286 CHAPTER 10. THE ABSTRACT LEBESGUE INTEGRAL
Proof: By assumption, g+ h′ = g′+ h. Then from the Lebesgue integral’s righteousalgebraic desires, Theorem 10.6.1,
∫gdµ +
∫h′dµ =
∫g′dµ +
∫hdµ which implies the
claimed result. ■
Lemma 10.7.5 Let Re(L1 (Ω)
)denote the vector space of real valued functions in
L1 (Ω) where the field of scalars is the real numbers. Then∫
dµ is linear on Re(L1 (Ω)
),
the scalars being real numbers.
Proof: First observe that from the definition of the positive and negative parts of a func-tion, ( f +g)+−( f +g)−= f++g+−( f−+g−) because both sides equal f +g. Thereforefrom Lemma 10.7.4 and the definition, it follows from Theorem 10.6.1 that∫
f +gdµ ≡∫
( f +g)+− ( f +g)− dµ =∫
f++g+dµ−∫
f−+g−dµ
=∫
f+dµ +∫
g+dµ−(∫
f−dµ +∫
g−dµ
)=∫
f dµ +∫
gdµ.
what about taking out scalars? First note that if a is real and nonnegative, then (a f )+ = a f+
and (a f )− = a f− while if a < 0, then (a f )+ = −a f− and (a f )− = −a f+. These claimsfollow immediately from the above definitions of positive and negative parts of a function.Thus if a < 0 and f ∈ L1 (Ω) , it follows from Theorem 10.6.1 that∫
a f dµ ≡∫
(a f )+ dµ−∫
(a f )− dµ =∫
(−a) f−dµ−∫
(−a) f+dµ
= −a∫
f−dµ +a∫
f+dµ = a(∫
f+dµ−∫
f−dµ
)≡ a
∫f dµ.
The case where a≥ 0 works out similarly but easier. ■Now here is the main result.
Theorem 10.7.6 ∫dµ is linear on L1 (Ω) and L1 (Ω) is a complex vector space. If
f ∈ L1 (Ω) , then Re f , Im f , and | f | are all in L1 (Ω) . Furthermore, for f ∈ L1 (Ω) ,∫f dµ ≡
∫(Re f )+ dµ−
∫(Re f )− dµ + i
[∫(Im f )+ dµ−
∫(Im f )− dµ
]≡
∫Re f dµ + i
∫Im f dµ
and the triangle inequality holds, ∣∣∣∣∫ f dµ
∣∣∣∣≤ ∫ | f |dµ. (10.4)
Also, for every f ∈ L1 (Ω) it follows that for every ε > 0 there exists a simple function ssuch that |s| ≤ | f | and
∫| f − s|dµ < ε.
Proof: First consider the claim that the integral is linear. It was shown above that theintegral is linear on Re
(L1 (Ω)
). Then letting a+ ib,c+ id be scalars and f ,g functions in
L1 (Ω) ,
(a+ ib) f +(c+ id)g = (a+ ib)(Re f + i Im f )+(c+ id)(Reg+ i Img)