208 CHAPTER 9. THE LEBESGUE INTEGRAL

Definition 9.7.3 Let (Ω,F ,µ) be a measure space. Then a complex valued mea-surable function f is in L1 (Ω) if

∫| f |dµ < ∞. For a function in L1 (Ω) , the integral is

defined as follows.∫f dµ ≡

∫(Re f )+ dµ−

∫(Re f )− dµ + i

[∫(Im f )+ dµ−

∫(Im f )− dµ

]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 functionsand 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 9.7.4 Let g,h,g′,h′ be nonnegative measurable functions in L1 (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 righteousalgebraic desires, Theorem 9.6.1,

∫gdµ +

∫h′dµ =

∫g′dµ +

∫hdµ which implies the

claimed result. ■

Lemma 9.7.5 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 9.7.4 and the definition, it follows from Theorem 9.6.1 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 9.6.1 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 9.7.6 ∫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

208 CHAPTER 9. THE LEBESGUE INTEGRALDefinition 9.7.3 Ler (Q,-F,u) be a measure space. Then a complex valued mea-surable function f is in L!(Q) if [\f|du <%. For a function in L' (Q), the integral isdefined as follows.[tans [ Rep) au [ (Ref an +i fampt an | aim)~an|I will show that with this definition, the integral is linear and well defined. First notethat it is clearly well defined because all the above integrals are of nonnegative functionsand are each equal to a nonnegative real number because for h equal to any of the functions,|h| < |fland f\fldp <e.Here is a lemma which will make it possible to show the integral is linear.Lemma 9.7.4 Let g,h,g',h' be nonnegative measurable functions in L' (Q) and supposethat g—h=g'—h'.Then f[ gdu— fhdu= f edu —fh'du.Proof: By assumption, g +h! = g'+h. Then from the Lebesgue integral’s righteousalgebraic desires, Theorem 9.6.1, fgdu+ fh'du = f e’du + fhdp which implies theclaimed result.Lemma 9.7.5 Let Re (L' (Q)) denote the vector space of real valued functions in L' (Q)where the field of scalars is the real numbers. Then f du is linear on Re(L!(Q)), thescalars being real numbers.Proof: First observe that from the definition of the positive and negative parts of a func-tion, (f +g)’ —(f+g) =ft+get—(f~ +87) because both sides equal f +g. Thereforefrom Lemma 9.7.4 and the definition, it follows from Theorem 9.6.1 that[tf sdu [tay -(+8) du= [pr +etau— [fp +enan[rau [eran - ( [rans [e-au) ~ [sau [ edu.what about taking out scalars? First note that if a is real and nonnegative, then (af yt =aftand (af) =af~ while if a <0, then (af)* = —af~ and (af)” = —af*. These claimsfollow immediately from the above definitions of positive and negative parts of a function.Thus if a < 0 and f € L! (Q), it follows from Theorem 9.6.1 thatfatdu = [apy au- [apy du= | (-a fae | (-a) frau-af Pana pan=al [rau | ran) =a | fa.The case where a > 0 works out similarly but easier. HiNow here is the main result.Theorem 9.7.6 [dy is linear on L! (Q) and L! (Q) is a complex vector space. Iff €L' (Q), then Re f,Im f, and | f| are all in L' (Q). Furthermore, for f € L' (Q),[fa[ower aw [wer ani] famsyau— [ams ay][Refdu+i [1m fay