Kenneth Kuttler

54 CHAPTER 4. THE RIEMANN STIELTJES INTEGRAL

Proof: This follows from Theorem 4.4.4 and Definition 4.4.6. For example, assume

c ∈ (a,b) .

Then from Theorem 4.4.4,∫ c

af (x) dF +

∫ b

cf (x) dF =

∫ b

af (x) dF

and so by Definition 4.4.6,∫ c

af (x) dF =

∫ b

af (x) dF−

∫ b

cf (x) dF

=∫ b

af (x) dF +

∫ c

bf (x) dF.

The other cases are similar.The following properties of the integral have either been established or they follow

quickly from what has been shown so far.

If f ∈ R([a,b]) then if c ∈ [a,b] , f ∈ R([a,c]) , (4.4.10)∫ b

aα dF = α (F (b)−F (a)) , (4.4.11)∫ b

a(α f +βg)(x) dF = α

∫ b

af (x) dF +β

∫ b

ag(x) dF, (4.4.12)∫ b

af (x) dF +

∫ c

bf (x) dF =

∫ c

af (x) dF, (4.4.13)∫ b

af (x) dF ≥ 0 if f (x)≥ 0 and a < b, (4.4.14)∣∣∣∣∫ b

af (x) dF

∣∣∣∣≤ ∣∣∣∣∫ b

a| f (x)| dF

∣∣∣∣ . (4.4.15)

The only one of these claims which may not be completely obvious is the last one. To showthis one, note that

| f (x)|− f (x)≥ 0, | f (x)|+ f (x)≥ 0.

Therefore, by 4.4.14 and 4.4.12, if a < b,∫ b

a| f (x)| dF ≥

∫ b

af (x) dF

and ∫ b

a| f (x)| dF ≥−

∫ b

af (x) dF.

Therefore, ∫ b

a| f (x)| dF ≥

∣∣∣∣∫ b

af (x) dF

∣∣∣∣ .If b < a then the above inequality holds with a and b switched. This implies 4.4.15.

54 CHAPTER 4. THE RIEMANN STIELTJES INTEGRALProof: This follows from Theorem 4.4.4 and Definition 4.4.6. For example, assumec € (a,b).Then from Theorem 4.4.4,Cc b b| f (x) ar + | f (x) dF =| f (x) dFa Cc aand so by Definition 4.4.6,[rear= [roar-[peyar= [rears [ roarThe other cases are similar.The following properties of the integral have either been established or they followquickly from what has been shown so far.If f € R((a,b]) then if c € [a,b], f € R([a,c]), (4.4.10)[car =a(F (6)—F(a)), (4.4.11)~b “b b| (af + Bg) (x) dF =a | f(x) aP +B | a(x) dF, (4.4.12)b c C[ rears | f(s)dF = [ f(oar. (4.4.13)[ro dF > Oif f (x) >Oanda<b, (4.4.14)b b[ fever < / Ir()|aF}. (4.4.15)The only one of these claims which may not be completely obvious is the last one. To showthis one, note thatIf (x)|—f (x) = 90, |F@&)| +f) = 0.Therefore, by 4.4.14 and 4.4.12, ifa <b,b b[\reojar= [roarJa aandb b| \re@lar=- [seara aTherefore,b b[vejar >| rovarl,a aIf b < a then the above inequality holds with a and b switched. This implies 4.4.15.