7.2. UNIFORM CONVERGENCE AND THE INTEGRAL 177
Suppose f ,g are piecewise continuous. Then let {zi}ni=1 include all the partition points
of both of these functions. Then, since it was just shown that no harm is done by includingmore partition points,
∫ ba α f (t)+βg(t)dt ≡
n
∑i=1
∫ zi
zi−1
(α fi (s)+βgi (s))ds =n
∑i=1
α
∫ zi
zi−1
fi (s)ds+n
∑i=1
β
∫ zi
zi−1
gi (s)ds
= α
n
∑i=1
∫ zi
zi−1
fi (s)ds+β
n
∑i=1
∫ zi
zi−1
gi (s)ds = α
∫ b
af (t)dt +β
∫ b
ag(t)dt
Also, the claim that∫ b
a f dt =∫ c
a f dt +∫ b
c f dt is obtained exactly as before by consideringall partition points on each integral preserving the order of the limits in the small intervalsdetermined by the partition points.
Definition 7.1.7 Let I be an interval. Then XI (t) is 1 if t ∈ I and 0 if t /∈ I.Then a step function will be of the form ∑
nk=1 ckXIk (t) where Ik = [ak−1,ak] is an inter-
val and {Ik}nk=1 are non-overlapping intervals whose union is an interval [a,b] so b−a =
∑nk=1 (ak −ak−1). Then, as explained above,∫ b
a
n
∑k=1
ckXIk (t)dt =n
∑k=1
ck
∫ ak
ak−1
1dt =n
∑k=1
ck (ak −ak−1) .
Is this as general as a complete treatment of Riemann integration? No it is not. The1800’s version of the integral is presented in the next section which is due to Riemann andDarboux. However, this version is sufficiently general to include all cases which are typi-cally of interest. It is also enough to build a theory of ordinary differential equations. Notethat Proposition 7.1.4 says that
∫ ba f (t)dt = G(b)−G(a) whenever G′ = f . This proposi-
tion also proves the fundamental theorem of calculus discovered by Newton and Leibniz,(∫ ta f (s)ds
)′= f (t) if f is continuous. Unlike what was done by Newton and Leibniz, this
approach also includes a rigorous definition of what is meant by the integral. To summarize,here is the procedure for finding an integral of a piecewise continuous function.
Procedure 7.1.8 Let f be continuous on [a,b]. To find∫ b
a f (t)dt, find an an-tiderivative of f ,F. Then
∫ ba f (t)dt = F (b)− F (a). If f is piecewise continuous and
equals the continuous function fk on (zk−1,zk) where fk is continuous on [zk−1,zk] anda = z0 < z1 < z2 < · · ·< zn = b, then
∫ ba f (x)dx ≡ ∑
nk=1
∫ zkzk−1
fk (x)dx.
The main assertion of the above Proposition 7.1.4 is that for any f continuous, thereexists a unique solution to the initial value problem F ′ (t) = f (t) , along with F (a) = 0 andit is F (t) =
∫ ta f (x)dx.
7.2 Uniform Convergence and the IntegralIt turns out that uniform convergence is very agreeable in terms of the integral. The follow-ing is the main result.
Theorem 7.2.1 Let fn be continuous and converging uniformly to f on [a,b] ,a < b.Then it follows f is also continuous and∫ b
af dx = lim
n→∞
∫ b
afndx