7.7. THE LEBESGUE INTEGRAL, L1 167

Theorem 7.7.5 ∫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) . Also, 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. (7.4)

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)

= cRe(g)−b Im( f )−d Im(g)+aRe( f )+ i(bRe( f )+ c Im(g)+a Im( f )+d Re(g))

It follows from the definition that∫(a+ ib) f +(c+ id)gdµ =

∫(cRe(g)−b Im( f )−d Im(g)+aRe( f ))dµ

+i∫

(bRe( f )+ c Im(g)+a Im( f )+d Re(g)) (7.5)

Also, from the definition,

(a+ ib)∫

f dµ +(c+ id)∫

gdµ = (a+ ib)(∫

Re f dµ + i∫

Im f dµ

)+(c+ id)

(∫Regdµ + i

∫Imgdµ

)which equals

= a∫

Re f dµ−b∫

Im f dµ + ib∫

Re f dµ + ia∫

Im f dµ

+c∫

Regdµ−d∫

Imgdµ + id∫

Regdµ−d∫

Imgdµ.

Using Lemma 7.7.4 and collecting terms, it follows that this reduces to 7.5. Thus theintegral is linear as claimed.

Consider the claim about approximation with a simple function. Letting h equal anyof

(Re f )+ ,(Re f )− ,(Im f )+ ,(Im f )− , (7.6)