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) .