The functions considered here have values in ℂ, which is a vector space. A function f with
values in ℂ is of the form f = Ref + iImf where Ref and Imf are real valued functions. In
fact
f + f- f − f-
Ref = -----, Im f =-----.
2 2i
We first define the integral of real valued functions and then the integral of a complex valued function will
be of the form
∫ ∫ ∫
fdμ = Re(f)dμ + i Im (f)dμ
Definition 6.7.1Let
(Ω,S,μ)
be a measure space and suppose f : Ω → ℂ. Then f is said to bemeasurable if bothRef andImf are measurable real valued functions.
As is always the case for complex numbers,
|z|
^{2} =
(Re z)
^{2} +
(Im z)
^{2}. Also, for g a real valued
function, one can consider its positive and negative parts defined respectively as
= g^{+} + g^{−} and g = g^{+}− g^{−} and both g^{+} and g^{−} are measurable nonnegative functions if g
is measurable. This follows because of Theorem 5.1.5. The mappings x → x^{+},x → x^{−} are
clearly continuous. Thus g^{+} is the composition of a continuous function with a measurable
function.
Then the following is the definition of what it means for a complex valued function f to be in
L^{1}
(Ω )
.
Definition 6.7.2Let
(Ω,ℱ,μ)
be a measure space.Then a complex valued measurable function f is inL^{1}
I will show that with this definition, the integral is linear and well defined. First note that it is
clearly well defined because all the above integrals are of nonnegative functions and are each
equal to a nonnegative real number because for h equal to any of the functions,
|h|
≤
|f|
and
∫
|f|
dμ < ∞.
Here is a lemma which will make it possible to show the integral is linear.
Lemma 6.7.3Let g,h,g^{′},h^{′}be nonnegative measurable functions in L^{1}
(Ω)
and supposethat
g− h = g′ − h ′.
Then
∫ ∫ ∫ ′ ∫ ′
gdμ − hdμ = gdμ − h dμ.
Proof:By assumption, g + h^{′} = g^{′} + h. Then from the Lebesgue integral’s righteous algebraic desires,
Theorem 6.6.1,
∫ ∫ ∫ ∫
gdμ + h′dμ = g′dμ + hdμ
which implies the claimed result. ■
Lemma 6.7.4LetRe
(L1(Ω ))
denote the vector space of real valued functions in L^{1}
(Ω )
wherethe field of scalars is the real numbers. Then∫dμ is linear onRe
(L1(Ω))
, the scalars being realnumbers.
Proof: First observe that from the definition of the positive and negative parts of a function,
( )
(f +g)+ − (f + g)− = f+ + g+ − f− +g−
because both sides equal f + g. Therefore from Lemma 6.7.3 and the definition, it follows from Theorem
6.6.1 that
. Then from what was shown
above about the integral being linear,
||∫ || ∫ ∫ ∫ ∫
|| f dμ|| = θ fdμ = θfdμ = Re (θf)dμ ≤ |f|dμ.
It is routine to verify that for f,g measurable, meaning real and imaginary parts are measurable, then
any complex linear combination is also measurable. This follows right away from Theorem 5.1.5 and
looking at the real and imaginary parts of this complex linear combination. Also
∫ ∫
|af + bg|dμ ≤ |a||f|+ |b||g|dμ < ∞. ■
The following corollary follows from this. The conditions of this corollary are sometimes taken as a
definition of what it means for a function f to be in L^{1}
(Ω )
.
Corollary 6.7.6f ∈ L^{1}(Ω) if and only if there exists a sequence of complex simple functions,
{sn}
suchthat
s (ω) → f (ω ) for all ω ∈ Ω
n ∫ (6.7)
limm,n→ ∞ (|sn − sm |) = 0
(6.7)
When f ∈ L^{1}
(Ω )
,
∫ ∫
fdμ ≡ lnim→∞ sn. (6.8)
(6.8)
Proof: From the above theorem, if f ∈ L^{1} there exists a sequence of simple functions
{sn}
such
that
∫
|f − s |dμ < 1∕n, s (ω) → f (ω ) for all ω
n n
Then
∫ ∫ ∫
|sn − sm |dμ ≤ |sn − f|dμ + |f − sm|dμ ≤ 1-+ 1-.
n m
Next suppose the existence of the approximating sequence of simple functions. Then f is
measurable because its real and imaginary parts are the limit of measurable functions. By Fatou’s
lemma,
∫ ∫
|f |dμ ≤ lim nin→f∞ |sn|dμ < ∞
because
| |
||∫ ∫ || ∫
| |sn|dμ − |sm |dμ| ≤ |sn − sm|dμ
which is given to converge to 0. Hence
{∫ |sn|dμ}
is a Cauchy sequence and is therefore, bounded.
In case f ∈ L^{1}
(Ω )
, letting
{sn}
be the approximating sequence, Fatou’s lemma implies
|∫ ∫ | ∫ ∫
|| f dμ− s dμ||≤ |f − s |dμ ≤ lim inf |s − s |dμ < ε
| n | n m→ ∞ m n