5.9. AN APPROACH TO THE INTEGRAL 139

= |(Pn (b)−Pm (b))− (Pn (a)−Pm (a))|=∣∣∣∣∫ b

a(pn− pm)dx

∣∣∣∣≤ ∥pn− pm∥|b−a|

Thus the limit exists because{∫ b

a pndx}

nis a Cauchy sequence and R is complete.

From 5.3, 1. holds for a polynomial p(x). Let ∥pn− f∥→ 0. Then by definition,∫ b

af dx≡ lim

n→∞

∫ b

apndx = pn (xn)(b−a) (5.4)

for some xn in the open interval determined by (a,b) . By compactness, there is a fur-ther subsequence, still denoted with n such that xn → x ∈ [a,b] . Then fixing m such that∥ f − pn∥< ε whenever n≥ m, assume n > m. Then ∥pm− pn∥ ≤ ∥pm− f∥+∥ f − pn∥<2ε and so

| f (x)− pn (xn)| ≤ | f (x)− f (xn)|+ | f (xn)− pm (xn)|+ |pm (xn)− pn (xn)|

≤ | f (x)− f (xn)|+∥ f − pm∥+∥pm− pn∥< | f (x)− f (xn)|+3ε

Now if n is still larger, continuity of f shows that | f (x)− pn (xn)|< 4ε. Since ε is arbitrary,pn (xn)→ f (x) and so, passing to the limit with this subsequence in 5.4 yields 1.

Now consider 2. It holds for polynomials p(x) obviously. So let ∥pn− f∥→ 0. Then∫ c

apndx+

∫ b

cpndx =

∫ b

apndx

Pass to a limit as n→ ∞ and use the definition to get 2. Also note that∫ b

b f (x)dx = 0follows from the definition.

Next consider 3. Let h ̸= 0 and let x be in the open interval determined by a and b. Thenfor small h,

F (x+h)−F (x)h

=1h

∫ x+h

xf (t)dt = f (xh)

where xh is between x and x+h. Let h→ 0. By continuity of f , it follows that the limit ofthe right side exists and so

limh→0

F (x+h)−F (x)h

= limh→0

f (xh) = f (x)

If x is either end point, the argument is the same except you have to pay attention to thesign of h so that both x and x+h are in [a,b]. Thus F is continuous on [a,b] and F ′ existson (a,b) so if G is an antiderivative,∫ b

af (t)dt ≡ F (b) = F (b)−F (a) = G(b)−G(a)

The claim that the integral is linear is obvious from this. Indeed, if F ′ = f ,G′ = g,∫ b

a(α f (t)+βg(t))dt = αF (b)+βG(b)− (αF (a)+βG(a))

= α (F (b)−F (a))+β (G(b)−G(a))

= α

∫ b

af (t)dt +β

∫ b

ag(t)dt