Measure, Integration & Real Analysis, 2021a

Section 6B Vector Spaces 157
You can easily show that if f , g : X → C are S-measurable functions such that
∫ | f | dμ < ∞ and
∫ |g| dμ < ∞, then
∫
∫
( f + g) dμ =
∫
f dμ +
g dμ.
Similarly, the definition of complex multiplication leads to the conclusion that
∫
∫
α f dμ = α f dμ
for all α ∈ C (see Exercise 8).
The inequality in the result below concerning integration of complex-valued
functions does not follow immediately from the corresponding result for real-valued
functions. However, the small trick used in the proof below does give a reasonably
simple proof.
6.22 bound on the absolute value of an integral
Suppose (X, S, μ) is a measure space and f : X → C is an S-measurable
function such that ∫ | f | dμ < ∞. Then
∫
∫
∣ f dμ∣ ≤ | f | dμ.
Proof The result clearly holds if ∫ f dμ = 0. Thus assume that ∫ f dμ ̸= 0.
Let
α = |∫ f dμ|
∫ . f dμ
Then
∫
∣
∫
f dμ∣ = α
∫
f dμ =
α f dμ
∫
=
∫
Re(α f ) dμ + i
Im(α f ) dμ
∫
=
Re(α f ) dμ
∫
≤
|α f | dμ
∫
=
| f | dμ,
where the second equality holds by Exercise 8, the fourth equality holds because
| ∫ f dμ| ∈R, the inequality on the fourth line holds because Re z ≤|z| for every
complex number z, and the equality in the last line holds because |α| = 1.
Because of the result above, the Bounded Convergence Theorem (3.26) and the
Dominated Convergence Theorem (3.31) hold if the functions f 1 , f 2 ,...and f in the
statements of those theorems are allowed to be complex valued.
Measure, Integration & Real Analysis, by Sheldon Axler