Let
X
{\displaystyle X}
Λ
{\displaystyle \Lambda }
C
c
(
X
)
{\displaystyle C_{c}(X)}
σ
{\displaystyle \sigma }
Σ
{\displaystyle \Sigma }
X
{\displaystyle X}
μ
{\displaystyle \mu }
Λ
f
=
∫
X
f
d
μ
{\displaystyle \Lambda f=\displaystyle \int _{X}fd\mu }
f
∈
C
c
(
X
)
{\displaystyle f\in C_{c}(X)}
μ
(
K
)
<
∞
{\displaystyle \mu (K)<\infty }
K
∈
Σ
{\displaystyle K\in \Sigma }
If
E
∈
Σ
{\displaystyle E\in \Sigma }
μ
(
E
)
<
∞
{\displaystyle \mu (E)<\infty }
μ
(
E
)
=
inf
{
μ
(
V
)
|
E
⊂
V
,
V
open
}
{\displaystyle \mu (E)=\inf\{\mu (V)|E\subset V,V{\text{ open}}\}}
If
E
∈
Σ
{\displaystyle E\in \Sigma }
μ
(
E
)
<
∞
{\displaystyle \mu (E)<\infty }
μ
(
E
)
=
sup
{
μ
(
K
)
|
E
⊃
K
,
K
compact
}
{\displaystyle \mu (E)=\sup\{\mu (K)|E\supset K,K{\text{ compact}}\}}
The measure space
(
X
,
Σ
,
μ
)
{\displaystyle (X,\Sigma ,\mu )}
Proof
Recall the Urysohn's lemma:
If
X
{\displaystyle X}
V
{\displaystyle V}
K
{\displaystyle K}
K
⊂
V
{\displaystyle K\subset V}
there exists
f
:
X
→
[
0
,
1
]
{\displaystyle f:X\to [0,1]}
f
∈
C
c
(
X
)
{\displaystyle f\in C_{c}(X)}
f
(
K
)
=
{
1
}
{\displaystyle f(K)=\{1\}}
supp
f
⊂
V
{\displaystyle {\text{supp}}f\subset V}
short as
K
≺
f
≺
V
{\displaystyle K\prec f\prec V}
We shall first prove that if such a measure exists, then it is unique. Suppose
μ
1
,
μ
2
{\displaystyle \mu _{1},\mu _{2}}
It suffices to show that
μ
1
(
K
)
=
μ
2
(
K
)
{\displaystyle \mu _{1}(K)=\mu _{2}(K)}
K
{\displaystyle K}
Let
K
{\displaystyle K}
ϵ
>
0
{\displaystyle \epsilon >0}
V
⊂
X
{\displaystyle V\subset X}
V
⊃
K
{\displaystyle V\supset K}
μ
1
(
V
)
<
μ
1
(
K
)
+
ϵ
{\displaystyle \mu _{1}(V)<\mu _{1}(K)+\epsilon }
Urysohn's lemma implies that there exists
f
∈
C
c
(
X
)
{\displaystyle f\in C_{c}(X)}
K
≺
f
≺
V
{\displaystyle K\prec f\prec V}
(1) implies that
μ
2
(
K
)
≤
∫
X
f
d
μ
2
=
Λ
f
=
∫
X
f
d
μ
1
{\displaystyle \mu _{2}(K)\leq \displaystyle \int _{X}fd\mu _{2}=\Lambda f=\int _{X}fd\mu _{1}}
μ
1
(
V
)
<
μ
1
(
K
)
+
ϵ
{\displaystyle \mu _{1}(V)<\mu _{1}(K)+\epsilon }
μ
2
(
K
)
<
μ
1
(
K
)
+
ϵ
{\displaystyle \mu _{2}(K)<\mu _{1}(K)+\epsilon }
μ
1
(
K
)
<
μ
2
(
K
)
+
ϵ
{\displaystyle \mu _{1}(K)<\mu _{2}(K)+\epsilon }
μ
2
(
K
)
=
μ
1
(
K
)
{\displaystyle \mu _{2}(K)=\mu _{1}(K)}
Suppose
V
{\displaystyle V}
X
{\displaystyle X}
μ
(
V
)
=
sup
{
Λ
f
|
f
≺
V
}
{\displaystyle \mu (V)=\sup\{\Lambda f|f\prec V\}}
If
V
1
⊂
V
2
{\displaystyle V_{1}\subset V_{2}}
μ
(
V
1
)
≤
μ
(
V
2
)
{\displaystyle \mu (V_{1})\leq \mu (V_{2})}
E
{\displaystyle E}
X
{\displaystyle X}
μ
(
E
)
=
inf
{
μ
(
V
)
|
E
⊂
V
,
V
open
}
{\displaystyle \mu (E)=\inf\{\mu (V)|E\subset V,V{\text{ open}}\}}
Define
Σ
F
=
{
E
⊂
X
|
μ
(
E
)
<
∞
,
μ
(
E
)
=
sup
{
μ
(
K
)
:
K
⊂
E
,
K
compact
}
}
{\displaystyle \Sigma _{F}=\{E\subset X|\mu (E)<\infty ,\mu (E)=\sup\{\mu (K):K\subset E,K{\text{ compact}}\}\}}
Let
Σ
=
{
E
⊂
X
|
E
∩
K
∈
Σ
F
for all
K
compact in
X
}
{\displaystyle \Sigma =\{E\subset X|E\cap K\in \Sigma _{F}{\text{ for all }}K{\text{ compact in }}X\}}
monotonicity of
μ
{\displaystyle \mu }
X
{\displaystyle X}
E
⊂
X
{\displaystyle E\subset X}
μ
(
E
)
=
0
{\displaystyle \mu (E)=0}
It is obvious that
E
∈
Σ
F
{\displaystyle E\in \Sigma _{F}}
E
∈
Σ
{\displaystyle E\in \Sigma }
{
X
,
Σ
,
μ
}
{\displaystyle \{X,\Sigma ,\mu \}}
Suppose
{
E
i
}
i
=
1
∞
{\displaystyle \displaystyle \{E_{i}\}_{i=1}^{\infty }}
X
{\displaystyle X}
μ
(
⋃
E
i
)
≤
∑
i
=
1
∞
μ
(
E
i
)
{\displaystyle \displaystyle \mu \left(\bigcup E_{i}\right)\leq \sum _{i=1}^{\infty }\mu (E_{i})}
Proof
Let
V
1
,
V
2
{\displaystyle V_{1},V_{2}}
X
{\displaystyle X}
μ
(
V
1
∪
V
2
)
≤
μ
(
V
1
)
+
μ
(
V
2
)
{\displaystyle \mu (V_{1}\cup V_{2})\leq \mu (V_{1})+\mu (V_{2})}
Given
ϵ
>
0
{\displaystyle \epsilon >0}
g
∈
C
c
(
X
)
{\displaystyle g\in C_{c}(X)}
g
≺
(
V
1
∪
V
2
)
{\displaystyle g\prec (V_{1}\cup V_{2})}
supp
g
=
K
⊂
(
V
1
∪
V
2
)
{\displaystyle \displaystyle {\text{supp }}g=K\subset (V_{1}\cup V_{2})}
μ
(
V
1
∪
V
2
)
−
ϵ
≤
Λ
g
{\displaystyle \mu (V_{1}\cup V_{2})-\epsilon \leq \Lambda g}
μ
(
V
1
∪
V
2
)
=
sup
{
Λ
g
|
g
≺
V
1
∪
V
2
}
{\displaystyle \mu (V_{1}\cup V_{2})=\sup\{\Lambda g|g\prec V_{1}\cup V_{2}\}}
Now by Urysohn's lemma we can find
h
i
{\displaystyle h_{i}}
i
=
1
,
2
,
…
{\displaystyle i=1,2,\ldots }
h
i
≺
V
i
{\displaystyle h_{i}\prec V_{i}}
h
1
+
h
2
=
1
{\displaystyle h_{1}+h_{2}=1}
K
{\displaystyle K}
h
i
∈
C
c
(
X
)
{\displaystyle h_{i}\in C_{c}(X)}
Thus,
h
i
g
≺
V
i
{\displaystyle h_{i}g\prec V_{i}}
h
1
g
+
h
2
g
=
g
{\displaystyle h_{1}g+h_{2}g=g}
K
{\displaystyle K}
Λ
{\displaystyle \Lambda }
Λ
f
≤
Λ
g
{\displaystyle \Lambda f\leq \Lambda g}
f
≤
g
{\displaystyle f\leq g}
Λ
g
=
Λ
h
1
g
+
Λ
h
2
g
≤
μ
(
V
1
)
+
μ
(
V
2
)
.
{\displaystyle \Lambda g=\Lambda h_{1}g+\Lambda h_{2}g\leq \mu (V_{1})+\mu (V_{2}).}
μ
(
V
1
∪
V
2
)
≤
μ
(
V
1
)
+
μ
(
V
2
)
+
ϵ
{\displaystyle \mu (V_{1}\cup V_{2})\leq \mu (V_{1})+\mu (V_{2})+\epsilon }
ϵ
>
0
{\displaystyle \epsilon >0}
μ
(
V
1
∪
V
2
)
≤
μ
(
V
1
)
+
μ
(
V
2
)
.
{\displaystyle \mu (V_{1}\cup V_{2})\leq \mu (V_{1})+\mu (V_{2}).}
{
E
i
}
i
=
1
∞
{\displaystyle \{E_{i}\}_{i=1}^{\infty }}
Σ
{\displaystyle \Sigma }
V
i
{\displaystyle V_{i}}
ϵ
>
0
{\displaystyle \epsilon >0}
μ
(
V
i
)
<
μ
(
E
i
)
+
ϵ
2
i
{\displaystyle \mu (V_{i})<\mu (E_{i})+{\frac {\epsilon }{2^{i}}}}
E
=
⋃
E
i
⊂
⋃
V
i
=
V
{\displaystyle E=\bigcup E_{i}\subset \bigcup V_{i}=V}
V
{\displaystyle V}
f
≺
V
{\displaystyle f\prec V}
μ
(
V
)
≥
μ
(
E
)
{\displaystyle \mu (V)\geq \mu (E)}
μ
(
V
)
<
Λ
f
{\displaystyle \mu (V)<\Lambda f}
μ
(
V
)
≤
∑
μ
(
V
i
)
≤
∑
μ
(
E
i
)
+
ϵ
.
{\displaystyle \mu (V)\leq \sum \mu (V_{i})\leq \sum \mu (E_{i})+\epsilon .}
If
K
{\displaystyle K}
K
∈
Σ
F
{\displaystyle K\in \Sigma _{F}}
μ
(
K
)
=
inf
{
Λ
f
:
K
≺
f
}
{\displaystyle \mu (K)=\inf\{\Lambda f:K\prec f\}}
Proof
It suffices to show that
μ
(
K
)
<
∞
{\displaystyle \mu (K)<\infty }
K
{\displaystyle K}
Let
0
<
α
<
1
{\displaystyle 0<\alpha <1}
K
≺
f
{\displaystyle K\prec f}
V
α
=
{
x
:
f
(
x
)
>
α
}
{\displaystyle V_{\alpha }=\{x:f(x)>\alpha \}}
V
α
{\displaystyle V_{\alpha }}
K
⊂
V
α
{\displaystyle K\subset V_{\alpha }}
Then, by Urysohn's lemma, there exists
g
∈
C
c
(
X
)
{\displaystyle g\in C_{c}(X)}
K
≺
g
≺
V
α
{\displaystyle K\prec g\prec V_{\alpha }}
α
g
≤
f
{\displaystyle \alpha g\leq f}
V
α
{\displaystyle V_{\alpha }}
Since
Λ
g
≥
μ
(
K
)
{\displaystyle \Lambda g\geq \mu (K)}
[ 1] [ 2]
μ
(
K
)
≤
1
α
Λ
f
{\displaystyle \mu (K)\leq {\frac {1}{\alpha }}\Lambda f}
As
Λ
f
<
∞
{\displaystyle \Lambda f<\infty }
μ
(
K
)
<
∞
{\displaystyle \mu (K)<\infty }
K
∈
Σ
F
{\displaystyle K\in \Sigma _{F}}
ϵ
>
0
{\displaystyle \epsilon >0}
By definition, there exists open
V
⊃
K
{\displaystyle V\supset K}
μ
(
K
)
>
μ
(
V
)
−
ϵ
{\displaystyle \mu (K)>\mu (V)-\epsilon }
f
{\displaystyle f}
K
≺
f
≺
V
{\displaystyle K\prec f\prec V}
μ
(
V
)
≥
Λ
f
{\displaystyle \mu (V)\geq \Lambda f}
μ
(
K
)
>
Λ
f
+
ϵ
{\displaystyle \mu (K)>\Lambda f+\epsilon }
Hence,
μ
(
K
)
=
inf
{
Λ
f
:
K
≺
f
}
{\displaystyle \mu (K)=\inf\{\Lambda f:K\prec f\}}
Every open set
V
{\displaystyle V}
μ
(
V
)
=
sup
{
μ
(
K
)
:
K
⊂
V
,
K
compact
}
{\displaystyle \mu (V)=\sup\{\mu (K):K\subset V,K{\text{ compact}}\}}
If
V
{\displaystyle V}
V
⊂
X
{\displaystyle V\subset X}
μ
(
X
)
<
∞
{\displaystyle \mu (X)<\infty }
V
∈
Σ
F
{\displaystyle V\in \Sigma _{F}}
Proof
Let
V
{\displaystyle V}
α
>
0
{\displaystyle \alpha >0}
α
<
μ
(
V
)
{\displaystyle \alpha <\mu (V)}
K
{\displaystyle K}
μ
(
K
)
>
α
{\displaystyle \mu (K)>\alpha }
By definition of
μ
{\displaystyle \mu }
f
∈
C
c
(
X
)
{\displaystyle f\in C_{c}(X)}
f
≺
V
{\displaystyle f\prec V}
Λ
f
>
α
{\displaystyle \Lambda f>\alpha }
Let
K
=
supp
f
{\displaystyle K={\text{supp}}f}
K
⊂
V
{\displaystyle K\subset V}
Let
W
{\displaystyle W}
K
⊂
W
{\displaystyle K\subset W}
F
≺
W
{\displaystyle F\prec W}
Λ
f
≤
μ
(
W
)
{\displaystyle \Lambda f\leq \mu (W)}
Λ
f
≤
μ
(
K
)
{\displaystyle \Lambda f\leq \mu (K)}
μ
(
K
)
>
α
{\displaystyle \mu (K)>\alpha }
Suppose
μ
(
E
i
)
{\displaystyle \mu (E_{i})}
Σ
F
{\displaystyle \Sigma _{F}}
E
=
⋃
i
=
1
∞
E
i
{\displaystyle E=\displaystyle \bigcup _{i=1}^{\infty }E_{i}}
μ
(
E
)
=
∑
i
=
1
∞
μ
(
E
i
)
{\displaystyle \mu (E)=\displaystyle \sum _{i=1}^{\infty }\mu (E_{i})}
Proof
If
μ
(
E
)
=
∞
{\displaystyle \mu (E)=\infty }
If
μ
(
E
)
{\displaystyle \mu (E)}
E
∈
Σ
F
{\displaystyle E\in \Sigma _{F}}
μ
{\displaystyle \mu }
Σ
F
{\displaystyle \Sigma _{F}}
K
1
,
K
2
{\displaystyle K_{1},K_{2}}
K
1
,
K
2
∈
Σ
F
{\displaystyle K_{1},K_{2}\in \Sigma _{F}}
K
1
∪
K
2
∈
Σ
F
{\displaystyle K_{1}\cup K_{2}\in \Sigma _{F}}
μ
(
K
1
∪
K
2
)
=
μ
(
K
1
)
+
μ
(
K
2
)
{\displaystyle \mu (K_{1}\cup K_{2})=\mu (K_{1})+\mu (K_{2})}
X
{\displaystyle X}
V
1
{\displaystyle V_{1}}
V
2
{\displaystyle V_{2}}
K
1
⊂
V
1
{\displaystyle K_{1}\subset V_{1}}
K
2
⊂
V
2
{\displaystyle K_{2}\subset V_{2}}
Hence, by Urysohn's lemma, there exists
g
∈
C
c
(
X
)
{\displaystyle g\in C_{c}(X)}
g
≺
K
1
∪
K
2
{\displaystyle g\prec K_{1}\cup K_{2}}
Λ
g
≤
μ
(
K
1
∪
K
2
)
+
ϵ
{\displaystyle \Lambda g\leq \mu (K_{1}\cup K_{2})+\epsilon }
Now,
supp
f
g
=
K
{\displaystyle {\text{supp}}fg=K}
K
≺
f
g
{\displaystyle K\prec fg}
K
≺
(
1
−
f
)
g
{\displaystyle K\prec (1-f)g}
μ
(
K
1
)
+
μ
(
K
2
)
<
Λ
g
≤
μ
(
K
1
+
K
2
)
+
ϵ
{\displaystyle \mu (K_{1})+\mu (K_{2})<\Lambda g\leq \mu (K_{1}+K_{2})+\epsilon }
Assume
μ
(
E
)
<
∞
{\displaystyle \mu (E)<\infty }
E
i
∈
Σ
F
{\displaystyle E_{i}\in \Sigma _{F}}
H
i
⊂
E
i
{\displaystyle H_{i}\subset E_{i}}
μ
(
H
i
)
>
μ
(
E
i
)
−
ϵ
2
i
{\displaystyle \mu (H_{i})>\mu (E_{i})-{\frac {\epsilon }{2^{i}}}}
Let
K
N
=
H
1
∪
H
2
∪
…
H
N
{\displaystyle K_{N}=H_{1}\cup H_{2}\cup \ldots H_{N}}
K
N
{\displaystyle K_{N}}
Thus,
μ
(
E
)
≥
μ
(
K
N
)
=
∑
i
=
1
N
μ
(
H
i
)
≥
∑
i
=
1
N
μ
(
E
i
)
−
ϵ
{\displaystyle \mu (E)\geq \mu (K_{N})=\displaystyle \sum _{i=1}^{N}\mu (H_{i})\geq \sum _{i=1}^{N}\mu (E_{i})-\epsilon }
and hence,
μ
(
E
)
≥
∑
i
=
1
∞
μ
(
E
i
)
{\displaystyle \mu (E)\geq \displaystyle \sum _{i=1}^{\infty }\mu (E_{i})}
μ
(
E
)
≤
∑
i
=
1
∞
μ
(
E
i
)
{\displaystyle \mu (E)\leq \displaystyle \sum _{i=1}^{\infty }\mu (E_{i})}
Thus,
μ
(
E
)
=
∑
i
=
1
∞
μ
(
E
i
)
{\displaystyle \mu (E)=\displaystyle \sum _{i=1}^{\infty }\mu (E_{i})}
If
E
∈
Σ
F
{\displaystyle E\in \Sigma _{F}}
ϵ
>
0
{\displaystyle \epsilon >0}
K
{\displaystyle K}
V
{\displaystyle V}
K
⊂
E
⊂
V
{\displaystyle K\subset E\subset V}
μ
(
V
∖
K
)
<
ϵ
{\displaystyle \mu (V\setminus K)<\epsilon }
Proof
(
V
∖
K
)
{\displaystyle (V\setminus K)}
E
∈
Σ
F
{\displaystyle E\in \Sigma _{F}}
K
{\displaystyle K}
V
{\displaystyle V}
K
⊂
E
⊂
V
{\displaystyle K\subset E\subset V}
μ
(
V
)
−
ϵ
2
<
μ
(
E
)
<
μ
(
K
)
+
ϵ
2
{\displaystyle \mu (V)-{\frac {\epsilon }{2}}<\mu (E)<\mu (K)+{\frac {\epsilon }{2}}}
Now,
μ
(
V
∖
K
)
<
μ
(
V
)
<
μ
(
E
)
+
ϵ
2
<
∞
{\displaystyle \mu (V\setminus K)<\mu (V)<\mu (E)+{\frac {\epsilon }{2}}<\infty }
Thus,
μ
(
V
∖
K
)
<
ϵ
{\displaystyle \mu (V\setminus K)<\epsilon }
Σ
F
{\displaystyle \Sigma _{F}}
X
{\displaystyle X}
Proof
Let
A
,
B
∈
Σ
F
{\displaystyle A,B\in \Sigma _{F}}
ϵ
>
0
{\displaystyle \epsilon >0}
There exist compact
K
1
,
K
2
{\displaystyle K_{1},K_{2}}
V
1
,
V
2
{\displaystyle V_{1},V_{2}}
K
1
⊂
A
⊂
V
1
{\displaystyle K_{1}\subset A\subset V_{1}}
K
2
⊂
A
⊂
V
2
{\displaystyle K_{2}\subset A\subset V_{2}}
μ
(
V
1
∖
K
1
)
,
μ
(
V
2
∖
K
2
)
<
ϵ
{\displaystyle \mu (V_{1}\setminus K_{1}),\mu (V_{2}\setminus K_{2})<\epsilon }
A
∖
B
=
(
V
1
∖
K
1
)
∪
(
K
1
∖
V
2
)
∪
(
V
2
∖
K
2
)
{\displaystyle A\setminus B=(V_{1}\setminus K_{1})\cup (K_{1}\setminus V_{2})\cup (V_{2}\setminus K_{2})}
As
K
1
∖
V
2
{\displaystyle K_{1}\setminus V_{2}}
K
{\displaystyle K}
μ
(
A
∖
B
)
≤
μ
(
V
1
∖
K
1
)
∪
μ
(
K
1
∖
V
2
)
∪
μ
(
V
2
∖
K
2
)
=
μ
(
K
1
∖
V
2
)
+
2
ϵ
{\displaystyle \mu (A\setminus B)\leq \mu (V_{1}\setminus K_{1})\cup \mu (K_{1}\setminus V_{2})\cup \mu (V_{2}\setminus K_{2})=\mu (K_{1}\setminus V_{2})+2\epsilon }
thus,
μ
(
A
∖
B
{\displaystyle \mu (A\setminus B}
A
∖
B
∈
Σ
F
{\displaystyle A\setminus B\in \Sigma _{F}}
Now write
A
∪
B
=
(
A
∖
B
)
∪
B
{\displaystyle A\cup B=(A\setminus B)\cup B}
and
A
∩
B
=
A
∖
(
A
∖
B
)
{\displaystyle A\cap B=A\setminus (A\setminus B)}
Σ
{\displaystyle \Sigma }
σ
{\displaystyle \sigma }
Proof
Let
C
{\displaystyle C}
Then,
C
∩
K
{\displaystyle C\cap K}
K
{\displaystyle K}
Therefore
C
∩
K
∈
Σ
F
{\displaystyle C\cap K\in \Sigma _{F}}
C
∈
Σ
{\displaystyle C\in \Sigma }
Σ
{\displaystyle \Sigma }
X
∈
Σ
{\displaystyle X\in \Sigma }
A
∈
Σ
{\displaystyle A\in \Sigma }
A
c
∩
K
⊂
K
{\displaystyle A^{c}\cap K\subset K}
A
c
∩
K
=
K
∖
(
K
∖
A
)
{\displaystyle A^{c}\cap K=K\setminus (K\setminus A)}
A
c
∈
Σ
{\displaystyle A^{c}\in \Sigma }
Now let
A
=
⋃
i
=
1
∞
A
i
{\displaystyle A=\displaystyle \bigcup _{i=1}^{\infty }A_{i}}
A
i
∈
Σ
{\displaystyle A_{i}\in \Sigma }
We know that
A
i
∩
K
∈
Σ
F
{\displaystyle A_{i}\cap K\in \Sigma _{F}}
K
{\displaystyle K}
Let
B
1
=
A
1
∩
K
{\displaystyle B_{1}=A_{1}\cap K}
B
n
=
A
n
∩
K
∖
(
B
1
∪
B
2
…
∪
B
n
−
1
)
{\displaystyle B_{n}=A_{n}\cap K\setminus (B_{1}\cup B_{2}\ldots \cup B_{n-1})}
A
∩
K
=
⋃
i
=
1
∞
{\displaystyle A\cap K=\displaystyle \bigcup _{i=1}^{\infty }}
μ
(
B
i
)
<
∞
{\displaystyle \mu (B_{i})<\infty }
A
∩
K
∈
Σ
{\displaystyle A\cap K\in \Sigma }
Σ
F
=
{
E
∈
Σ
:
μ
(
E
)
<
∞
}
{\displaystyle \Sigma _{F}=\{E\in \Sigma :\mu (E)<\infty \}}
Proof
Let
E
⊂
Σ
F
{\displaystyle E\subset \Sigma _{F}}
E
∩
K
∈
Σ
F
{\displaystyle E\cap K\in \Sigma _{F}}
K
⊂
X
{\displaystyle K\subset X}
Now, let
E
∈
Σ
{\displaystyle E\in \Sigma }
μ
(
E
)
<
∞
{\displaystyle \mu (E)<\infty }
ϵ
>
0
{\displaystyle \epsilon >0}
V
{\displaystyle V}
E
⊂
V
{\displaystyle E\subset V}
μ
(
V
)
<
∞
{\displaystyle \mu (V)<\infty }
V
∈
Σ
F
{\displaystyle V\in \Sigma _{F}}
K
⊂
V
{\displaystyle K\subset V}
μ
(
V
∖
K
)
<
∞
{\displaystyle \mu (V\setminus K)<\infty }
E
∈
Σ
{\displaystyle E\in \Sigma }
E
∩
K
∈
Σ
F
{\displaystyle E\cap K\in \Sigma _{F}}
H
⊂
E
∩
K
{\displaystyle H\subset E\cap K}
μ
(
E
∩
K
)
<
μ
(
H
)
+
ϵ
{\displaystyle \mu (E\cap K)<\mu (H)+\epsilon }
E
⊂
(
E
∩
K
)
∪
(
V
∖
K
)
{\displaystyle E\subset (E\cap K)\cup (V\setminus K)}
μ
(
E
)
≤
μ
(
H
)
+
2
ϵ
{\displaystyle \mu (E)\leq \mu (H)+2\epsilon }
As
ϵ
>
0
{\displaystyle \epsilon >0}
For
f
∈
C
c
(
X
)
{\displaystyle f\in C_{c}(X)}
Λ
f
=
∫
X
f
d
μ
{\displaystyle \Lambda f=\displaystyle \int _{X}fd\mu }
Proof
Without loss of generality, we may assume that
f
{\displaystyle f}
It is obvious from the definition of
μ
{\displaystyle \mu }
Λ
f
≥
∫
X
f
d
μ
{\displaystyle \Lambda f\geq \displaystyle \int _{X}fd\mu }
Let
K
=
supp
f
{\displaystyle K={\text{supp}}f}
f
{\displaystyle f}
f
(
K
)
{\displaystyle f(K)}
f
(
K
)
⊂
[
a
,
b
]
{\displaystyle f(K)\subset [a,b]}
a
,
b
∈
R
{\displaystyle a,b\in \mathbb {R} }
ϵ
>
0
{\displaystyle \epsilon >0}
P
=
{
a
=
y
0
<
y
1
<
…
<
y
n
=
b
}
{\displaystyle {\mathcal {P}}=\{a=y_{0}<y_{1}<\ldots <y_{n}=b\}}
ϵ
{\displaystyle \epsilon }
[
a
,
b
]
{\displaystyle [a,b]}
Let
E
i
=
{
x
∈
X
|
y
i
−
1
<
f
(
x
)
<
y
i
}
∩
K
{\displaystyle E_{i}=\{x\in X|y_{i-1}<f(x)<y_{i}\}\cap K}
f
{\displaystyle f}
K
{\displaystyle K}
E
i
{\displaystyle E_{i}}
i
{\displaystyle i}
K
=
⋃
i
=
1
n
E
i
{\displaystyle K=\displaystyle \bigcup _{i=1}^{n}E_{i}}
μ
(
E
)
=
inf
{
μ
(
V
)
|
V
⊃
E
,
V
open
}
{\displaystyle \mu (E)=\inf\{\mu (V)|V\supset E,V{\text{ open}}\}}
Hence, we can find open sets
V
i
⊃
E
i
{\displaystyle V_{i}\supset E_{i}}
μ
(
V
i
)
<
μ
(
E
i
)
+
ϵ
n
{\displaystyle \mu (V_{i})<\mu (E_{i})+{\frac {\epsilon }{n}}}
f
(
x
)
<
y
i
+
ϵ
{\displaystyle f(x)<y_{i}+\epsilon }
x
∈
V
i
{\displaystyle x\in V_{i}}
K
⊂
V
1
∪
V
2
∪
…
∪
V
n
{\displaystyle K\subset V_{1}\cup V_{2}\cup \ldots \cup V_{n}}
V
i
{\displaystyle V_{i}}
h
i
∈
C
c
(
X
)
{\displaystyle h_{i}\in C_{c}(X)}
h
i
≺
V
i
{\displaystyle h_{i}\prec V_{i}}
∑
i
=
1
n
h
i
=
1
{\displaystyle \displaystyle \sum _{i=1}^{n}h_{i}=1}
K
{\displaystyle K}
Hence, there exist functions
h
i
≺
V
i
{\displaystyle h_{i}\prec V_{i}}
∑
i
=
1
n
h
i
=
1
{\displaystyle \displaystyle \sum _{i=1}^{n}h_{i}=1}
K
{\displaystyle K}
Thus,
∑
i
=
1
n
h
i
(
x
)
f
(
x
)
=
f
(
x
)
{\displaystyle \displaystyle \sum _{i=1}^{n}h_{i}(x)f(x)=f(x)}
x
∈
X
{\displaystyle x\in X}
μ
(
K
)
≤
∑
i
=
1
n
Λ
h
i
{\displaystyle \displaystyle \mu (K)\leq \sum _{i=1}^{n}\Lambda h_{i}}
h
i
f
<
(
y
i
+
ϵ
)
h
i
{\displaystyle h_{i}f<(y_{i}+\epsilon )h_{i}}
V
i
{\displaystyle V_{i}}
Thus,
Λ
f
=
∑
i
=
1
n
Λ
h
i
f
≤
∑
i
=
1
n
Λ
h
i
(
y
i
+
ϵ
)
=
(
∑
i
=
1
n
(
y
i
+
ϵ
)
Λ
h
i
+
Λ
∑
i
=
1
n
|
a
|
h
i
)
−
Λ
∑
i
=
1
n
|
a
|
h
i
{\displaystyle \Lambda f=\displaystyle \sum _{i=1}^{n}\Lambda h_{i}f\leq \sum _{i=1}^{n}\Lambda h_{i}(y_{i}+\epsilon )=\left(\sum _{i=1}^{n}(y_{i}+\epsilon )\Lambda h_{i}+\Lambda \sum _{i=1}^{n}|a|h_{i}\right)-\Lambda \sum _{i=1}^{n}|a|h_{i}}
=
∑
i
=
1
n
(
y
i
+
ϵ
+
|
a
|
)
Λ
h
i
−
Λ
∑
i
=
1
n
|
a
|
h
i
{\displaystyle =\displaystyle \sum _{i=1}^{n}(y_{i}+\epsilon +|a|)\Lambda h_{i}-\Lambda \sum _{i=1}^{n}|a|h_{i}}
≤
∑
i
=
1
n
(
y
i
+
ϵ
+
|
a
|
)
(
μ
(
E
i
)
+
ϵ
n
)
−
Λ
∑
i
=
1
n
|
a
|
h
i
{\displaystyle \leq \displaystyle \sum _{i=1}^{n}(y_{i}+\epsilon +|a|)\left(\mu (E_{i})+{\frac {\epsilon }{n}}\right)-\Lambda \sum _{i=1}^{n}|a|h_{i}}
=
∑
i
=
1
n
y
i
μ
(
E
i
)
+
∑
i
=
1
n
ϵ
(
μ
(
E
i
)
+
ϵ
n
+
|
a
|
n
)
≤
∑
i
=
1
n
(
y
i
−
ϵ
)
μ
(
E
i
)
+
∑
i
=
1
n
ϵ
(
2
μ
(
E
i
)
+
ϵ
n
+
|
a
|
n
)
{\displaystyle =\displaystyle \sum _{i=1}^{n}y_{i}\mu (E_{i})+\sum _{i=1}^{n}\epsilon \left(\mu (E_{i})+{\frac {\epsilon }{n}}+{\frac {|a|}{n}}\right)\leq \sum _{i=1}^{n}(y_{i}-\epsilon )\mu (E_{i})+\sum _{i=1}^{n}\epsilon \left(2\mu (E_{i})+{\frac {\epsilon }{n}}+{\frac {|a|}{n}}\right)}
≤
∑
i
=
1
n
f
(
x
i
)
μ
(
E
i
)
+
∑
i
=
1
n
ϵ
(
2
μ
(
E
i
)
+
ϵ
n
+
|
a
|
n
)
{\displaystyle \leq \displaystyle \sum _{i=1}^{n}f(x_{i})\mu (E_{i})+\sum _{i=1}^{n}\epsilon \left(2\mu (E_{i})+{\frac {\epsilon }{n}}+{\frac {|a|}{n}}\right)}
ϵ
>
0
{\displaystyle \epsilon >0}
Λ
f
≤
∫
X
f
d
μ
{\displaystyle \Lambda f\leq \displaystyle \int _{X}fd\mu }
↑ https://4dspace.mtts.org.in/expository-article-download.php?ai=160 ↑ https://laurent.claessens-donadello.eu/pdf/giulietta.pdf , search for the label THOooTWZWooHqGDAx. 
![{\displaystyle f:X\to [0,1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fbce99d737c06662e393d2d3dba7505a1e6c6eb6)
![{\displaystyle f(K)\subset [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/237c40b77803e063e9ff336f4172ca1085c9ada8)
![{\displaystyle [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c4b788fc5c637e26ee98b45f89a5c08c85f7935)
