10.2. NONNEGATIVE SIMPLE FUNCTIONS 281

10.2 Nonnegative Simple FunctionsTo begin with, here is a useful lemma.

Lemma 10.2.1 If f (λ ) = 0 for all λ > a, where f is a decreasing nonnegative function,then

∫∞

0 f (λ )dλ =∫ a

0 f (λ )dλ .

Proof: From the definition,∫∞

0f (λ )dλ = lim

R→∞

∫ R

0f (λ )dλ = sup

R>1

∫ R

0f (λ )dλ = sup

R>1sup

M

∫ R

0f (λ )∧Mdλ

= supM

supR>1

∫ R

0f (λ )∧Mdλ = sup

MsupR>1

∫ a

0f (λ )∧Mdλ

= supM

∫ a

0f (λ )∧Mdλ ≡

∫ a

0f (λ )dλ . ■

Now the Lebesgue integral for a nonnegative function has been defined, what does itdo to a nonnegative simple function? Recall a nonnegative simple function is one whichhas finitely many nonnegative real values which it assumes on measurable sets. Thus asimple function can be written in the form s(ω) = ∑

ni=1 ciXEi (ω) where the ci are each

nonnegative, the distinct values of s.

Lemma 10.2.2 Let s(ω) = ∑pi=1 aiXEi (ω) be a nonnegative simple function where the

Ei are distinct but the ai might not be. Thus the values of s are the ai. Then∫sdµ =

p

∑i=1

aiµ (Ei) . (10.1)

Proof: Without loss of generality, assume 0≡ a0 < a1≤ a2≤ ·· · ≤ ap and that µ (Ei)<∞, i > 0. Here is why. If µ (Ei) = ∞, then letting a ∈ (ai−1,ai) , by Lemma 10.2.1, the leftside is∫ ap

0µ ([s > λ ])dλ ≥

∫ ai

a0

µ ([s > λ ])dλ

≡ supM

∫ ai

0µ ([s > λ ])∧Mdλ ≥ sup

Msup

MMµ (Ei)ai = ∞

and so both sides of 10.1 are equal to ∞. Thus it can be assumed for each i,µ (Ei) < ∞.Then it follows from Lemma 10.2.1 and Lemma 10.1.2,∫

0µ ([s > λ ])dλ =

∫ ap

0µ ([s > λ ])dλ =

p

∑k=1

∫ ak

ak−1

µ ([s > λ ])dλ

=p

∑k=1

(ak−ak−1)p

∑i=k

µ (Ei) =p

∑i=1

µ (Ei)i

∑k=1

(ak−ak−1) =p

∑i=1

aiµ (Ei) ■

Note that this is the same result as in Problem 12 on Page 270 but here there is no questionabout the definition of the integral of a simple function being well defined.

Lemma 10.2.3 If a,b≥ 0 and if s and t are nonnegative simple functions, then∫as+btdµ = a

∫sdµ +b

∫tdµ .

10.2. NONNEGATIVE SIMPLE FUNCTIONS 28110.2 Nonnegative Simple FunctionsTo begin with, here is a useful lemma.Lemma 10.2.1 /f f(A) =0/forall A > a, where f is a decreasing nonnegative function,then fo f(A)dA = fo f (A) da.Proof: From the definition,“R[ fayaa = tim Ff ()da = sup | f(A)dA =supsup [ f(A) AMdARr JO R>1- R>1 M JORsupsup | f(A) AMdA =supsup "Fa )AMdrM R>1/0 M R>1sup [ f(A)AMdA =[ f(A)dA.Now the Lebesgue integral for a nonnegative function has been defined, what does itdo to a nonnegative simple function? Recall a nonnegative simple function is one whichhas finitely many nonnegative real values which it assumes on measurable sets. Thus asimple function can be written in the form s(@) = Y_, ci; Ze, (@) where the c; are eachnonnegative, the distinct values of s.Lemma 10.2.2 Let s(@) =Y?_, a; 2x, (@) be a nonnegative simple function where theE; are distinct but the a; might not be. Thus the values of s are the a;. ThenP[sau =) au (Ej). (10.1). i=lProof: Without loss of generality, assume 0 = ap < aj < a2 <--- <p and that pl (Ej) <oo, i > 0. Here is why. If 4 (£;) =, then letting a € (aj_1,a;) , by Lemma 10.2.1, the leftside is[Pule>ayaa = [ mis>anaa0sup [a ( ([s > A]) |) \AMdA > supsupMu (E i) aj = 00and so both sides of 10.1 are equal to oo. Thus it can be assumed for each i, (E;) < ©.Then it follows from Lemma 10.2.1 and Lemma 10.1.2,[outo>apar= ["u(s>anar=¥ | w(ls> ayaak=17%-1vPpi ag — Ag—1 Yue y= Sule) ¥, (a —an-1) = Yaw (Ed)k=l i=Note that this is the same result as in Problem 12 on Page 270 but here there is no questionabout the definition of the integral of a simple function being well defined.Lemma 10.2.3 If a,b > 0 and if s and t are nonnegative simple functions, then[ostord =a [sdu+b [ tay.