INTRODUCTION
Gabor (1964) has introduced the concept of product lattice.
In the recent past (Royden, 1981) has made an effort on
the concept of function lattice. Tanaka (2009) has established
a Hahn Decomposition Theorem of Signed Lattice Measure and introduced the concept
of lattice σalgebra σ(L). Recently, Anil Kumar
et al. (2011) made a categorization of Class of Measurable Borel
Lattices. Also, Anil Kumar et al. (2011) made
an investigation on Lattice Boolean Valued Measurable functions and defined
the concepts of lattice measurable space, lattice measurable set, σlattice
and δlattice.
In this study we establish the general frame work for the study of the characterization of super lattice measurable sets. Here some concepts in measure theory can be generalized by means of a lattice σalgebra σ(L). We establish super lattice, super lattice measurable set, elementary lattice, monotone class. We studied the characterization of super lattice measurable sets.
In this study, we establish union, intersection, difference of two super lattice measurable sets. Also we establish that class of elementary lattice is closed under union, intersection and difference. Finally we confirm that the product lattice σalgebra is the smallest monotone class contains all elementary lattices.
PRELIMINARIES
Here, we shall briefly review the wellknown facts about lattice theory specified
by Birkhoff (1967).
(L, ∧, ∨) is called a lattice if it is enclosed under operations ∧ and ∨ and satisfies, for any elements x, y, z, in 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 ∧ (y∨x)
= x. Hereafter, the lattice (L, ∧, ∨) will often be written as L
for simplicity. A lattice (L, ∧, ∨) is called distributive if, for
any x, y, z, in L 
• 
(L4) distributive law holds: x∨ (y∧z) = (x∨y) ∧
(x∨z) and x∧ (y∨z) = (x∧y) ∨ (x∧z) 
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 x^{c} such that:
• 
(L5) the law of excluded middle: x∨x^{c} = 1 
• 
(L6) the law of noncontradiction: x∧x^{c} = 0 
Let L be a lattice and €: L→L be an operator. Then € is called a lattice
complement in L if the following conditions are satisfied.
• 
(L5) and (L6): ∀x∈L, x∨x^{c} = 1 and
x∧x^{c} = 0 
• 
(L7) the law of contrapositive: ∀x, y∈L, x<y
implies x^{c} >y^{c} 
• 
(L8) the law of double negation: ∀x∈L, (x^{c})^{c}
= x 
Throughout this paper, we consider lattices as complete lattices which obey
(L1), (L8) except for (L6) the law of noncontradiction. Unless otherwise stated,
X is the entire set and L is a lattice of any subsets of X.
Definition 1: If a lattice L satisfies the following conditions, then it is called a lattice σAlgebra:
• 
If h_{n}∈L for n = 1, 2, 3....., then
hn∈L 
We denote σ(L), as the lattice σAlgebra generated by L.
Note 1: By definition 1, it is clear that σ(L) is closed under finite unions and finite intersections.
Definition 2: Let σ(L) be a lattice σalgebra of sub sets of a set X. A function μ: σ(L)>[0, ∞] is called a positive lattice measure defined on σ(L) if:
• 

• 

where, {A_{n}} is a disjoint countable collection of members of σ(L) and μ(A)<∞ for at least one A∈ σ(L).
Definition 3: The ordered pair (X, σ(L)) is said to be lattice measurable space.
Definition 4: A lattice A is said to be lattice measurable set if A belongs to σ(L).
Definition 5: A function lattice is a collection L^{1} of extended real valued functions defined on a lattice L^{1} with respect to usual partial ordering on functions. That is if f, g∈L^{1} then f∨g∈L^{1}, f∨g∈L^{1}.
Definition 6: If f and g are extended real valued lattice measurable functions defined on L^{f}, then f∨g, f∧g are defined by (f∨g) (x) = sup {f(x), g(x)} and (f∧g) (x) = inf {f(x), g(x)} for any x∈L.
Definition 7: Let E be a lattice then the complement of E is defined as E^{c} = {x∈E^{c}/x∉E}.
Note 2: (E^{c})^{c} = E.
Definition 8: A countable union of lattice measurable sets is called a σlattice.
Definition 9: A countable intersection of lattice measurable sets is called a δlattice.
Definition 10: Let X and Y be two lattices then their Cartesian product denoted by XxY is defined as XxY = {(x, y)/x∈X, y∈Y}.
Definition 11: If A<X, B<Y then AxB<XxY. Any lattice of the form AxB is called super lattice in XxY.
Remark 1 (Rudin, 1987): Let (X, S), (Y, T) be lattice
measurable spaces. Then S is a lattice σalgebra in X and T is a lattice
σalgebra in Y.
Definition 12: If A∈S and B∈T then the lattice of the form AxB is called super lattice measurable set.
Definition 13: If Q =R_{1}∨R_{2}∨...∨R_{n} where each R_{i} is a super lattice measurable set and R_{i}∧R_{j} = φ for i≠j, then Q is called elementary lattice. The class of all elementary lattices is denoted by L_{E}.
Definition 14: SxT is defined to be smallest lattice σ algebra in XxY which contains every super lattice measurable set.
Definition 15: If A_{i}, B_{i}∈ σ(L) such that A_{i}<A_{i+1}, B_{i}>B_{i+1} for i = 1, 2, 3, …. and
then A∈ σ(L) and B∈ σ(L), this lattice σalgebra σ(L)
is a monotone class.
Example 1: XxY is a monotone class.
Definition 16: Let E<XxY where x∈X, y∈Y, we define x section lattice of E by E_{x} = {y/ (x, y)∈E} and ysection lattice of E_{y} = {x/(x, y)∈E}.
Note 3: E_{x}<Y and E_{y}<X.
CHARACTERIZATION OF CLASS OF SUPER LATTICE MEASURABLE SETS
Result 1: The union of two super lattice measurable sets is a super lattice measurable set.
Proof: Let (A_{1}xB_{1}), (A_{2}xB_{2}) be two super lattice measurable sets, clearly A_{1}, A_{2}∈S implies A_{1}A_{2}∈S, A_{1}∧ A_{2}∈S and A_{1}∨A_{2}∈S (Since S is a lattice σalgebra). Also B_{1}, B_{2}∈T implies B_{1}B_{2}∈T, B_{1}∧B_{2}∈T and B_{1}∨B_{2}∈T (Since T is a lattice σalgebra). Now (A_{1}xB_{1})∨(A_{2}xB_{2}) = (A_{1}∨A_{2})x(B_{1}∨B_{2}). By definition of σlattice (A_{1}∨A_{2}), (B_{1}∨B_{2}) are lattice measurable sets implies (A_{1}∨A_{2})x(B_{1}∨B_{2}) is a super lattice measurable set. Therefore (A_{1}xB_{1})∨(A_{2}xB_{2}) is a super lattice measurable set.
Result 2: The intersection of two super lattice measurable sets is a super lattice measurable set.
Proof: Let (A_{1}xB_{1}), (A_{2}xB_{2}) be two super lattice measurable sets, clearly A_{1}, A_{2}∈S implies A_{1}A_{2}∈S, A_{1}∧ A_{2}∈S and A_{1}∧A_{2}∈S (Since S is a lattice σ algebra). Also B_{1}, B_{2}∈T implies B_{1}B_{2}∈T, B_{1}∧B_{2}∈T and B_{1}∨B_{2}∈T (Since T is a lattice σ algebra). Now (A_{1}xB_{1})∧(A_{2}xB_{2}) = (A_{1}∧A_{2}) x(B_{1}∧B_{2}). By definition of δ lattice (A_{1}∧A_{2}), (B_{1}∧B_{2}) are lattice measurable sets implies (A_{1}∧A_{2})x(B_{1}∧B_{2}) is a super lattice measurable set. Therefore (A_{1}xB_{1})∧(A_{2}xB_{2}) is a super lattice measurable set.
Result 3: The difference of two super lattice measurable sets is a super lattice measurable set.
Proof: Let (A_{1}xB_{1}), (A_{2}xB_{2}) be two super lattice measurable sets, clearly A_{1}, A_{2}∈S implies A_{1}A_{2}∈S, A_{1}∧ A_{2}∈S and A_{1}∨A_{2}∈S (Since S is a lattice σ algebra). Also B_{1}, B_{2}∈T implies B_{1}B_{2}∈T, B_{1}∧B_{2}∈T and B_{1}∨B_{2}∈T (Since T is a lattice σalgebra) implies (A_{1}A_{2})xB_{1} is a super lattice measurable set. (A_{1}∧A_{2})x( B_{1}B_{2}) is a super lattice measurable set implies ((A_{1}A_{2})xB_{1})∨((A_{1}∧A_{2})x( B_{1}B_{2})) is a super lattice measurable set (By definition of σlattice) implies (A_{1}xB_{1})(A_{2}xB_{2}) is a super lattice measurable set.
Theorem 1: If E∈S xT, then E_{x}∈T and E_{y}∈S for every x∈ X and y∈Y.
Proof: Let K be the class of all E∈SxT such that E_{x}∈T for every x∈X that is K = {E∈SxT/ E_{x}∈T for every x∈X}. Let F = AxB be a super lattice measurable set that is A∈S, B∈T. Also F_{x} = B if x∈A and F_{x} = φ if x∉A implies F_{x}∈T for every x∈X. Therefore F∈K. That is every super lattice measurable set belongs to K. In particular XxY∈K.
Let E∈K. Then y∈(E^{c})_{x} if and only if (x, y)∈E^{c} if and only if (x, y)∉E if and only if y∉E_{x} if and only if y∈(E_{x})^{c}. Therefore (E^{c})_{x} = (E_{x})^{c}. Since E_{x}∈T and since T is a lattice σ algebra we have E^{c}_{x}∈T. Therefore E^{c}∈K.
Let E_{i}∈K (i = 1, 2, 3, ….) and let:
then Y∈E_{x} if and only if (x, y)∈E if and only if (x, y)∈E_{c}
for some i if and only if y∈(E_{i})_{x}. Therefore:
Since T is a lattice σ algebra,
(E_{i})_{x}∈T implies E_{x}∈T. Therefore E∈T.
From
K is a lattice σalgebra. Since K<SxT we get K = SxT. Hence for any
E∈SxT, E_{x}∈T for every x∈X. In a similar way we can
prove E_{y}∈S for every y∈Y.
Lemma 1: To prove SxT is a monotone class.
Proof: Let A_{i}, B_{i}∈SxT, A_{i}<A_{i+1}, B_{i}>B_{i+1} for i = 1, 2, 3, …. and:
Since SxT is a lattice σ algebra implies A∈SxT and:
implies B∈SxT. Since SxT is a lattice σ algebra. Therefore SxT is
a monotone class.
Lemma 2: To prove L_{E} is closed under intersection, difference and unions.
Proof: Let A_{1}xB_{1}, A_{2}xB_{2} be two super lattice measurable sets. Clearly (A_{1}xB_{1})∧(A_{2}xB_{2}) = (A_{1}∧A_{2})x(B_{1}∧B_{2}), we get (A_{1}∧A_{2})x(B_{1}∧B_{2}) is super lattice measurable set (Since by definition of δlattice and every δlattice is lattice measurable). Also (A_{1}xB_{1})(A_{2}xB_{2}) = ((A_{1}A_{2})x B_{1})∨((A_{1} ∧A_{2})x(B_{1}B_{2})), we get the difference of two super lattice measurable sets is a union of two disjoint super lattice measurable sets. (Since by σlattice and every σlattice is lattice measurable). [Note that A_{1}, A_{2}∈S implies A_{1}A_{2}∈S and A_{1}∧A_{2}∈S since S is a lattice σalgebra. Also B_{1}B_{2}∈T, B_{1}∈T implies (A_{1}A_{2})xB_{1}, (A_{1}∧A_{2})x(B_{1}B_{2}) are super lattice measurable sets and they are disjoint since (A_{1}∧A_{2})∧(A_{1} A_{2}) = φ]. Hence (A_{1}xB_{1})(A_{2}xB_{2})∈L_{E}.
Part 1: Closed under intersection. Let P, Q∈L_{E} implies P = R_{1}∨R_{2}∨.....∨R_{n}, Q = R^{1}_{1}∨R^{1}_{2}∨.....∨R^{1}_{m} where R_{i}∧R_{j} = φ for i≠j, R^{1}_{i}∧R^{1}_{j} = φ for i≠j and R_{i}’s∧R_{j}’s are super lattice measurable sets (By result3..). Now P∧Q = (R_{1}∧R^{1}_{1})∨(R_{1}∧R^{1}_{2})∨….∨(R_{1}∧R^{1}_{m})∨.....∨(R_{n}∧R^{1}_{1})∨......∨(R_{n} ∧R^{1}_{m}). Here each R_{i}∧R_{j} is super lattice measurable set (By result 3.2) and clearly these are disjoint. Therefore P∧Q∈ L_{E}.
Part 2: Closed under difference. Now PQ = P∧Q^{c} = (R_{1}∨R_{2}∨......∨R_{n})∧(R^{1}_{1}∨R^{1}_{2}∨......∨R^{1}_{m})^{c}
= (R_{1}∨R_{2}∨......∨R_{n})∧(R^{1c}_{1}∧R^{1c}_{2}∧.....∧R^{1c}_{m})
= (R_{1}∧(R^{1c}_{1}∧R^{1c}_{2}∧.....∧R^{1c}_{m}))∨R_{2}∧
(R^{1c}_{1}∧R^{1c}_{2}∧......∧R^{1c}_{m})∨......∨R_{n}∧(R^{1c}_{1}∧R^{1c}_{2}∧......∧R^{1c}_{m}))
= {([(R_{1}R^{1}_{1})R^{1}_{2}]…..)R^{1}_{m}}∨.....∨{([(_{RnR11})R^{1}_{2}]......)R^{1}_{m}}.
Since the difference of two super lattice measurable sets is the disjoint union of two super lattice measurable set (By result 3), we get right hand side is the disjoint union of super lattice measurable sets. Hence PQ∈L_{E}.
Part 3: Closed under union. Now P∨Q = (PQ)∨Q and (P Q)∧Q = φ, we get P∨Q∈L_{E}. Therefore L_{E} is closed under intersection, difference and unions.
Theorem 2: SxT is the smallest monotone class which contains all elementary lattices.
Proof: Let σ(L) be the smallest monotone class which contains L_{E}. This can be exists, let M be the family of all monotone class containing L_{E}. Since XxY∈M implies M is nonempty. Let F be the intersection of all members of M that is F = ∧M. Then L_{E}<F. We prove F is monotone class. Let A_{i}. B_{i}∈F, A_{i}<A_{i+1}, B_{i}>B_{i+1} for i = 1, 2, 3, …. and
Then A_{i} , B_{i} belongs to every member of M and since every
member of M is a monotone class, A, B belongs to every member of M therefore
A, B∈F. Therefore F is a monotone class. By lemma1, SxT is a monotone class
also it is obvious that L_{E}<SxT and since σ(L) be the smallest
monotone class which contains L_{E} we get L:
Now for any lattice P<XxY define K(P) = {Q<XxY/PQ∈ σ(L), QP∈ σ(L) and P∨Q∈ σ(L)}. Clearly Q∈K(P) iff PQ, Q P, P∨Q∈ σ(L) iff QP, PQ, Q∨P∈ σ(L) iff P∈K(Q) that I s:
Let A_{i}, B_{i}∈K(P), A_{i}<A_{i+1}, B_{i}>B_{i+1} for i = 1, 2, 3, …. and:
then A_{i}P, PA_{i}, P∨A_{i}, B_{i}P,
PB_{i}, P∨B_{i} all belongs to σ(L) for every i, also:
(Since σ(L) is a monotone class).
Since PA_{i}∈ σ(L) and PA_{i}>PA_{i+1}
we get:
(Since σ(L) is a monotone class) implies PA∈ σ(L). Again:
(Since each P∨A_{i}∈ σ(L) and σ(L) is a monotone
class). Hence A∈K(P). Similarly B∈K(P). Therefore K(P) is a monotone
class. Fix P∈L_{E}. Let Q be any lattice of L_{E}. Then
PQ, QP, P∨Q∈ L_{E} (By lemma2). Hence PQ, QP, P∨Q∈
σ(L) (Since L_{E}< σ(L). Therefore Q∈K(P) that is
Q∈K(P) for every Q∈L_{E} that is L_{E} <K(P) and
by Eq. 3 K(P) is a monotone class. But σ(L) is the smallest
monotone class containing L_{E}. Therefore σ(L)<K(P). Again
fix Q∈ σ(L) for any P∈L_{E}. By Eq. 5
we have Q∈K(P). Therefore by Eq. 2 P∈K(Q). Therefore
by Eq. 4 L_{E}∈K(Q) and hence by Eq.
5 σ(L)<K(Q). Thus σ(L)<K(Q) for every Q∈ σ(L).
Let P, Q∈ σ(L) then Q∈K(P). Therefore QP, PQ, P∨Q∈
σ(L). Therefore σ(L) is closed under difference and since:
Let Q∈ σ(L) then Q^{c} = (XxY)Q∈ σ(L). By Eq.
6. Let P_{i}∈ σ(L) for i = 1, 2, 3, ….. Let P = ∨P_{i}
and Q_{n} = P_{1}∨P_{2}∨....∨P_{n}.
Since by Eq. 6 σ(L) is closed under finite union Q_{n}∈
σ(L). Since Q_{n}<Q_{n+1} and P = ∨Q_{n},
the monotonicity of σ(L) shows that P∈ σ(L). Thus σ(L) is
a lattice σalgebra. Also L_{E}< σ(L) ∈S xT. But SxT
is the smallest lattice σalgebra in XxY contains every super lattice measurable
set. Here σ(L) is a lattice σalgebra containing L_{E} and
hence every super lattice measurable set. Therefore σ(L) = SxT.