Let
be a
-finite measure space. Suppose
is a positive simple measurable function, with
;
are disjoint.
Define
Let
be measurable, and let
.
Define
Now let
be any measurable function. We say that
is integrable if
and
are integrable and if
. Then, we write
The class of measurable functions on
is denoted by
For
, we define
to be the collection of all measurable functions
such that
A property is said to hold almost everywhere if the set of all points where the property does not hold has measure zero.
Let
be a measure space and let
be measurable on
. Then
- If
, then ![{\displaystyle \displaystyle \int _{X}fd\mu \leq \int _{X}gd\mu }](https://wikimedia.org/api/rest_v1/media/math/render/svg/044fde012d4c441d7c45a6f23459a6c906b1aaa2)
- If
,
, then ![{\displaystyle \displaystyle \int _{A}fd\mu \leq \int _{B}fd\mu }](https://wikimedia.org/api/rest_v1/media/math/render/svg/5080a93d153862832e0f950debf2d7fa99f4daae)
- If
and
then ![{\displaystyle \displaystyle \int _{X}cfd\mu =c\int _{X}fd\mu }](https://wikimedia.org/api/rest_v1/media/math/render/svg/b18b7e0883e867770c391a4b7e52e8127b271a14)
- If
,
, then
, even if ![{\displaystyle f(E)=\{\infty \}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3622195a1c905001906f0cbd43789520af3f706b)
- If
,
, then
, even if ![{\displaystyle \mu (E)=\infty }](https://wikimedia.org/api/rest_v1/media/math/render/svg/1b38efab7c5b1d13495b79ee52d5c12ed992d169)
Proof
Suppose
and
are measurable for all
such that
for every ![{\displaystyle x\in X}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3e580967f68f36743e894aa7944f032dda6ea01d)
almost everywhere on ![{\displaystyle X}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68baa052181f707c662844a465bfeeb135e82bab)
Then,
Proof
is an increasing sequence in
, and hence,
(say). We know that
is measurable and that
. That is,
Hence,
Let
Define
;
. Observe that
and
Suppose
. If
then
implying that
. If
, then there exists
such that
and hence,
.
Thus,
, therefore
. As this is true if
, we have that
. Thus,
.
Let
be measurable functions. Then,
Proof
For
define
. Observe that
are measurable and increasing for all
.
As
,
. By monotone convergence theorem,
and as
, we have the result.
Let
be a complex measure space. Let
be a sequence of complex measurable functions that converge pointwise to
;
, with
Suppose
for some
then
and
as
Proof
We know that
and hence
, that is,
Therefore, by Fatou's lemma,
As
,
implying that
- Suppose
is measurable,
with
such that
. Then
almost everywhere ![{\displaystyle E}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4232c9de2ee3eec0a9c0a19b15ab92daa6223f9b)
- Let
and let
for every
. Then,
almost everywhere on ![{\displaystyle X}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68baa052181f707c662844a465bfeeb135e82bab)
- Let
and
then there exists constant
such that
almost everywhere on ![{\displaystyle E}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4232c9de2ee3eec0a9c0a19b15ab92daa6223f9b)
Proof
- For each
define
. Observe that ![{\displaystyle A_{n}\uparrow E}](https://wikimedia.org/api/rest_v1/media/math/render/svg/90d373195af16d7b1a284aa72ca91b989d7f7d27)
but
Thus
for all
, by continuity,
almost everywhere on ![{\displaystyle E}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4232c9de2ee3eec0a9c0a19b15ab92daa6223f9b)
- Write
, where
are non-negative real measurable.
Further as
are both non-negative, each of them is zero. Thus, by applying part I, we have that
vanish almost everywhere on
. We can similarly show that
vanish almost everywhere on
.