# Slightly Regular Measures and Measureable Sets

Authors: James Camacho
Journal: Journal of Mathematics
 (2016)

---

## Abstract

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


---

## Body

## 1. Introduction

LetX be an nonempty set and l a lattice of subsets of X such that ϕ and X belong to l. 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 X be a nonempty set and l a lattice of subsets of X such that ϕ and X belong to l. A
(
l
) denotes the algebra generated by l, and M
(
l
) denotes those nontrivial, finitely additive measures on A(l
). M
R
(
l
) denotes those elements of M
(
l
) that are l-regular. Note μ is l-regular if we have μ
(
E
)
=
s
u
p
μ
(
L
)
∣
L
⊂
E
,
E
∈
A
(
l
)
,
L
∈
l. Let M
σ
(
l
) be the set of those elements of M
(
l
) that are σ-smooth on l. Note μ is σ-smooth on l if L
n
,
L
∈
l and L
n
↓
ϕ implies μ
(
L
n
)
→
0. M
σ
(
l
) is the class of the strongly σ-smooth measures on l. μ is strongly σ-smooth on l if L
n
∈
l and L
n
↓
L implies μ
(
L
)
=
inf
⁡
μ
L
n
=
lim
⁡
μ
(
L
n
). M
σ
(
l
) are those elements of M
(
l
) that are σ-smooth on A
(
l
) and are, equivalently, countably additive. Note that μ is σ-smooth on A
(
l
) if L
n
,
L
∈
A
(
l
) and L
n
↓
ϕ imply μ
(
L
n
)
→
0. We say that l is a δ-lattice if ⋂
i
L
i
∈
l whenever L
i
∈
l.ForE
⊂
X, we define two outer measures: for μ
∈
M
(
l
), we have μ
′
(
E
)
=
inf
⁡
μ
(
L
′
)
∣
E
⊂
L
′
,
L
∈
l, and for μ
∈
M
σ
(
l
) we have (1)
μ
′
′
E
=
inf
⁡
∑
i
μ
L
i
′
∣
E
⊂
⋃
i
L
i
′
,
L
i
∈
l
.We have(2)
μ
≤
μ
′
l
,
μ
=
μ
′
l
′
μ
=
μ
′
l
′
,
μ
′
′
≤
μ
l
′
,
μ
′
′
≤
μ
′
.Note ifμ
∈
M
σ
(
l
), we have μ
≤
μ
′
(
l
), for μ
∈
M
σ
(
l
); then, μ
′
=
μ
(
l
′
), and if μ
∈
M
R
σ
(
l
), then μ
′
=
μ
(
l
). We also have for our two outer measures μ
∈
M
w
l
⊆
M
l if μ
L
′
=
sup
⁡
μ
′
B
∣
B
⊆
L
′
,
B
∈
l for L
∈
l and μ
∈
M
v
l
⊆
M
σ
l if μ
L
′
=
sup
⁡
μ
′
′
B
∣
B
⊆
L
′
,
B
∈
l for L
∈
l. M
w
l is the set of all weakly regular measures on A
l, and M
v
l is the set of all vaguely regular measures on A
l, and also μ
∈
M
S
l
⊆
M
l if ⋂
i
L
i
=
L, where L
,
L
i
∈
l; then, μ
(
L
)
=
inf
⁡
μ
(
L
i
).Note. It is not difficult to show that M
R
σ
l
⊂
M
S
l
⊂
M
σ
l
⊂
M
σ
(
l
)
⊂
M
σ
(
l
). The proof of M
S
l
⊂
M
σ
(
l
) is as follows.IfL
i
↓
ϕ, then, ⋂
i
L
i
=
ϕ, where L
i
∈
l; then, μ
(
ϕ
)
=
inf
⁡
μ
(
L
i
) as μ
∈
M
S
l. Therefore, μ
(
ϕ
)
+
ϵ
>
μ
(
L
i
), where ϵ
>
0. Since μ
(
ϕ
)
=
0, then μ
(
L
i
)
<
ϵ so μ
∈
M
σ
l, and the proof of M
S
l
⊂
M
σ
(
l
) is given below.Given thatμ is a finitely subadditive outer measure, we define the notion that μ is regular if for every A
⊂
X there exists E
∈
S
μ such that A
⊂
E and μ
(
A
)
=
μ
(
E
). In addition, if μ is a regular outer measure, E
i
↑, and E
i
⊂
X, then μ
(
l
i
m
E
i
)
=
l
i
m
μ
(
E
i
), and if μ is a regular finitely additive outer measure, we have E
∈
S
μ if and only if u
(
L
)
+
μ
(
L
′
)
=
u
(
X
).We have ifμ
∈
M
σ
l that μ
∈
M
S
l which implies μ
′
′
=
μ
′ on l
′ and if μ
′
′
=
μ
′ on l
′ and μ
′
′ is regular then μ
∈
M
S
l (see [3]).We now define two more set functions.Letμ
∈
M
l. For any E
⊆
X, we have μ
i
E
=
sup
⁡
μ
L
∣
L
⊂
E
,
L
∈
l which is an inner measure.Letμ
∈
M
σ
l. For any E
⊆
X, we have μ
j
E
=
μ
X
-
μ
′
′
E
′.Ifμ
′
′ is regular, then μ
j is an inner measure.Forμ
∈
M
σ
l and E
⊆
X, we get(3)
μ
i
E
≤
μ
j
E
≤
sup
⁡
μ
′
′
⋂
1
∞
L
i
,
⋂
L
i
⊂
E
,
∣
L
i
∈
l
≤
μ
′
′
E
≤
μ
′
E
1
.For the proof, see [4].


## 3. Results

The following theorem is proved in [2] using I
S
l which is a subset of M
S
l consisting of zero-one valued measures.Theorem 1.
Ifμ
,
ν
∈
M
S
l, μ
≤
ν
(
l
), and μ
(
X
)
=
ν
(
X
), then μ
=
ν.Proof.
ForL
∈
l
⊆
X, we have that μ
≤
ν
l implies ν
≤
μ
l
′ so μ
L
′
≥
ν
(
L
′
). Now, μ
L
′
+
ϵ
≥
ν
(
L
′
)
+
ϵ
=
ν
′
′
(
L
′
)
+
ϵ as μ
∈
M
S
l, which implies μ
′
′
=
μ
′ on l
′ and μ
=
μ
′ on l
′ always. Now μ
L
′
+
ϵ
≥
ν
(
L
′
)
+
ϵ
=
ν
′
′
(
L
′
)
+
ϵ
>
∑
ν
(
L
n
′
)
>
ν
(
L
n
′
) by the definition of ν
′
′.
Therefore,μ
L
′
+
ϵ
≥
ν
(
L
n
′
). It follows that -
μ
L
′
-
ϵ
≤
-
ν
(
L
n
′
) and μ
(
X
)
-
μ
L
′
-
ϵ
≤
ν
(
X
)
-
ν
(
L
n
′
) as μ
(
X
)
=
ν
(
X
).
Therefore,μ
L
-
ϵ
≤
ν
(
L
n
) giving us μ
L
≥
ν
(
L
n
). This implies that μ
L is an upper bound of ν
(
L
n
). So sup
⁡
ν
(
L
n
)
≤
μ
L and it follows that inf
⁡
μ
(
L
n
)
≤
inf
⁡
ν
L
n
≤
sup
⁡
ν
(
L
n
)
≤
μ
L as μ
≤
ν
(
l
). Using μ
,
ν
∈
M
S
l, we get μ
(
L
)
≤
ν
(
L
)
≤
sup
⁡
ν
(
L
n
)
≤
μ
L as inf
⁡
μ
(
L
n
)
=
μ
L and inf
⁡
ν
(
L
n
)
=
ν
L. Therefore, μ
(
L
)
≤
ν
(
L
)
≤
μ
L giving us μ
=
ν.Theorem 2.
Ifμ
∈
M
S
l and if there exists ν
∈
M
R
σ
l such that μ
≤
ν
(
l
) and μ
(
X
)
=
ν
(
X
), then μ
=
ν.Proof.
SinceM
R
σ
l
⊂
M
S
l, then ν
∈
M
R
σ
l implies ν
∈
M
S
l. Therefore, μ
=
ν by Theorem 1.Theorem 3.
Ifμ
∈
M
S
l, μ
≤
ν
(
l
), l is a δ-lattice, μ
(
X
)
=
μ
′
(
X
), and ν
∈
M
σ
l, then ν
∈
M
S
l and therefore μ
=
ν.Proof.
IfL
i
↓
L, then, ⋂
i
L
i
=
L and so L
=
⋂
i
L
i
⊆
L
i. It follows that ν
L
≤
ν
L
i so ν
L is a lower bound of ν
L
i. Therefore, ν
L
≤
inf
⁡
ν
L
i. Now inf
⁡
ν
L
i
≤
sup
⁡
ν
L
i so ν
L
≤
inf
⁡
ν
L
i
≤
sup
⁡
ν
L
i. Now μ
′
′
L
′
+
ϵ
>
μ
L
i
′ by definition of μ
′
′ so -
μ
′
′
L
′
-
ϵ
<
-
μ
L
i
′, and since μ
X
=
μ
′
X and μ
′
=
μ
′
′ on l
′, we have μ
X
-
μ
′
L
′
-
ϵ
<
μ
X
-
μ
L
i
′ giving us μ
′
L
-
ϵ
<
μ
L
i.
Sinceμ
≤
μ
′
(
l
), we have μ
L
-
ϵ
<
μ
′
L
-
ϵ
<
μ
L
i giving μ
L
-
ϵ
<
μ
L
i and μ
L
-
ϵ
<
μ
L
i
<
ν
L
i as μ
≤
ν
(
l
) which implies μ
L
≥
ν
L
i so μ
L is an upper bound of ν
L
i giving μ
L
≥
sup
⁡
ν
L
i. So we have sup
⁡
ν
L
i
≤
μ
L
≤
ν
L as μ
≤
ν
(
l
). Remember we showed early in the proof that ν
L
≤
inf
⁡
ν
L
i
≤
sup
⁡
ν
L
i and therefore we have ν
L
≤
inf
⁡
ν
L
i
≤
sup
⁡
ν
L
i
≤
μ
L
≤
ν
L which gives us ν
L
=
inf
⁡
ν
L
i. So ν
∈
M
S
l. Therefore, μ
=
ν by Theorem 1.Theorem 4.
Ifμ
∈
M
S
l
∩
M
σ
l, then μ
=
μ
′
′
(
l
).Proof.
We know sinceμ
∈
M
σ
(
l
), we have μ
≤
μ
′
(
l
). Let ϵ
>
0, L
i
↓
L, and L. L
i
∈
l so then μ
′
′
L
′
+
ϵ
>
∑
i
μ
(
L
i
′
), where L
′
⊂
⋃
i
L
i
′ for L
i
∈
L. We have L
=
⋂
(
L
i
∪
L
) and L
i
∪
L
↓
L as L
i
↓
L and as μ
∈
M
S
l; then, μ
(
L
)
+
ϵ
>
μ
(
L
N
∪
L
) for a fixed N. Now since μ
≤
μ
′
(
l
), we have μ
′
(
L
)
+
ϵ
>
μ
(
L
)
+
ϵ
>
μ
(
L
N
∪
L
) and by the monotone property of μ, we have μ
′
(
L
)
+
ϵ
>
μ
(
L
)
+
ϵ
>
μ
(
L
N
∪
L
)
>
μ
(
L
). So μ
′
(
l
)
≤
μ
(
l
) as μ
′
(
L
)
+
ϵ
>
μ
(
L
) and since μ
≤
μ
′
(
l
) we get μ
=
μ
′
′
(
l
).Note for the next resultsS
μ
′ is the collection of μ
′-measureable sets, and S
μ
′
′ is the collection of μ
′
′-measureable sets.Theorem 5.
Ifμ
∈
M
S
l
∩
M
σ
l, l is a δ-lattice, and μ
(
X
)
=
μ
′
(
X
), then l
⊂
S
μ
′.Proof.
We haveμ
j
=
μ
′
=
μ
′
′ on l; as for L
∈
l
⊆
X, we have μ
j
L
=
μ
X
-
μ
′
′
L
′. Since μ
∈
M
S
l, then μ
′
′
=
μ
′ on l
′ so μ
j
L
=
μ
X
-
μ
′
L
′
=
μ
′
L as μ
′
L
′
+
μ
′
(
L
′
)
=
μ
′
(
X
). Therefore, μ
j
=
μ
′ on l.Fromμ
i
E
≤
μ
j
E
≤
sup
⁡
μ
′
′
⋂
1
∞
L
i
,
⋂
L
i
⊂
E
,
∣
L
i
∈
l
≤
μ
′
′
E
≤
μ
′
E
1, we have with E
=
L that μ
i
L
≤
μ
j
L
≤
s
u
p
μ
′
′
⋂
1
∞
L
i
,
⋂
L
i
⊂
L
,
∣
L
i
∈
l
≤
μ
′
′
L
≤
μ
′
L. Therefore, μ
i
L
≤
μ
j
L
≤
sup
⁡
μ
′
′
⋂
1
∞
L
i
,
⋂
L
i
⊂
L
,
∣
L
i
∈
l
≤
μ
′
′
L
≤
μ
′
L
=
μ
j
(
L
) as μ
j
=
μ
′ on l. Therefore, μ
j
=
μ
′
=
μ
′
′ on l, and using μ
i
E
≤
μ
j
E
≤
sup
⁡
μ
′
′
⋂
1
∞
L
i
,
⋂
L
i
⊂
E
,
∣
L
i
∈
l
≤
μ
′
′
E
≤
μ
′
E
1 for E
∈
l, we get μ
′
E
=
sup
⁡
μ
′
′
⋂
1
∞
L
i
,
⋂
L
i
⊂
E
,
∣
L
i
∈
l. Also since l is a δ-lattice, we get ⋂
1
∞
L
i
=
L
∈
l so we get μ
′
E
=
s
u
p
μ
′
′
L
,
L
⊂
E
,
∣
L
,
E
∈
l. Now μ
=
μ
′
′
(
l
) from Theorem 4 so μ
′
E
=
sup
⁡
μ
L
,
L
⊂
E
,
∣
L
,
E
∈
l and it follows from [3] that E
∈
S
μ
′. Since E
∈
l, we have l
⊂
S
μ
′.Theorem 6.
Ifμ
∈
M
S
l
∩
M
σ
l, l is a δ-lattice, and μ
(
X
)
=
μ
′
(
X
), then l
⊂
S
μ
′
′.Proof.
Sinceμ
∈
M
S
l and l is a δ-lattice, then S
μ
′
=
S
μ
′
′ from Theorem 4.8 part (c) of [3]. Now since μ
∈
M
S
l
∩
M
σ
l, l is a δ-lattice, and μ
(
X
)
=
μ
′
(
X
), then l
⊂
S
μ
′ from Theorem 5. So since S
μ
′
=
S
μ
′
′, we get l
⊂
S
μ
′
′.Theorem 7.
Ifμ
∈
M
S
l
∩
M
σ
l, l is a δ-lattice, and μ
(
X
)
=
μ
′
(
X
), then μ
∈
M
σ
(
l
).Proof.
From Theorem6, we have l
⊂
S
μ
′
′, so μ
′
′ is countably additive on A
l. Now since μ
∈
M
S
l
∩
M
σ
l, then μ
=
μ
′
′
(
l
) from Theorem 4 so μ is countably additive on A
l. Therefore, μ
∈
M
σ
(
l
) by definition.Theorem 8.
IfE
∈
S
μ
′, then μ
′
=
μ
i.Proof.
From Lemma4.1 of [3], E
∈
S
μ
′ if and only if μ
′
E
=
sup
⁡
μ
L
,
L
⊂
E
,
∣
L
,
E
∈
l. Now μ
i
E
=
sup
⁡
μ
L
∣
L
⊂
E
,
L
∈
l; therefore, μ
′
=
μ
i.Theorem 9.
IfE
∈
S
μ
′, then μ
′
=
μ
i
=
μ
j
=
μ
′
′.Proof.
IfE
∈
S
μ
′, then μ
′
=
μ
i from Theorem 8. Now using μ
i
E
≤
μ
j
E
≤
sup
⁡
μ
′
′
⋂
1
∞
L
i
,
⋂
L
i
⊂
E
,
∣
L
i
∈
l
≤
μ
′
′
E
≤
μ
′
E
1 for E
∈
S
μ
′ we get μ
′
=
μ
i
=
μ
j
=
μ
′
′.Theorem 10.
Ifμ
∈
M
S
l
∩
M
σ
l, μ
′
=
μ
i, and l is a δ-lattice, then E
∈
S
μ
′.Proof.
Sinceμ
′
=
μ
i, μ
i
E
≤
μ
j
E
≤
s
u
p
⁡
{
μ
′
′
⋂
1
∞
L
i
,
⋂
L
i
⊂
E
,
∣
L
i
∈
l
}
≤
μ
′
′
E
≤
μ
′
E
1 becomes μ
′
E
=
sup
⁡
{
μ
′
′
⋂
1
∞
L
i
,
⋂
L
i
⊂
E
,
∣
L
i
∈
l
}. As l is a δ-lattice, we get ⋂
1
∞
L
i
=
L
∈
l so we get μ
′
E
=
sup
⁡
μ
′
′
L
,
L
⊂
E
,
∣
L
∈
l and since μ
∈
M
S
l
∩
M
σ
l we have by Theorem 4 that μ
=
μ
′
′
(
l
) so μ
′
E
=
sup
⁡
μ
L
,
L
⊂
E
,
∣
L
∈
l, and from [3] this implies E
∈
S
μ
′.Theorem 11.
Ifμ
∈
M
S
l and l is a δ-lattice, then S
μ
′
′
∩
l
=
L
∈
l
∣
μ
L
=
μ
′
′
L.Proof.
LetL
∈
S
μ
′
′
∩
l; then, L
∈
S
μ
′
∩
l as S
μ
′
=
S
μ
′
′ from [3], and in [3] we have that if μ
∈
M
l, then S
μ
′
∩
l
=
L
∈
l
∣
μ
L
=
μ
′
L. So L
∈
S
μ
′
∩
l implies μ
L
=
μ
′
L. However, μ
∈
M
S
l, and l is a δ-lattice implying μ
′
′
=
μ
′ everywhere in [3] so μ
L
=
μ
′
′
L.
Now letμ
L
=
μ
′
′
L for L
∈
l; then, μ
L
=
μ
′
L as μ
′
′
=
μ
′ everywhere. Therefore, L
∈
S
μ
′
∩
l and as S
μ
′
=
S
μ
′
′ we have L
∈
S
μ
′
′
∩
l.We now end with a sort of converse to Theorem4.Theorem 12.
Ifμ
∈
M
σ
l, μ
(
X
)
=
μ
′
′
(
X
), μ
=
μ
′
′
(
l
), l
⊂
S
μ, and μ
′
′ is regular, then μ
∈
M
S
l.Proof.
Nowμ
(
L
′
)
=
u
(
X
)
-
u
(
L
)
=
μ
′
′
(
X
)
-
μ
′
′
(
L
) as μ
(
X
)
=
μ
′
′
(
X
) and μ
=
μ
′
′
(
l
). Therefore, μ
(
L
′
)
=
μ
′
′
(
X
)
-
μ
′
′
(
L
) giving us μ
(
L
′
)
=
μ
′
′
(
L
′
) as μ
′
′ is regular, so we have μ
=
μ
′
′
(
l
′
). Then, it follows μ
∈
M
S
l from [3].



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*Source: 1456039-2016-10-03.xml*