5.6. EXERCISES 133

this is not necessary) is said to be Riemann Stieltjes integrable if there is a number, Isuch that for all ε > 0 there exists δ such that if ∥P∥< δ , then∣∣∣∣∣ n

∑i=1

f (τ i)(g(ti)−g(ti−1))− I

∣∣∣∣∣< ε

for any Riemann Stieltjes sum defined as the above in which τ i ∈ [ti−1, ti]. This I isdenoted as

∫ ba f (t)dg(t) and we will say that f ∈ R([a,b] ,g). Show that if g is of

bounded variation and f is continuous, then∫ b

a f (t)dg(t) exists. Note the differencebetween this and

∫ ba f (g(t))dg(t) which is a case of line integrals considered in this

chapter and how either includes the ordinary Riemann integral∫ b

a f (t)dt.

8. Suppose∫ b

a f dg exists. Explain the following: Let P ≡ {x0, · · · ,xn} and let ti ∈[xi−1,xi].

f g(b)− f g(a)−n

∑i=1

g(ti)( f (xi)− f (xi−1))

=n

∑i=1

f (xi)g(xi)− f (xi−1)g(xi−1)−n

∑i=1

g(ti)( f (xi)− f (xi−1))

=n

∑i=1

f (xi)(g(xi)−g(ti))+n

∑i=1

f (xi−1)(g(ti)−g(xi−1))

and if ∥P∥ is small enough, this is a Riemann sum for∫ b

a f dg which is closer to∫ ba f dg than ε . Use to explain why if

∫ ba f dg exists, then so does

∫ ba gd f and

∫ ba f dg+∫ b

a gd f = f g(b)− f g(a). Note how this says roughly that d ( f g) = f dg+gd f . As anexample, suppose g(t) = t and t→ f (t) is decreasing. In particular, it is of boundedvariation. Thus

∫ ba gd f exists. It follows then that

∫ ba f dg =

∫ ba f (t)dt exists.

9. Let f be increasing and g continuous on [a,b]. Then there exists c ∈ [a,b] such that

∫ b

agd f = g(c)( f (b)− f (c)) .

Hint: First note g Riemann Stieltjes integrable because it is continuous. Since g iscontinuous, you can let m = min{g(x) : x ∈ [a,b]} and M = max{g(x) : x ∈ [a,b]}Then m

∫ ba d f ≤

∫ ba gd f ≤M

∫ ba d f Now if f (b)− f (a) ̸= 0, you could divide by it

and conclude m ≤∫ b

a gd ff (b)− f (a) ≤ M. You need to explain why

∫ ba d f = f (b)− f (a).

Next use the intermediate value theorem to get the term in the middle equal to g(c)for some c. What happens if f (b)− f (a) = 0? Modify the argument and fill in thedetails to show the conclusion still follows.

10. Suppose g is increasing and f is continuous and of bounded variation. Then it followsthat g ∈ R([a,b] , f ) . Show there exists c ∈ [a,b] such that

∫ b

agd f = g(a)

∫ c

ad f +g(b)

∫ b

cd f

5.6. EXERCISES 13310.this is not necessary) is said to be Riemann Stieltjes integrable if there is a number, Jsuch that for all € > 0 there exists 6 such that if ||P|| < 6, thenys (a) (g (ti) —g(ti-1)) -I] <e€i=lfor any Riemann Stieltjes sum defined as the above in which 7; € [t;-1,¢;]. This / isdenoted as [” f (t)dg(t) and we will say that f € R({a,b|,g). Show that if g is ofbounded variation and f is continuous, then [/ iM f (t) dg (t) exists. Note the differencebetween this and f° S (g (t)) dg (t) which is a case of line integrals considered in thischapter and how either includes the ordinary Riemann integral [ ° f (t) dt.. Suppose f° fdg exists. Explain the following: Let P = {xo,---,x,} and let 4; €[xi-1, i].Ms:fg (b)—f8(@) — La (ti) (fi) — fi)i=]Ms:Ms:Ff (xi) (Xi) — f i-1) 8 (i-1) — D8 (ti) (F (Xi) — f @i-1))I i=lnn= L(x) (g (xi) — 8 (ti) + LX F(%-1) (8 (ti) —g @i-1))and if ||P|| is small enough, this is a Riemann sum for f? fdg which is closer tofe fdg than €. Use to explain why if ? fdg exists, then so does fe gdf and ? fdg+f? gd f = fg (b)— fg (a). Note how this says roughly that d(fg) = fdg+gdf. Asanexample, suppose g(t) =f and t — f(t) is decreasing. In particular, it is of boundedvariation. Thus [ ° gdf exists. It follows then that [' ° fdg = f ° f (t) dt exists.Let f be increasing and g continuous on [a,b]. Then there exists c € [a,b] such thatb[ s4f=8F®)-F(0).Hint: First note g Riemann Stieltjes integrable because it is continuous. Since g iscontinuous, you can let m = min {g (x) : x € [a,b]} and M = max {g (x) :x € [a,b]}Then mf? df < [? gdf <M J? df Now if f(b) — f (a) 40, you could divide by itband conclude m < ao <M. You need to explain why f° df = f(b) —f (a).Next use the intermediate value theorem to get the term in the middle equal to g(c)for some c. What happens if f(b) — f (a) = 0? Modify the argument and fill in thedetails to show the conclusion still follows.Suppose g is increasing and f is continuous and of bounded variation. Then it followsthat g € R({a,b], f) . Show there exists c € [a,b] such that[ear=ea) [arse [ar