Slightly Regular Measures and Measureable Sets

Abstract

Outer and inner measures of a measure are defined and used to prove results involving them on a lattice and its complement . The results concern slightly regular measures and sets such as which is the collection of -measureable sets.

1. Introduction

Let be an nonempty set and a lattice of subsets of such that and belong to . Outer and inner measures of a measure are used to generate results on a lattice as in [1], giving more results for slightly regular measures which were studied in [2, 3]. In Section 2, these outer and inner measures along with slightly regular measures are introduced with relevant notations, definitions, and properties. Some results are presented in Section 3 for a measure on a lattice. Here, the results concern mostly slightly regular measures and measureable sets.

2. Background and Notations

We follow standard notation and definitions, according to [1–9]. Let be a nonempty set and a lattice of subsets of such that and belong to . denotes the algebra generated by , and denotes those nontrivial, finitely additive measures on (. denotes those elements of that are -regular. Note is -regular if we have . Let be the set of those elements of that are -smooth on . Note is -smooth on if and implies . is the class of the strongly -smooth measures on . is strongly -smooth on if and implies . are those elements of that are -smooth on and are, equivalently, countably additive. Note that is -smooth on if and imply . We say that is a -lattice if whenever .

For , we define two outer measures: for , we have , and for we have

We have

Note if , we have , for ; then, , and if , then . We also have for our two outer measures if for and if for . is the set of all weakly regular measures on , and is the set of all vaguely regular measures on , and also if , where ; then, .

Note. It is not difficult to show that . The proof of is as follows.

If , then, , where ; then, as . Therefore, , where . Since , then so , and the proof of is given below.

Given that is a finitely subadditive outer measure, we define the notion that is regular if for every there exists such that and . In addition, if is a regular outer measure, , and , then , and if is a regular finitely additive outer measure, we have if and only if .

We have if that which implies on and if on and is regular then (see [3]).

We now define two more set functions.

Let . For any , we have which is an inner measure.

Let . For any , we have .

If is regular, then is an inner measure.

For and , we get

For the proof, see [4].

3. Results

The following theorem is proved in [2] using which is a subset of consisting of zero-one valued measures.

Theorem 1. If , , and , then .

Proof. For , we have that implies so . Now, as , which implies on and on always. Now by the definition of .
Therefore, . It follows that and as .
Therefore, giving us . This implies that is an upper bound of . So and it follows that as . Using , we get as and . Therefore, giving us .

Theorem 2. If and if there exists such that and , then .

Proof. Since , then implies . Therefore, by Theorem 1.

Theorem 3. If , , is a -lattice, , and , then and therefore .

Proof. If , then, and so . It follows that so is a lower bound of . Therefore, . Now so . Now by definition of so , and since and on , we have giving us .
Since , we have giving and as which implies so is an upper bound of giving . So we have as . Remember we showed early in the proof that and therefore we have which gives us . So . Therefore, by Theorem 1.

Theorem 4. If , then .

Proof. We know since , we have . Let , , and . so then , where for . We have and as and as ; then, for a fixed . Now since , we have and by the monotone property of , we have . So as and since we get .

Note for the next results is the collection of -measureable sets, and is the collection of -measureable sets.

Theorem 5. If , is a -lattice, and , then .

Proof. We have on ; as for , we have . Since , then on so as . Therefore, on .

From , we have with that . Therefore, as on . Therefore, on , and using for , we get . Also since is a -lattice, we get so we get . Now from Theorem 4 so and it follows from [3] that . Since , we have .

Theorem 6. If , is a -lattice, and , then .

Proof. Since and is a -lattice, then from Theorem part (c) of [3]. Now since , is a -lattice, and , then from Theorem 5. So since , we get .

Theorem 7. If , is a -lattice, and , then .

Proof. From Theorem 6, we have , so is countably additive on . Now since , then from Theorem 4 so is countably additive on . Therefore, by definition.

Theorem 8. If , then .

Proof. From Lemma of [3], if and only if . Now ; therefore, .

Theorem 9. If , then .

Proof. If , then from Theorem 8. Now using for we get .

Theorem 10. If , , and is a -lattice, then .

Proof. Since , becomes . As is a -lattice, we get so we get and since we have by Theorem 4 that so , and from [3] this implies .

Theorem 11. If and is a -lattice, then .

Proof. Let ; then, as from [3], and in [3] we have that if , then . So implies . However, , and is a -lattice implying everywhere in [3] so .
Now let for ; then, as everywhere. Therefore, and as we have .

We now end with a sort of converse to Theorem 4.

Theorem 12. If , , , , and is regular, then .

Proof. Now as and . Therefore, giving us as is regular, so we have . Then, it follows from [3].

Competing Interests

The author declares that there is no conflict of interests regarding the publication of this paper.