304 CHAPTER 13. THE GENERALIZED RIEMANN INTEGRAL

Proof: Let η = ε

|β |+|α|+1 and choose gauges, δ g and δ f such that if P is δ g fine,∣∣∣∣S (P,g)−∫IgdF

∣∣∣∣< η

and that if P is δ f fine, ∣∣∣∣S (P, f )−∫

If dF

∣∣∣∣< η .

Now let δ = min(δ g,δ f

). Then if P is δ fine the above inequalities both hold. Therefore,

from the definition of S (P, f ),

S (P,α f +βg) = αS (P, f )+βS (P,g)

and so ∣∣∣∣S (P,α f +βg)−(

β

∫IgdF +α

∫I

f dF)∣∣∣∣≤ ∣∣∣∣βS (P,g)−β

∫IgdF

∣∣∣∣+

∣∣∣∣αS (P, f )−α

∫I

f dF∣∣∣∣≤ |β |η + |α|η < ε.

Since ε > 0 is arbitrary, this shows the number β∫

I gdF +α∫

I f dF qualifies in the defini-tion of the generalized Riemann integral and so α f +βg ∈ R∗ [a,b] and∫

I(α f +βg)dF = β

∫IgdF +α

∫I

f dF.

The following lemma is also very easy to establish from the definition.

Lemma 13.1.10 If f ≥ 0 and f ∈ R∗ [a,b], then∫

I f dF ≥ 0. Also, if f has complexvalues and is in R∗ [I], then both Re f and Im f are in R∗ [I].

Proof: To show the first part, let ε > 0 be given and let δ be a gauge function such thatif P is δ fine then |S ( f ,P)−

∫I f dF | ≤ ε. Since F is increasing, it is clear that S ( f ,P)≥ 0.

Therefore,∫

I f dF ≥ S ( f ,P)− ε ≥ −ε and since ε is arbitrary, it follows∫

I f dF ≥ 0 asclaimed.

To verify the second part, note that by Proposition 13.1.5 there exists a gauge, δ suchthat if P,Q are δ fine then |S ( f ,P)−S ( f ,Q)|< ε . But

|S (Re f ,P)−S (Re f ,Q)| = |Re(S ( f ,P))−Re(S ( f ,Q))|≤ |S ( f ,P)−S ( f ,Q)|

and so the conditions of Proposition 13.1.5 are satisfied and you can conclude Re f ∈ R∗ [I] .Similar reasoning applies to Im f .

Corollary 13.1.11 If | f | , f ∈ R∗ [a,b], where f has values in C, then∣∣∣∣∫If dF

∣∣∣∣≤ ∫I| f |dF.