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-
Re f = -----, Im f =-----.
2 2i
Definition 8.7.1Let
(Ω,S,μ )
be a measure space and suppose f : Ω → ℂ.Then f is said to be measurable if bothRef andImf are measurable real valuedfunctions.
Of course there is another definition of measurability which says that inverse images of
measurable sets are measurable. This is equivalent to this new definition.
Lemma 8.7.2Let f : Ω → ℂ. Then f is measurable if and only ifRef,Imf areboth real valued measurable functions. Also if f,g are complex measurable functions anda,b are complex scalars, then af + bg is also measurable.
Proof: ⇒Suppose first that f is measurable. Recall that ℂ is considered as ℝ^{2} with
(x,y)
being identified with x + iy. Thus the open sets of ℂ can be obtained with either of the two
equivlanent norms
|z|
≡
∘ ---------------
(Rez)2 + (Im z)2
or
∥z∥
_{
∞} = max
(Re z,Im z)
. Therefore, if f is
measurable
Re f−1(a,b) ∩Im f− 1(c,d) = f −1((a,b)+ i(c,d)) ∈ ℱ
In particular, you could let
(c,d)
= ℝ and conclude that Ref is measurable because in this
case, the above reduces to the statement that Ref^{−1}
(a,b)
∈ℱ. Similarly Imf is
measurable.
⇐ Next, if each of Ref and Imf are measurable, then
f−1((a,b)+ i(c,d)) = Re f−1(a,b)∩Im f −1(c,d) ∈ ℱ
and so, since every open set is the countable union of sets of the form
(a,b)
+ i
(c,d)
, it
follows that f is measurable.
Now consider the last claim. Let
h : ℂ× ℂ → ℂ
be given by h
(z,w )
≡ az + bw. Then h is continuous. If f,g are complex valued measurable
functions, consider the complex valued function,
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 8.7.4Let g,h,g^{′},h^{′}be nonnegative measurable functions in L^{1}
(Ω)
and supposethat
g − h = g′ − h′.
Then
∫ ∫ ∫ ′ ∫ ′
gdμ − hdμ = gdμ − hdμ.
Proof:By assumption, g + h^{′} = g^{′} + h. Then from the Lebesgue integral’s righteous
algebraic desires, Theorem 8.6.1,
∫ ∫ ∫ ∫
′ ′
gdμ + h dμ = g dμ+ hdμ
which implies the claimed result. ■
Lemma 8.7.5LetRe
(L1(Ω ))
denote the vector space of real valued functions inL^{1}
(Ω )
where the field of scalars is the real numbers. Then∫dμ is linear onRe
(L1(Ω))
,the scalars being real numbers.
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 8.7.4 and the definition, it follows
from Theorem 8.6.1 that
, then it is known that for a,b scalars, it follows that af + bg is measurable.
See Lemma 8.7.2. 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 8.7.7f ∈ L^{1}(Ω) if and only if there exists a sequence of complex simplefunctions,
{sn}
such that
sn(ω) → f (∫ω) for all ω ∈ Ω
limm,n →∞ (|sn − sm |) = 0 (8.7)
(8.7)
When f ∈ L^{1}
(Ω )
,
∫ ∫
fdμ ≡ lim sn. (8.8)
n→ ∞
(8.8)
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
∫ ∫ ∫
|s − s |dμ ≤ |s − f|dμ + |f − s |dμ ≤ 1-+ 1-.
n m n m 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
{s }
n
be the approximating sequence, Fatou’s lemma
implies
|∫ ∫ | ∫ ∫
|| fdμ − s dμ||≤ |f − s|dμ ≤ lim inf |s − s |dμ < ε
| n | n m→ ∞ m n