Measurable Iso-Functions
Svetlin G. Georgiev
Department of Mathematics, Sorbonne University, Paris, France
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To cite this article:
Svetlin G. Georgiev. Measurable Iso-Functions. American Journal of Modern Physics. Special Issue: Issue I: Foundations of Hadronic Mathematics. Vol. 4, No. 5-1, 2015, pp. 24-34. doi: 10.11648/j.ajmp.s.2015040501.13
Abstract: In this article are given definitions definition for measurable is-functions of the first, second, third, fourth and fifth kind. They are given examples when the original function is not measurable and the corresponding iso-function is measurable and the inverse. They are given conditions for the isotopic element under which the corresponding is-functions are measurable. It is introduced a definition for equivalent iso-functions. They are given examples when the iso-functions are equivalent and the corresponding real functions are not equivalent. They are deducted some criterions for measurability of the iso-functions of the first, second, third, fourth and fifth kind. They are investigated for measurability the addition, multiplication of two iso-functions, multiplication of iso-function with an iso-number and the powers of measurable iso-functions. They are given definitions for step iso-functions, iso-step iso-functions, characteristic iso-functions, iso-characteristic iso-functions. It is investigate for measurability the limit function of sequence of measurable iso-functions. As application they are formulated the iso-Lebesgue’s theorems for iso-functions of the first, second, third, fourth and fifth kind. These iso-Lebesgue’s theorems give some information for the structure of the iso-functions of the first, second, third, fourth and fifth kind.
Keywords: Measurable Iso-Sets, Measurable Is-Functions, Is-Lebesgue Theorems
1. Introduction
Genious idea is the Santilli’s generalization of the basic unit of quantum mechanics into an integro-differential operator which is as positive-definite as +1 and it depends of local variables and it is assumed to be the inverse of the isotopic element
and it is called Santilli isounit. Santilli introduced a generalization called lifting of the conventional associative product into the form
Called isoproduct for which
For every element of the field of real numbers, complex numbers and quaternions. The Santilli isonumbers are defi ned as follows: for given real number or complex number or quaternion ,
With isoproduct
.
If , the corresponding isoelement of will be denoted withor .
With we will denote the field of the is-numbers for which and basic isounit .
In [1], [3]-[12] are defined isocontinuous isofunctions and isoderivative of isofunction and in [1] are proved some of their properties. If is an isoset in , the class of isofunctions is denoted by and the class of isodifferentiable isofunctions is denoted by , with the same basic isounit , it is supposed
.
Here is the corresponding real set of . If is an independent variable, then the corresponding lift is , if is real-valued function on , then the corresponding lift of first kind is defined as follows
and we will denote it by .
In accordance with [1], the isodifferential is defined as follows
.
Then
_{,}
_{.}
In accordance with [1], the first is-derivative of the is-function is defined as follows
When , then
Our aim in this article is to be investigated some aspects of theory of measurable iso-functions. The paper is organized as follows. In Section 2 are defined measurable iso-functions and they are deducted some of their properties. In Section 3 is investigated the structure of the measurable iso-functions.
2. The Definition and the Simplest Properties of Measurable Is-Functions
We suppose that is a given point set, , for every be a given constant, bea given real-valued function. With we will denote the corresponding is-function of the first, second, third, fourth and fifth kind. More precisely,
1. , when is an is-function of the first kind.
2. , when for when is an is-function of the second kind.
3. , for when is an is-function of the third kind.
4. when for , when is an is-function of the fourth kind.
5. , for when is an is-function of the fifth kind.
For with we will denote the set
We define the symbols , , , and etc., in the same way.
If the set on which the is-function is defined is designated by a letter C or D, we shall write >a) or >a).
Definition 2.1. The is-function is said to be measurable if
1. The set A is measurable.
2. The set is measurable for all .
Theorem 2.3. Let be a measurable is-function defined on the set A. If B is a measurable subset of A, then the is-function , considered only for, is measurable.
Proof. Let be arbitrarily chosen and fixed. We will prove that
(1)
Really, let be arbitrarily chosen. Then and (x)>a. Since , we have that . From and it follows that . Because was arbitrarily chosen and for it we get that it is an element of the set , we conclude that
(2)
Let now be arbitrarily chosen. Then and . Hence and . Therefore Because was arbitrarily chosen and we get that it is an element of we conclude that
.
From the last relation and from (2) we prove the relation (1).
Since the iso-function is a measurable function on the set A, we have that is a measurable set. As the intersection of two measurable sets is a measurable set, we have that is a measurable set. Consequently, using (1), the set >a) is measurable set. In this way we have
1. B is a measurable set,
2. is a measurable set for all .
Therefore the iso-function , considered only for , is a measurable is-function.
Theorem 2.4. Let be defined on the set A, which is the union of a finite or denumerable number of measurable sets , . If is measurable on each of the sets , then it is also measurable on A.
Proof. Let be arbitrarily chosen. We will prove that
(3)
Let be arbitrarily chosen. Then and . Since and , there exists k such that . Therefore and . Hence, and . Because was arbitrarily chosen and for it we get that it is an element of , we conclude that
(4)
Let now be arbitrarily chosen. Then there exists l such that >a). From here and Hence, and Consequently . Because was arbitrarily chosen and for it we get that it is an element of we conclude that
From the last relation and from (4) we prove the relation (3).
Since the union of finite or denumerable number of measurable sets is a measurable set, using that the sets are measurable, we obtain that A and are measurable sets. Therefore is a measurable is-function.
Definition 2.5. Two is-functions and , defined on the same setr A, are said to be equivalent if
=0.
We will write
.
Remark 2.6. There is a possibility and in the same time.
Let
Then
.
On the oth-1+er hand,
We have that
Or
Remark 2.7. There is a possibility and in the same time . Let
Then
.
On the other hand,
=.
Then
Therefore
Hence,
Consequently
.
Proposition 2.8. The fuAnctions f and g are equivalent if and only if the functions and are equivalent
Proof. We have
Definition 2.9. Let some property P holds for all the points of the set A, except for the points of a subset B of the set A. If , we say that the property P holds almost everywhere on the set A, or for almost all points of A.
Definition 2.10. We say that two is-functions defined on the set A are equivalent if they are equal almost everywhere on the set A.
Theorem 2.11. If is a measurable is-function defined on the set A, and if , then the is-function is also measurable.
Proof. Let
Because we have that
or
Since every function, definite on a set with measure zero is measurable on it, we have that the is-function is measurable on the set B.
We note that the is-functions and are identical on D and since the is0-function is measurable on D, we get that the is-function is measurable on D.
Consequently the is-function is measurable on
Theorem 2.12. If the is-function , defined on the set A, is measurable, then the sets
),
Are measurable for all .
Proof. We will prove that
(5)
Really, let be arbitrarily chosen. Then and . Hence, for every we have . Therefore
.
Because was arbitrarily chosen and for it we obtain
We conclude that
(6)
Let now be arbitrarily chosen. Then for every natural number n. From here and
For all natural number n. Consequently
or
and . Since was arbitrarily chosen and we get that , we conclude
From the last relation and from (6) we obtain the relation (5).
Because the intersection of denumerable measurable sets is a measurable set, using the relation (5) and the fact that all sets are measurable for all natural numbers n, we conclude that the set is a measurable set.
The set is a measurable set because
The set is measurable set since
The set is measurable since
Remark 2.13. We note that if at least one of the sets
),
Is measurable for all , then the iso-function is measurable on the set A.
Really, let is measurable for all . Then, using the relation
(7)
we obtain that the set is measurable for all .
If is measurable for all , then using the relation
we get that the set is measurable for all .
If is measurable for all , then using the relation
,
We conclude that the set is measurable for all .
Theorem 2.14. If for all points of a measurable set A, then the is-function is measurable.
Proof. For all we have that
.
Since the sets A and are measurable sets, then is measurable for all . Therefore the is-function is measurable.
Definition 2.15. An is-function defined on the closed interval [a, b] is said to be a step is-function if there is a finite number of points
Such that is a constant on , .
Proposition 2.16. A step is-function is measurable.
Proof. Let (x) is a step is-function on the closed interval [a, b]. Let also,
be such that is a constant on , . From the previous theorem we have that is measurable on , . We note that
the sets , , are sets with measure zero. Therefore the is-function
is measurable on , . From here, using that
We conclude that the is-function is measurable on [a, b].
Theorem 2.17. If the is-function , defined on the set A is measurable and then the is-functions
1. ,
2.
3.
4.
5.
are also measurable.
Proof. Let be arbitrarily chosen. The assertion follows from the following relations.
1.
2. if c>0, if c<0.
3. if a<0, if
4. if a<0, if .
5. if a>0, if a<0, if a=0.
Definition 2.18. An is-function , defined on the closed interval [pa, b], is said to be is-step is-function, if there is a finite number of points
Theorem 2.19. Let >0 for every and is measurable on [a, b]. Let also, is an iso-step is-function on [a, b]. Then is measurable on [a, b].
Proof. Let
be
From the last theorem it follows that is a measurable is-function on , Fromn-1 here and from
Since {b} is a set with measure zero, we conclude that the is-step is-function is measurable on [a, b].
Definition 2.20. Let M be a subset of the closed interval [a, b]. The function for and for is called the characteristic function of the set M.
Theorem 2.21. If the set M is a measurable subset of the closed interval A=[a, b], then the characteristic function is measurable on [a, b].
Proof. The assertion follows from the following relations. if if if a<0.
Definition 2.22. Let M be a subset of the set A=[a, b]. The iso- function if and if , will be called characteristic is-function of the set M.
Theorem 2.23. Let be a measurable function on A=[a, b], M be a measurable subset of A. Then the characteristic is-function of the set M is measurable.
Proof. Let be arbitrarily chosen. Then
,
From here, using that the sets and are measurable sets, we conclude that is a measurable set. Because the constant a was arbitrarily chosen, we have that the characteristic function is a measurable is-function.
Theorem 2.24. Let f and are continuous functions on the closed set A. Then the is-function is measurable.
Proof. Let be arbitrarily chosen. Since every closed set is a measurable set, we conclude that the set A is a measurable set.
We will prove that the set is a closed set.
Let be a sequence of elements of the set such that
Since is a subset of the set A we have that . Because the set A is a closed set, we obtain that . From the definition of the set we have that
Hence, when , using that f and are continuous functions on the set A, we get
i.e., . Therefore the set is a closed set. From here, the set is a measurable set. Because the difference of two measurable sets is a measurable set, we have that the set
Is a measurable set.
Since was arbitrarily chosen, we obtain that the is-function of the first kind is measurable.
Theorem 2.25. Let f and are continuous functions on the closed set A. The the is-functions
are measurable on A.
Theorem 2.26. If two measurable is-functions and are defined on the set A, then the set is measurable.
Proof. We enumerate all rational numbers
.
We will prove that
(8)
Let
Be arbitrarily chosen. Then
There exists a rational number such that
Therefore
i.e.,
).
Consequently
)
And
Because was arbitrarily chosen and for it we get
we conclude that
(9)
Let no
be arbitrarily chosen. Then there exists a natural k so that
.
Hence,
Then
or
Therefore
Because
Was arbitrarily chosen and for it we get that , we conclude that
From the last relation and from the relation (9) we get the relation (8).
Since and are measurable iso-functions on A, we have that the sets
are measurable sets for every natural k, whereupon the sets
Are measurable sets for every natural k.
Therefore, using the relation (8), we obtain that the set is a a measurable set.
Theorem 2.27. Let and be finite measurable is-functions on the set A. Then each of the is-functions
1.
2.
3.
4. if on A,
Is measurable.
Proof.
1. Let be arbitrarily chosen. Since is measurable, then is measurable. From here and from the last theorem it follows that the set
Is measurable. Because was arbitrarily chosen, we conclude that the function is measurable.
2. Since is a measurable is-function, we have that the function is a measurable is-function. From here and from 1) we conclude that the is-function
Is measurable.
3. We note that
(10)
Since and are measurable iso-functions, using 1) and 2) we have that
Are measurable is-functions. Hence the is-functions
Are measurable, whereupon
Are measurable. From here, using 1) and (10), we conclude that is measurable.
4. Since is measurable and on A, we have that the is-function is measurable. From here and from 3) the is-function
Is measurable.
Theorem 2.28. Let be a sequence of measurable is-functions defined on the set A. If
(11)
Exists for every then the is-function is measurable.
Proof. Let be arbitrarily chosen. For we define the sets
We will prove that
(12)
Let
Be arbitrarily chosen. Then
Hence, there is enough large natural number such that
Using (11), there are enough large natural numbers k and m such that
i.e.,
From here, it follows that there is enough large n so that for every , i.e., and then
Since was arbitrarily chosen and we get that it is an element of the set , we conclude that
(13)
Let now be arbitrarily chosen.
Then, there are so that
or
.
Hence,
or
Therefore
Since was arbitrarily chosen and for it we obtain , we conclude that
From the last relation and from (13) it follows the relation (12).
Since are measurable, we have that the sets are measurable for every hence are measurable for every and then, using (12), the set is measurable. Consequently the is-function is measurable.
Theorem 2.29. be a sequence of measurable is-functions defined on the set A. If
(14)
Exists for almost everywhere then the is-function is measurable.
Proof. Let B be the subset of A so that the relation (14) holds for every . From the previous theorem it follows that the is-function is measurable on the set B.
We note that
Therefore the is-function is measurable on . Hence, the is-function is measurable on A.
Let
Then
1.
2.
If
3.
If
4.
If
5.
If
.
3. The Structure of the Measurable Is-Functions
Theorem 3.1. (is-Lebesgue theorem for is-functions of the first kind) Let there be given a sequence of measurable functions on a set A, all of which are finite almost everywhere. Let also, be a sequence of measurable functions on the set A,
For all natural numbers n and for all where and are positive constants. Suppose that
Almost everywhere on the set A, and is finite almost everywhere on A,
For all Then
For all
Proof. We will note that the limit functions f(x) and are measurable and the sets under considerations are measurable.
Let
Since
,
using the properties of the measurable sets, we have that
Let
We have that
.
Hence,
Let us assume that Then, using the definition of the set , we have
Since
we have that
and their limit
are finite.
Therefore there is an enough large natural n such that
for every . Then where and from here
Consequently
Because , from the last relation, we have that i.e.,
,
and since
,
or
.
As in above one can prove the following results for the other kinds of is-functions.
Theorem 3.2. (is-Lebesgue theorem for is-functions of the second kind) Let there be given a sequence of measurable functions on a set A, all of which are finite almost everywhere. Let also, be a sequence of measurable functions on the set A,
For all natural numbers n and for all where and are positive constants. Suppose that
Almost everywhere on the set A, and is finite almost everywhere on A,
For all Then
for all
Theorem 3.3. (is-Lebesgue theorem for is-functions of the third kind) Let there be given a sequence of measurable functions on a set A, all of which are finite almost everywhere. Let also, be a sequence of measurable functions on the set A,
For all natural numbers n and for all where and are positive constants. Suppose that
Almost everywhere on the set A, and is finite almost everywhere on A,
For all Then
for all
Theorem 3.4. (is-Lebesgue theorem for is-functions of the fourth kind) Let there be given a sequence of measurable functions on a set A, all of which are finite almost everywhere. Let also, be a sequence of measurable functions on the set A,
For all natural numbers n and for all where and are positive constants. Suppose that
Almost everywhere on the set A, and is finite almost everywhere on A,
For all Then
For all
Theorem 3.5. (is-Lebesgue theorem for is-functions of the fifth kind) Let there be given a sequence of measurable functions on a set A, all of which are finite almost everywhere. Let also, be a sequence of measurable functions on the set A,
For all natural numbers n and for all where and are positive constants. Suppose that
Almost everywhere on the set A, and is finite almost everywhere on A,
For all Then
for all
References