = [0, 1] | |
| Thus λ is not lattice σ-finite. | |
| Thus the lattice σ-finiteness of λ, so μ can not be omitted. | |
Result 3: The condition that f is lattice measurable with respect to S x T can not be dropped.
Proof: Consider X = Y = [0, 1], μ = λ = Lebesgue lattice measure on [0, 1], S = T = class of all Lebesgue lattice measurable sets in [0, 1].
Let us assume the following consequence of continuum hypothesis: there exists
a one-to-one map θ from [0, 1] onto a well- ordered set W such that θ
(x) has at most countably many predecessors in W for each xε[0, 1].
Let Q = {(x, y)∈XxY: θ(x) precedes θ(y) in W }.
For each xε[0, 1] Qx = {y: (x, y)εQ}.
YεQx if and only if (x, y)εQ if and only if θ(x) precedes
θ(y) in W.
Since θ(x) has at most countably many precedessors in W, there will be only countably many y’s in [0, 1] such that θ(y) processes θ(x).
Hence, all but countably many y’s in [0, 1] are such that θ(x) precedes θ(y) that is, Qx contains all but countably many points of [0, 1].
For each yε[0, 1] Qy = {x: (x, y)εQ}.
That is, xεQy if and only if (x, y)εQ if and only if θ(x) precedes θ(y).
But θ(y) has at most countably many predecessors in W.
Hence, Qy contains at most countably many points of [0,1].
Let f = χQ.
Since Qx and Qy are Borel lattice measurable, we get that fx and fy are Borel lattice measurable and:
since for any fixed x, f (x, y) =
All since
Hence:
In this result. f is not lattice measurable w.r.t. lattice σ-algebra S x T.
Hence, the condition that f is lattice measurable with respect to S x T can not be dropped.
CONCLUSION
This manuscript illustrate the concept of product lattice measurable functions and their various characterizations. In particular these functions were defined over topological spaces. Also it has been introduced and advanced the characteristics of S- lattice measurable function and T-lattice measurable function. The concept of iterated integral of a product lattice measurable function has been defined and proved that the two iterated integrals of a product lattice measurable function are finite and equal. The condition that product lattice measurable function is lattice measurable is obtained and it has been derived that the condition that product lattice measurable function is lattice measurable cannot be dropped.