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
-- --
Ref = f-+-f, Im f = f-−-f.
2 2i
Definition 20.7.1Let
(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.
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 20.7.2Let f : Ω → ℂ. Then f is measurable if and only ifRef,Imf are both real valuedmeasurable functions. Also if f,g are complex measurable functions and a,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|
≡
∘ -----2--------2
(Re z) +(Im z)
or
∥z∥
_{
∞} = max
(Rez,Im z)
. Therefore, if f is
measurable
−1 −1 −1
Re f (a,b)∩Im f (c,d) = f ((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
−1 −1 −1
f ((a,b)+ i(c,d)) = Re f (a,b)∩ Im f (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 20.7.4Let g,h,g^{′},h^{′}be nonnegative measurable functions in L^{1}
(Ω)
and supposethat
g− h = g′ − h ′.
Then
∫ ∫ ∫ ∫
gdμ − hdμ = g′dμ − h′dμ.
Proof:By assumption, g + h^{′} = g^{′} + h. Then from the Lebesgue integral’s righteous algebraic desires,
Theorem 20.6.1,
∫ ∫ ′ ∫ ′ ∫
gdμ + h dμ = gdμ + hdμ
which implies the claimed result. ■
Lemma 20.7.5LetRe
(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 20.7.4 and the definition, it follows from Theorem
20.6.1 that
. Then from what was shown
above about the integral being linear,
|∫ | ∫ ∫ ∫ ∫
|| f dμ||= θ fdμ = θfdμ = Re (θf)dμ ≤ |f|dμ.
| |
If f,g ∈ L^{1}
(Ω )
, then it is known that for a,b scalars, it follows that af + bg is measurable. See Lemma
20.7.2. 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 20.7.7f ∈ L^{1}(Ω) if and only if there exists a sequence of complex simple functions,
{sn}
suchthat
sn(ω) → f (ω ) for all ω ∈ Ω
lim ∫ (|s − s |) = 0 (20.7)
m,n→ ∞ n m
(20.7)
When f ∈ L^{1}
(Ω )
,
∫ ∫
fdμ ≡ lim s . (20.8)
n→∞ n
(20.8)
Proof: From the above theorem, if f ∈ L^{1} there exists a sequence of simple functions
{sn}
such
that
∫
|f − sn|dμ < 1∕n, sn(ω) → f (ω ) for all ω
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 inf |sn|dμ < ∞
n→ ∞
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
This is a good time to observe the following fundamental observation which follows from a repeat of the
above arguments.
Theorem 20.7.8Suppose Λ
(f)
∈ [0,∞] for all nonnegative measurable functions and suppose that fora,b ≥ 0 and f,g nonnegative measurable functions,
Λ (af + bg) = aΛ (f)+ bΛ(g).
In other words, Λ wants to be linear. Then Λ has a unique linear extension to the set of measurablefunctions
{f measurable : Λ (|f|) < ∞ },
this set being a vector space. Since all the manipulations used apply just as well to continuous asmeasurable, you can replace the word “measurable” in the above with the word “continuous” and∫dμ with
Λ where Λ wants to be linear and acts like it is on the set of nonnegative functions.