9.3 Uniform Convergence And The Integral
It turns out that uniform convergence is very agreeable in terms of the integral. The following
is the main result.
Theorem 9.3.1 Let g be of bounded variation and let f_{n} be continuous and
converging uniformly to f on
. Then it follows f is also integrable and
∫ b ∫ b
fdg = lim fndg
a n→∞ a
 

Proof: The uniform convergence implies f is also continuous. See Theorem 6.9.7.
Therefore, ∫
_{a}^{b}fdg exists. Now let n be given large enough that
f − fn ≡ max f (x) − fn (x) < ε
x∈[a,b]
 

Next pick δ > 0 small enough that if
< δ, then
∫ b n 
 fdg− ∑ f (tk)(g(xk)− g(xk−1)) < ε
 a k=1 
∫ b ∑n 
 fndg− fn (tk)(g(xk)− g(xk−1)) < ε
 a k=1 
for any choice
t_{k} ∈. Pick such a
P and the same
t_{k} for both sums. Then
∫ b ∫ b  ∫ b ∑n 
 fdg − fndg ≤  fdg − f (tk)(g (xk) − g(xk− 1))
a a  a k=1  
∑n 
+ (f (tk)− fn(tk))(g(xk)− g(xk−1))
k=1 
∫ b ∑n 
+ a fndg− fn (tk)(g(xk)− g(xk−1))
k=1
n
< ε+ ∑ εg(xk)− g(xk−1)+ ε ≤ 2ε+ V[a,b](g)ε
k=1
 

Since ε is arbitrary, this shows that lim_{n→∞}
 
∫ bfdg− ∫ bfndg
a a
= 0.
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