Let
(
X
,
σ
,
μ
)
{\displaystyle (X,\sigma ,\mu )}
be a
σ
{\displaystyle \sigma }
-finite measure space. Suppose
s
{\displaystyle s}
is a positive simple measurable function, with
s
=
∑
i
=
1
3
y
i
χ
A
i
{\displaystyle s=\displaystyle \sum _{i=1}^{3}y_{i}\chi _{A_{i}}}
;
A
i
∈
σ
{\displaystyle A_{i}\in \sigma }
are disjoint.
Define
∫
X
s
d
μ
=
∑
y
i
μ
(
A
i
)
{\displaystyle \displaystyle \int _{X}s~d\mu =\sum y_{i}\mu (A_{i})}
Let
f
:
X
→
R
¯
{\displaystyle f:X\to {\overline {\mathbb {R} }}}
be measurable, and let
f
≥
0
{\displaystyle f\geq 0}
.
Define
∫
X
f
d
μ
=
sup
{
∫
X
s
d
μ
=
∑
y
i
μ
(
A
i
)
|
s
simple
,
s
≥
0
,
s
≤
f
}
{\displaystyle \displaystyle \int _{X}f~d\mu =\sup\{\int _{X}s~d\mu =\sum y_{i}\mu (A_{i})|s{\text{ simple }},s\geq 0,s\leq f\}}
Now let
f
{\displaystyle f}
be any measurable function. We say that
f
{\displaystyle f}
is integrable if
f
+
{\displaystyle f^{+}}
and
f
−
{\displaystyle f^{-}}
are integrable and if
∫
X
f
+
d
μ
,
∫
X
f
−
d
μ
<
∞
{\displaystyle \displaystyle \int _{X}f^{+}~d\mu ,\int _{X}f^{-}~d\mu <\infty }
. Then, we write
∫
X
f
d
μ
=
∫
X
f
+
d
μ
−
∫
X
f
−
d
μ
{\displaystyle \displaystyle \int _{X}f~d\mu =\int _{X}f^{+}~d\mu -\int _{X}f^{-}~d\mu }
The class of measurable functions on
X
{\displaystyle X}
is denoted by
L
1
(
X
)
{\displaystyle {\mathcal {L}}^{1}(X)}
For
0
<
p
<
∞
{\displaystyle 0<p<\infty }
, we define
L
p
{\displaystyle {\mathcal {L}}^{p}}
to be the collection of all measurable functions
f
{\displaystyle f}
such that
|
f
|
p
∈
L
1
{\displaystyle |f|^{p}\in {\mathcal {L}}^{1}}
A property is said to hold almost everywhere if the set of all points where the property does not hold has measure zero.
Let
(
X
,
σ
,
μ
)
{\displaystyle (X,\sigma ,\mu )}
be a measure space and let
f
,
g
{\displaystyle f,g}
be measurable on
X
{\displaystyle X}
. Then
If
f
≤
g
{\displaystyle f\leq g}
, then
∫
X
f
d
μ
≤
∫
X
g
d
μ
{\displaystyle \displaystyle \int _{X}fd\mu \leq \int _{X}gd\mu }
If
A
,
B
∈
σ
{\displaystyle A,B\in \sigma }
,
A
⊂
B
{\displaystyle A\subset B}
, then
∫
A
f
d
μ
≤
∫
B
f
d
μ
{\displaystyle \displaystyle \int _{A}fd\mu \leq \int _{B}fd\mu }
If
f
≥
0
{\displaystyle f\geq 0}
and
c
≥
0
{\displaystyle c\geq 0}
then
∫
X
c
f
d
μ
=
c
∫
X
f
d
μ
{\displaystyle \displaystyle \int _{X}cfd\mu =c\int _{X}fd\mu }
If
E
∈
σ
{\displaystyle E\in \sigma }
,
μ
(
E
)
=
0
{\displaystyle \mu (E)=0}
, then
∫
E
f
d
μ
=
0
{\displaystyle \displaystyle \int _{E}fd\mu =0}
, even if
f
(
E
)
=
{
∞
}
{\displaystyle f(E)=\{\infty \}}
If
E
∈
σ
{\displaystyle E\in \sigma }
,
f
(
E
)
=
{
0
}
{\displaystyle f(E)=\{0\}}
, then
∫
E
f
d
μ
=
0
{\displaystyle \displaystyle \int _{E}fd\mu =0}
, even if
μ
(
E
)
=
∞
{\displaystyle \mu (E)=\infty }
Proof
Monotone Convergence Theorem
edit
Suppose
f
n
≥
0
{\displaystyle f_{n}\geq 0}
and
f
n
{\displaystyle f_{n}}
are measurable for all
n
{\displaystyle n}
such that
f
1
(
x
)
≤
f
2
(
x
)
≤
…
{\displaystyle f_{1}(x)\leq f_{2}(x)\leq \ldots }
for every
x
∈
X
{\displaystyle x\in X}
f
n
(
x
)
→
f
(
x
)
{\displaystyle f_{n}(x)\to f(x)}
almost everywhere on
X
{\displaystyle X}
Then,
∫
X
f
n
d
μ
→
∫
X
f
d
μ
{\displaystyle \displaystyle \int _{X}f_{n}d\mu \to \int _{X}fd\mu }
Proof
∫
X
f
n
d
μ
{\displaystyle \displaystyle \int _{X}f_{n}d\mu }
is an increasing sequence in
R
{\displaystyle \mathbb {R} }
, and hence,
∫
X
f
n
d
μ
→
α
∈
R
¯
{\displaystyle \displaystyle \int _{X}f_{n}d\mu \to \alpha \in {\overline {\mathbb {R} }}}
(say). We know that
f
{\displaystyle f}
is measurable and that
f
≥
f
n
∀
n
{\displaystyle f\geq f_{n}\forall n}
. That is,
∫
X
f
1
d
μ
≤
∫
X
f
2
d
μ
≤
…
∫
X
f
n
d
μ
≤
…
∫
X
f
d
μ
{\displaystyle \displaystyle \int _{X}f_{1}d\mu \leq \int _{X}f_{2}d\mu \leq \ldots \int _{X}f_{n}d\mu \leq \ldots \int _{X}fd\mu }
Hence,
α
≤
∫
X
f
d
μ
=
sup
{
∫
X
s
d
μ
:
s
is simple
,
0
≤
s
≤
1
}
{\displaystyle \displaystyle \alpha \leq \int _{X}fd\mu =\sup\{\int _{X}sd\mu :s{\text{ is simple }},0\leq s\leq 1\}}
Let
c
∈
[
0
,
1
]
{\displaystyle c\in [0,1]}
Define
E
n
=
{
x
|
f
n
(
x
)
≥
c
s
(
x
)
}
{\displaystyle E_{n}=\{x|f_{n}(x)\geq cs(x)\}}
;
n
=
1
,
2
…
{\displaystyle n=1,2\ldots }
. Observe that
E
1
⊂
E
2
⊂
…
{\displaystyle E_{1}\subset E_{2}\subset \ldots }
and
⋃
E
n
=
X
{\displaystyle \bigcup E_{n}=X}
Suppose
x
∈
X
{\displaystyle x\in X}
. If
f
(
x
)
=
0
{\displaystyle f(x)=0}
then
s
(
x
)
=
0
{\displaystyle s(x)=0}
implying that
x
∈
E
1
{\displaystyle x\in E_{1}}
. If
f
(
x
)
>
0
{\displaystyle f(x)>0}
, then there exists
n
{\displaystyle n}
such that
f
n
(
x
)
>
c
s
(
x
)
{\displaystyle f_{n}(x)>cs(x)}
and hence,
x
∈
E
n
{\displaystyle x\in E_{n}}
.
Thus,
⋃
E
n
=
X
{\displaystyle \bigcup E_{n}=X}
, therefore
∫
X
f
n
(
x
)
d
μ
≥
∫
E
n
f
n
d
μ
≥
c
∫
E
n
s
d
μ
{\displaystyle \displaystyle \int _{X}f_{n}(x)d\mu \geq \int _{E_{n}}f_{n}d\mu \geq c\int _{E_{n}}sd\mu }
. As this is true if
c
∈
[
0
,
1
]
{\displaystyle c\in [0,1]}
, we have that
α
≥
∫
X
f
d
μ
{\displaystyle \alpha \geq \displaystyle \int _{X}fd\mu }
. Thus,
∫
X
f
n
d
μ
→
∫
X
f
d
μ
{\displaystyle \displaystyle \int _{X}f_{n}d\mu \to \int _{X}fd\mu }
.
Dominated convergence theorem
edit
Let
(
X
,
σ
,
μ
)
{\displaystyle (X,\sigma ,\mu )}
be a complex measure space. Let
{
f
n
}
{\displaystyle \{f_{n}\}}
be a sequence of complex measurable functions that converge pointwise to
f
{\displaystyle f}
;
f
(
x
)
=
lim
n
→
∞
f
n
(
x
)
{\displaystyle f(x)=\displaystyle \lim _{n\to \infty }f_{n}(x)}
, with
x
∈
X
{\displaystyle x\in X}
Suppose
|
f
n
(
x
)
|
≤
g
(
x
)
{\displaystyle |f_{n}(x)|\leq g(x)}
for some
g
∈
L
1
(
X
)
{\displaystyle g\in {\mathcal {L}}^{1}(X)}
then
f
∈
L
1
{\displaystyle f\in {\mathcal {L}}^{1}}
and
∫
X
|
f
n
−
f
|
d
μ
→
0
{\displaystyle \displaystyle \int _{X}|f_{n}-f|d\mu \to 0}
as
n
→
∞
{\displaystyle n\to \infty }
Proof
We know that
|
f
|
≤
g
{\displaystyle |f|\leq g}
and hence
|
f
n
−
f
|
≤
2
g
{\displaystyle |f_{n}-f|\leq 2g}
, that is,
0
≤
2
g
−
|
f
n
−
f
|
{\displaystyle 0\leq 2g-|f_{n}-f|}
Therefore, by Fatou's lemma,
∫
X
2
g
d
μ
≤
lim inf
∫
X
(
2
g
−
|
f
n
−
f
|
)
d
μ
≤
∫
X
2
g
d
μ
+
lim inf
∫
X
(
−
|
f
n
−
f
|
)
d
μ
{\displaystyle \displaystyle \int _{X}2gd\mu \leq \liminf \int _{X}(2g-|f_{n}-f|)d\mu \leq \displaystyle \int _{X}2gd\mu +\liminf \int _{X}(-|f_{n}-f|)d\mu }
=
∫
X
2
g
d
μ
−
lim sup
∫
X
|
f
n
−
f
|
d
μ
{\displaystyle =\displaystyle \int _{X}2gd\mu -\limsup \int _{X}|f_{n}-f|d\mu }
As
g
∈
L
1
{\displaystyle g\in {\mathcal {L}}^{1}}
,
lim sup
n
→
∞
∫
X
|
f
n
−
f
|
d
μ
≤
{\displaystyle \displaystyle \limsup _{n\to \infty }\int _{X}|f_{n}-f|d\mu \leq }
implying that
lim sup
∫
X
|
f
n
−
f
|
d
μ
{\displaystyle \displaystyle \limsup \int _{X}|f_{n}-f|d\mu }
Suppose
f
:
X
→
[
0
,
∞
]
{\displaystyle f:X\to [0,\infty ]}
is measurable,
E
∈
σ
{\displaystyle E\in \sigma }
with
μ
(
E
)
>
0
{\displaystyle \mu (E)>0}
such that
∫
E
f
d
μ
=
0
{\displaystyle \displaystyle \int _{E}fd\mu =0}
. Then
f
(
x
)
=
0
{\displaystyle f(x)=0}
almost everywhere
E
{\displaystyle E}
Let
f
∈
L
1
(
X
)
{\displaystyle f\in {\mathcal {L}}^{1}(X)}
and let
∫
E
f
d
μ
=
0
{\displaystyle \displaystyle \int _{E}fd\mu =0}
for every
E
∈
σ
{\displaystyle E\in \sigma }
. Then,
f
=
0
{\displaystyle f=0}
almost everywhere on
X
{\displaystyle X}
Let
f
∈
L
1
(
X
)
{\displaystyle f\in {\mathcal {L}}^{1}(X)}
and
|
∫
X
f
d
μ
|
=
∫
X
|
f
|
d
μ
{\displaystyle \displaystyle \left|\int _{X}fd\mu \right|=\int _{X}|f|d\mu }
then there exists constant
α
{\displaystyle \alpha }
such that
|
f
|
=
α
f
{\displaystyle |f|=\alpha f}
almost everywhere on
E
{\displaystyle E}
Proof
For each
n
∈
N
{\displaystyle n\in \mathbb {N} }
define
A
n
=
{
x
∈
E
|
f
(
x
)
>
1
n
}
{\displaystyle A_{n}=\{x\in E|f(x)>{\frac {1}{n}}\}}
. Observe that
A
n
↑
E
{\displaystyle A_{n}\uparrow E}
but
1
n
μ
(
A
n
)
≤
∫
A
n
f
d
μ
≤
∫
E
f
d
μ
=
0
{\displaystyle {\frac {1}{n}}\mu (A_{n})\leq \displaystyle \int _{A_{n}}fd\mu \leq \int _{E}fd\mu =0}
Thus
μ
(
A
n
)
=
0
{\displaystyle \mu (A_{n})=0}
for all
n
{\displaystyle n}
, by continuity,
f
=
0
{\displaystyle f=0}
almost everywhere on
E
{\displaystyle E}
Write
∫
E
f
d
μ
=
∫
E
u
+
d
μ
−
∫
E
u
−
d
μ
+
i
(
∫
E
v
+
d
μ
−
∫
E
v
−
d
μ
)
{\displaystyle \displaystyle \int _{E}fd\mu =\int _{E}u^{+}d\mu -\int _{E}u^{-}d\mu +i\left(\int _{E}v^{+}d\mu -\int _{E}v^{-}d\mu \right)}
, where
u
+
,
u
−
,
v
+
,
v
−
{\displaystyle u^{+},u^{-},v^{+},v^{-}}
are non-negative real measurable. Further as
∫
E
u
+
d
μ
,
∫
E
(
−
u
−
)
d
μ
{\displaystyle \displaystyle \int _{E}u^{+}d\mu ,\int _{E}(-u^{-})d\mu }
are both non-negative, each of them is zero. Thus, by applying part I, we have that
u
+
,
u
−
{\displaystyle u^{+},u^{-}}
vanish almost everywhere on
E
{\displaystyle E}
. We can similarly show that
v
+
,
v
−
{\displaystyle v^{+},v^{-}}
vanish almost everywhere on
E
{\displaystyle E}
.