Kenneth Kuttler

338 CHAPTER 14. THE LEBESGUE INTEGRAL

Proof: Let gn(ω) = inf{ fk(ω) : k ≥ n}. Then g−1n ([a,∞]) = ∩∞

k=n f−1k ([a,∞]) ∈ F

Thus gn is measurable. Now the functions gn form an increasing sequence of nonnega-tive measurable functions. Thus g−1 ((a,∞)) = ∪∞

n=1g−1n ((a,∞)) ∈F so g is measurable

also. By monotone convergence theorem,∫

gdµ = limn→∞

∫gndµ ≤ liminfn→∞

∫fndµ.

The last inequality holding because∫

gndµ ≤∫

fndµ . (Note that it is not known whetherlimn→∞

∫fndµ exists.)

14.10 The Integral’s Righteous Algebraic DesiresThe monotone convergence theorem shows the integral wants to be linear. This is theessential content of the next theorem.

Theorem 14.10.1 Let f ,g be nonnegative measurable functions and let a,b benonnegative numbers. Then a f +bg is measurable and∫

(a f +bg)dµ = a∫

f dµ +b∫

gdµ. (14.7)

Proof: By Theorem 14.5.6 on Page 332 there exist increasing sequences of nonnegativesimple functions, sn→ f and tn→ g. Then a f +bg, being the pointwise limit of the simplefunctions asn+btn, is measurable. Now by the monotone convergence theorem and Lemma14.8.3,

∫(a f +bg)dµ =

limn→∞

∫asn +btndµ = lim

n→∞

(a∫

sndµ +b∫

tndµ

)= a

∫f dµ +b

∫gdµ.

14.11 Integrals of Real Valued FunctionsAs long as you are allowing functions to take the value +∞, you cannot consider somethinglike f +(−g) and so you can’t very well expect a satisfactory statement about the integralbeing linear until you restrict yourself to functions which have values in a vector space. Tobe linear, a function must be defined on a vector space. The integral of real valued functionsis next.

Definition 14.11.1 Let (Ω,F ,µ) be a measure space and let f : Ω→ R be mea-surable. Then it is said to be in L1 (Ω,µ) when

∫Ω| f (ω)|dµ < ∞.

Lemma 14.11.2 If g−h = ĝ− ĥ where g, ĝ,h, ĥ are measurable and nonnegative, withall integrals finite, then ∫

Ω

gdµ−∫

Ω

hdµ =∫

Ω

ĝdµ−∫

Ω

ĥdµ

Proof: From Theorem 14.10.1,∫ĝdµ +

∫hdµ =

∫(ĝ+h)dµ =

∫ (g+ ĥ

)dµ =

∫gdµ +

∫ĥdµ

and so, ∫ĝdµ−

∫ĥdµ =

∫gdµ−

∫hdµ

The functions you can integrate are those which have | f | integrable, and then you canmake sense of

∫f dµ for f having values in R or C although here, I will emphasize R.

338 CHAPTER 14. THE LEBESGUE INTEGRALProof: Let g,(@) = inf{f,(@) :k >n}. Then g,!([a,0]) =f; | (la,~]) © FThus g, is measurable. Now the functions g, form an increasing sequence of nonnega-tive measurable functions. Thus g~! ((a,0°)) = U%_,g,! ((a,0e)) € F so g is measurablealso. By monotone convergence theorem, f gdu = limy +0 f gndu < liminf, +. f frdU.The last inequality holding because { g,du < f f,dp. (Note that it is not known whetherlimy—yoo f fnd exists.) Wl14.10 The Integral’s Righteous Algebraic DesiresThe monotone convergence theorem shows the integral wants to be linear. This is theessential content of the next theorem.Theorem 14.10.1 Ler f,g be nonnegative measurable functions and let a,b benonnegative numbers. Then af + bg is measurable and[tafrbe)du=a [fay +b | gap. (14.7)Proof: By Theorem 14.5.6 on Page 332 there exist increasing sequences of nonnegativesimple functions, s, — f and t, - g. Then af + bg, being the pointwise limit of the simplefunctions as, + bt, is measurable. Now by the monotone convergence theorem and Lemma14.8.3, f (af +bg)du =n-ootim [ as, + bed = tim (4 foray +0 [ mds) =a | fay +b [ gdp. 114.11 Integrals of Real Valued FunctionsAs long as you are allowing functions to take the value +-co, you cannot consider somethinglike f + (—g) and so you can’t very well expect a satisfactory statement about the integralbeing linear until you restrict yourself to functions which have values in a vector space. Tobe linear, a function must be defined on a vector space. The integral of real valued functionsis next.Definition 14.11.1 Le (Q, F,U) be a measure space and let f : Q— R be mea-surable. Then it is said to be in L!(Q,) when Jo |f (@)|du < ».Lemma 14.11.2 if g—h= 8 —h where g,8,h,h are measurable and nonnegative, withall integrals finite, then[isdu- [indy =| edu | fayfo) Q Q QProof: From Theorem 14.10.1,[gdur [rau = [(@+nau= | (¢+hau= | eau+ [ha[edu fidu= [eau [rauThe functions you can integrate are those which have |f| integrable, and then you canmake sense of { fdu for f having values in R or C although here, I will emphasize R.and so,