
“real: chapter_08” — 2011/5/22 — 23:34 — page 52 — #52
8-52 Real Analysis
Here also we have assumed that f (b)>0. However if f (b) = 0 then
by hypothesis 0 ≤ f (x) ≤ f (b) = 0 and hence f (x) ≡ 0on[a, b]
and in this case
b
a
fdα = 0 = f (b)(α(b) − α(c)) still holds for any
c ∈[a, b]. This proves (i) and a similar proof is applicable for (ii).
SOLVED EXERCISES
1. Let f ∈ R[0, 1]. Define
a
n
=
1
n
n
k=1
f
k
n
(n = 1, 2, 3, ...)
Show that {a
n
} converges to
1
0
f (x)dx.
Solution: Using the theory of Riemann integrals, given >0 there
exists a δ>0 such that if P ={x
0
, x
1
, x
2
, ..., x
n
} is a partition on [0, 1]
with diameter d(P)<δ, then
n
k=1
f (t
k
)(x
k
− x
k−1
) −
1
0