Kenneth Kuttler

11.4. THE SPACE L1 251

Definition 11.4.4 f ∈ L1(Ω) means there exists a sequence of complex simple functions,{sn} such that

sn (ω)→ f (ω) for all ω ∈Ω

limm,n→∞ I (|sn− sm|) = limn,m→∞

∫|sn− sm|dµ = 0 (11.4.23)

ThenI ( f )≡ lim

n→∞I (sn) . (11.4.24)

Lemma 11.4.5 Definition 11.4.4 is well defined.

Proof: There are several things which need to be verified. First suppose 11.4.23. Thenby Lemma 11.4.3

|I (sn)− I (sm)|= |I (sn− sm)| ≤ I (|sn− sm|)and for m,n large enough this last is given to be small so {I (sn)} is a Cauchy sequence in Cand so it converges. This verifies the limit in 11.4.24 at least exists. It remains to consideranother sequence {tn} having the same properties as {sn} and verifying I ( f ) determinedby this other sequence is the same. By Lemma 11.4.3 and Fatou’s lemma, Theorem 11.3.18on Page 248,

|I (sn)− I (tn)| ≤ I (|sn− tn|) =∫|sn− tn|dµ

≤∫|sn− f |+ | f − tn|dµ

≤ lim infk→∞

∫|sn− sk|dµ + lim inf

k→∞

∫|tn− tk|dµ < ε

whenever n is large enough. Since ε is arbitrary, this shows the limit from using the tn isthe same as the limit from using sn. This proves the lemma.

What if f has values in [0,∞)? Earlier∫

f dµ was defined for such functions and nowI ( f ) has been defined. Are they the same? If so, I can be regarded as an extension of

∫dµ

to a larger class of functions.

Lemma 11.4.6 Suppose f has values in [0,∞) and f ∈ L1 (Ω) . Then f is measurable and

I ( f ) =∫

f dµ.

Proof: Since f is the pointwise limit of a sequence of complex simple functions, {sn}having the properties described in Definition 11.4.4, it follows f (ω) = limn→∞ Resn (ω)and so f is measurable. Also∫ ∣∣(Resn)

+− (Resm)+∣∣dµ ≤

∫|Resn−Resm|dµ ≤

∫|sn− sm|dµ

where x+ ≡ 12 (|x|+ x) , the positive part of the real number, x. 2Thus there is no loss of

generality in assuming {sn} is a sequence of complex simple functions having values in2The negative part of the real number x is defined to be x− ≡ 1

2 (|x|− x) . Thus |x|= x++x− and x = x+−x−..

11.4. THE SPACE L! 251Definition 11.4.4 f € L'(Q) means there exists a sequence of complex simple functions,{s,} such thatSn(@) > f(@) forall@a EQ. . 11.4.23lity seo! (|S —Sin|) = litt ms f |5n — Sn) dit = 0 ( )ThenI(f) = lim I (s,). (11.4.24)n—00Lemma 11.4.5 Definition 11.4.4 is well defined.Proof: There are several things which need to be verified. First suppose 11.4.23. Thenby Lemma 11.4.3IZ (Sn) —1(Sm)| = |L (Sn — Sm)| <1 (|Sn — Sm)and for m,n large enough this last is given to be small so {J (s,)} is a Cauchy sequence in Cand so it converges. This verifies the limit in 11.4.24 at least exists. It remains to consideranother sequence {t,,} having the same properties as {s,} and verifying J (f) determinedby this other sequence is the same. By Lemma 11.4.3 and Fatou’s lemma, Theorem 1 1.3.18on Page 248,(50) LO] <H(se—tal) = f Is —tnlte<fin—Fi+it—mlanslim inf Ji — s,|d + lim inf Jin —t|du<ek-00 k-00whenever n is large enough. Since € is arbitrary, this shows the limit from using the ¢, isthe same as the limit from using s,,. This proves the lemma.What if f has values in [0,cc)? Earlier { fd was defined for such functions and nowI(f) has been defined. Are they the same? If so, 7 can be regarded as an extension of [duto a larger class of functions.Lemma 11.4.6 Suppose f has values in |0,-0) and f € L' (Q). Then f is measurable and1(f)= f faw.Proof: Since f is the pointwise limit of a sequence of complex simple functions, {s,}having the properties described in Definition 11.4.4, it follows f (@) = limy_,. Res, (@)and so f is measurable. Also[ \(Resn)* = (Resn)* dy < | |Resy —Resn|dy < [isn — slawhere x* = 5 (|x|+.x), the positive part of the real number, x. *Thus there is no loss ofgenerality in assuming {s,} is a sequence of complex simple functions having values in°The negative part of the real number x is defined to be x~ = 5 (|x| —x). Thus |x| =x+ +27 andx=x* —37.