The Real And Complex Number Systems

9.16 Let {f n } be a sequence of real-valued continuous functions defined
on [0, 1] and assume that f n → f uniformly on [0, 1] . Prove or disprove
∫ 1−1/n
lim
n→∞
0
f n (x) dx =
∫ 1
0
f (x) dx.
Proof: By Theorem 9.8, we have
∫ 1
lim
n→∞
0
f n (x) dx =
∫ 1
0
f (x) dx. (*)
Note that {f n } is uniform bound, say |f n (x)| ≤ M for all x ∈ [0, 1] and all
n by Exercise 9.1. Hence,
∫ 1
∣ f n (x) dx
∣ ≤ M → 0. (**)
n
1−1/n
Hence, by (*) and (**), we have
∫ 1−1/n
lim
n→∞
0
f n (x) dx =
∫ 1
0
f (x) dx.
9.17 Mathematicinas from Slobbovia decided that the Riemann integral
was too complicated so that they replaced it by Slobbovian integral, defined
as follows: If f is a function defined on the set Q of rational numbers
in [0, 1] , the Slobbovian integral of f, denoted by S (f) , is defined to be the
limit
1
S (f) = lim
n→∞ n
n∑
f
k=1
( k
n)
,
whenever the limit exists. Let {f n } be a sequence of functions such that
S (f n ) exists for each n and such that f n → f uniformly on Q. Prove that
{S (f n )} converges, that S (f) exists, and S (f n ) → S (f) as n → ∞.
Proof: f n → f uniformly on Q, then given ε > 0, there exists a positive
integer N such that as n > m ≥ N, we have
|f n (x) − f (x)| < ε/3 (1)
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