7.3. THE RIEMANN DARBOUX INTEGRAL∗ 183

Proposition 7.3.16 Suppose H : R×R→ R satisfy

|H (x,y)−H (x̂, ŷ)| ≤ K (|x− x̂|+ |y− ŷ|)

Then if f ,g ∈ R([a,b]) it follows that H ( f ,g) ∈ R([a,b]) .

Proof: By hypothesis and the Riemann criterion, there is a partition P such that forh = f or g,U (h,P)−L(h,P)< ε

2K . Say P = x0 < x1 < · · ·< xn. Then consider

n

∑i=1

(Mi (H ( f ,g))−mi (H ( f ,g)))(xi − xi−1)

Say H ( f (zi) ,g(zi))+η > Mi (H ( f ,g)) and H ( f (wi) ,g(wi))−η < mi (H ( f ,g)) wherezi,wi are in [xi−1,xi]. Then

Mi (H ( f ,g))−mi (H ( f ,g)) ≤ H ( f (zi) ,g(zi))− (H ( f (wi) ,g(wi)))+2η

≤ K (| f (zi)− f (wi)|+ |g(zi)−g(wi)|)+2η

< K ((Mi ( f )−mi ( f ))+(Mi (g)−mi (g)))+2η

Since η is arbitrary, it follows that

Mi (H ( f ,g))−mi (H ( f ,g))≤ K ((Mi ( f )−mi ( f ))+(Mi (g)−mi (g)))

and soU (H ( f ,g) ,P)−L(H ( f ,g) ,P)< 2K

ε

2K= ε

Since ε is arbitrary, this verifies that H ( f ,g) ∈ R([a,b]).Note that H (x,y) = αx + βy satisfies the above conditions for α,β real numbers.

Therefore, if f ,g ∈ R([a,b]) , it follows that the linear combination α f +βg is integrable.Similarly α f and βg are integrable. If f ,g have values in some interval [a,b] and H :[a,b]× [a,b] → R is only continuous, it could be shown that if f ,g are integrable so isH ( f ,g) but this is more trouble.

Lemma 7.3.17 Let f ,g be integrable. Then∫ b

a ( f +g)(x)dx =∫ b

a f (x)dx+∫ b

a g(x)dx.

Proof: For x ∈ [xi−1,xi] , ( f +g)(x) ≥ mi ( f ) + mi (g) and so mi ( f +g) ≥ mi ( f ) +mi (g) . Similarly Mi ( f +g)≤ Mi ( f )+Mi (g). Therefore,

U ( f +g,P)≤U ( f ,P)+U (g,P) , L( f +g,P)≥ L( f ,P)+L(g,P)

Let U ( f ,P)−L( f ,P)< ε and U (g,P)−U (g,P)< ε . Then∫ b

a( f +g)(x)dx ∈ [L( f +g,P) ,U ( f +g,P)]

⊆ [L( f ,P)+L(g,P) ,U ( f ,P)+U (g,P)]∫ b

af (x)dx+

∫ b

ag(x)dx ∈ [L( f ,P)+L(g,P) ,U ( f ,P)+U (g,P)]

Thus both∫ b

a ( f +g)(x)dx and∫ b

a f (x)dx+∫ b

a g(x)dx are in an interval of length 2ε and

so∣∣∣∫ b

a ( f +g)(x)dx−(∫ b

a f (x)dx+∫ b

a g(x)dx)∣∣∣< 2ε. Since ε is arbitrary, this proves the

lemma.