188 CHAPTER 9. INTEGRATION

Proof: Assume that a< b in what follows. If not, simply switch a and b in the argument.Let {pn} be a sequence of polynomials for which ∥pn− f∥ → 0 and let P′n (x) = pn (x) forall x ∈ (a,b). By the mean value theorem,

|Pn (x)−Pn (a)− (Pm (x)−Pm (a))|=

|Pn (x)−Pm (x)− (Pn (a)−Pm (a))|= |(pn (t)− pm (t))(x−a)| ≤ ∥pn− pm∥|b−a|≤ (∥pn− f∥+∥ f − pm∥) |b−a|

The right side converges to 0 as n,m→ ∞ and so by completeness, there exists

F (x) = limn→∞

(Pn (x)−Pn (a)) ,

this for any choice of x. It remains to verify that F ′ (x) = f (x) . Say x ∈ [a,b) and let h > 0.Then by the mean value theorem,

Pn (x+h)−Pn (x)h

=(Pn (x+h)−Pn (a))− (Pn (x)−Pn (a))

h= pn (thn) (∗)

for some thn ∈ (x,x+h). By compactness, there is a subsequence, still denoted as thn forwhich limn→∞ thn = th ∈ [x,x+h]. Now

|pn (thn)− f (th)| ≤ |pn (thn)− f (thn)|+ | f (thn)− f (th)|≤ ∥pn− f∥+ | f (thn)− f (th)|

and so, letting n→ ∞, this shows, from continuity of f that |pn (thn)− f (th)| → 0. Takinga limit in ∗,

F (x+h)−F (x)h

= f (th) , th ∈ [x,x+h]

Now by continuity of f , we can take a limit of this as h→ 0 and obtain F ′ (x) = f (x) ,where F ′ (x) is a right derivative at x = a. For x ∈ (a,b], the situation is exactly the samefor when h is restrained to be negative.

F (x+h)−F (x)h

=−F (x− (−h))−F (x)−h

=F (x)−F (x− k)

k

where k≡−h and so for F ′ (x) the left derivative, it exists at each point of (a,b] and equalsf (x) by exactly similar arguments to the above. Thus at every point of (a,b) both the rightand left derivatives exist for F and both are equal to f so F is differentiable on (a,b). Also,the appropriate one sided derivatives for F exist at x ∈ {a,b} and are likewise f (x).

Definition 9.1.3 For the rest of this section, [a,b] will denote the closed intervalhaving end points a and b but a could be larger than b or smaller than b. It is written thisway to indicate that there is a direction of motion from a to b which will be reflected by thedefinition of the integral given below. It is an “oriented interval”. Then for f continuouson [a,b],

∫ ba f (x)dx≡ F (b)−F (a) where F is an antiderivative for f on [a,b].

Proposition 9.1.4 The integral is well defined for f continuous on [a,b].