4.4. PROPERTIES OF THE INTEGRAL 51
Then since mi (− f ) =−Mi ( f ) ,
ε >∫ b
a− f (x) dF−
n
∑i=1
mi (− f )∆Fi =∫ b
a− f (x) dF +
n
∑i=1
Mi ( f )∆Fi
which implies
ε >∫ b
a− f (x) dF +
n
∑i=1
Mi ( f )∆Fi ≥∫ b
a− f (x) dF +
∫ b
af (x) dF.
Thus, since ε is arbitrary, ∫ b
a− f (x) dF ≤−
∫ b
af (x) dF
whenever f ∈ R([a,b]) . It follows∫ b
a− f (x) dF ≤−
∫ b
af (x) dF =−
∫ b
a−(− f (x)) dF ≤
∫ b
a− f (x) dF
and this proves the lemma.
Theorem 4.4.3 The integral is linear,∫ b
a(α f +βg)(x) dF = α
∫ b
af (x) dF +β
∫ b
ag(x) dF.
whenever f ,g ∈ R([a,b]) and α,β ∈ R.
Proof: First note that by Theorem 4.3.1, α f +βg ∈ R([a,b]) . To begin with, considerthe claim that if f ,g ∈ R([a,b]) then∫ b
a( f +g)(x) dF =
∫ b
af (x) dF +
∫ b
ag(x) dF. (4.4.7)
Let P1,Q1 be such that
U ( f ,Q1)−L( f ,Q1)< ε/2, U (g,P1)−L(g,P1)< ε/2.
Then letting P≡ P1∪Q1, Lemma 4.1.2 implies
U ( f ,P)−L( f ,P)< ε/2, and U (g,P)−U (g,P)< ε/2.
Next note that
mi ( f +g)≥ mi ( f )+mi (g) , Mi ( f +g)≤Mi ( f )+Mi (g) .
Therefore,
L(g+ f ,P)≥ L( f ,P)+L(g,P) , U (g+ f ,P)≤U ( f ,P)+U (g,P) .
4.4. PROPERTIES OF THE INTEGRAL 51Then since m;(—f) = —M;(f),b n ob ne> [Foy dr—Yomi(—Nari= [fo ar+ YMA anwhich impliese> [rl ar+ym(nari> [ £0) ar+[ ro) dF.a i—1 a ai=Thus, since € is arbitrary,b b[ -rears-[ searwhenever f € R([a,b]) . It followsb b b b[ -fevars-f sejar=—[ -Crejar< [rearand this proves the lemma.Theorem 4.4.3 The integral is linear,b b b[ (af +Bs)(x) dF =a | f(x) dF +B g(x) dF.whenever f,g € R(|a,b]) and a, B ER.Proof: First note that by Theorem 4.3.1, af + Bg € R(|a,b]). To begin with, considerthe claim that if f, g € R([a,b]) then[uramar= [roars [oar (4.4.7)Let P;,Q1 be such thatU(f,Q1)-L(f,Q1) < €/2, U(g,Pi) -L(g,P\) < €/2.Then letting P= P, UQ;, Lemma 4.1.2 impliesU(f,P)—L(f,P) < €/2, and U(g,P)—U(g,P) < €/2.Next note thatm(f+g) =mi(f)+mi(g), Mi(f+s) <Mi(f)+Mi(g)-Therefore,L(g+f,P) 2L(f,P)+L(g,P), U(gtf,P) <U (f,P)+U (8,P).