The functions considered here have values in ℂ, a vector space.
Definition 9.4.1Let
(Ω,S,μ)
be a measure space and suppose f : Ω → ℂ. Then fis said to be measurable if bothRef andImf are measurable real valued functions.
Definition 9.4.2A complex simple function will be a function which is of theform
∑n
s(ω) = ckXEk (ω)
k=1
where c_{k}∈ ℂ and μ
(Ek)
< ∞. For s a complex simple function as above, define
∑n
I(s) ≡ ckμ(Ek).
k=1
Lemma 9.4.3The definition, 9.4.2is well defined. Furthermore, I is linearon the vector space of complex simple functions. Also the triangle inequalityholds,
whenever n is large enough. Since ε is arbitrary, this shows the limit from using the t_{n} is
the same as the limit from using s_{n}. This proves the lemma.
What if f has values in [0,∞)? Earlier ∫fdμ was defined for such functions and now
I
(f)
has been defined. Are they the same? If so, I can be regarded as an extension of
∫dμ to a larger class of functions.
Lemma 9.4.6Suppose f has values in [0,∞) and f ∈ L^{1}
(Ω)
. Then f is measurableand
∫
I(f) = fdμ.
Proof: Since f is the pointwise limit of a sequence of complex simple functions,
{sn}
having the properties described in Definition 9.4.4, it follows f
(ω )
= lim_{n→∞}Res_{n}
(ω)
and so f is measurable. Also
∫ | | ∫ ∫
||(Re sn)+ − (Re sm)+||dμ ≤ |Re sn − Re sm|dμ ≤ |sn − sm|dμ
where x^{+}≡
1
2
(|x|+ x)
, the positive part of the real number, x.
^{2}Thus
there is no loss of generality in assuming
{sn}
is a sequence of complex simple
functions having values in [0,∞). Then since for such complex simple functions,
I
whenever n is large enough. But by Fatou’s lemma, Theorem 9.3.18 on Page 670, the last
term is no larger than
∫
lim inf |s − s |dμ < ε
k→∞ n k
whenever n is large enough. Since ε is arbitrary, this shows I
(f)
= ∫fdμ as
claimed.
As explained above, I can be regarded as an extension of ∫dμ so from now on,
the usual symbol, ∫dμ will be used. It is now easy to verify ∫dμ is linear on
L^{1}
is the version most often used because it is easy to
verify the conditions for it.
Corollary 9.4.8Let
(Ω,S,μ)
be a measure space and let f : Ω → ℂ. Thenf ∈ L^{1}
(Ω)
if and only if f is measurable and∫
|f|
dμ < ∞.
Proof: Suppose f ∈ L^{1}
(Ω)
. Then from Definition 9.4.4, it follows both real and
imaginary parts of f are measurable. Just take real and imaginary parts of s_{n} and
observe the real and imaginary parts of f are limits of the real and imaginary parts of s_{n}
respectively. Why is ∫
|f|
dμ < ∞? It follows from Theorem 9.4.7. Recall why this
was so. Let
{s }
n
be a sequence of simple functions attached to f as in the
definition of what it means to be L^{1}. Then from the definition of I
(s)
for s
simple,
|I(|sn|− |sm|)| ≤ I (|sn − sm |)
which converges to 0. Since
{I(|s |)}
n
is a Cauchy sequence, it is bounded by a constant
C and also
{|s |}
n
is a sequence of simple functions of the right sort which converges
pointwise to
|f|
and so by definition,
∫
|f|dμ = I (f ) = lim I(|sn|) ≤ C.
n→∞
This shows the only if part.
The more interesting part is the if part. Suppose then that f is measurable and
∫
|f|
dμ < ∞. Suppose first that f has values in [0,∞). It is necessary to obtain the
sequence of complex simple functions. By Theorem 9.3.9, there exists a sequence of
nonnegative simple functions,
{sn}
such that s_{n}
(ω)
↑ f
(ω)
. Then by the monotone
convergence theorem,
∫ ∫
lim (2f − (f − s ))dμ = 2fdμ
n→∞ n
and so
∫
lim (f − sn)dμ = 0.
n→ ∞
Letting m be large enough, it follows ∫
(f − sm)
dμ < ε and so if n > m
∫ ∫
|sm − sn|dμ ≤ |f − sm|dμ < ε.
Therefore, f ∈ L^{1}
(Ω)
because
{sn}
is a suitable sequence.
The general case follows from considering positive and negative parts of real and
imaginary parts of f. These are each measurable and nonnegative and their integral is
finite so each is in L^{1}
(Ω)
by what was just shown. Thus
( )
f = Ref+ − Re f− + i Im f+ − Im f−
and so f ∈ L^{1}
(Ω)
. This proves the corollary.
Theorem 9.4.9(Dominated Convergence theorem)Let f_{n}∈ L^{1}(Ω) and suppose
f(ω) = lim fn(ω),
n→ ∞
and there exists a measurable function g, with values in
[0,∞],^{3}such that
∫
|fn(ω)| ≤ g(ω ) and g(ω)dμ < ∞.
Then f ∈ L^{1}(Ω) and
∫ ∫
fdμ = lnim→∞ fndμ.
Proof: f is measurable by Theorem 9.1.8. Since |f|≤ g, it follows that