6.1 Suppose α increases on [a, b], a ≤ x∗ ≤ b, α is continuous
at x∗ , f (x∗ ) = 1, and
Z
f (x) = 0 if x 6= x∗ . Prove that f ∈ R(α) and that f dα = 0.
Rudin’s Ex. 1
Proof Since α is continuous at the only point where f is discontinuous, by Theorem
6.10, we know that f ∈ R(α).
To evaluate the integral, we fix any partition
P : a = x0 ≤ x1 ≤ · · · ≤ xn−1 ≤ xn = b.
By the definition,
L(P, f, α) =
n X
inf
xi−1 ≤x≤xi
i=1
f (x) ∆αi = 0,
since f (x) = 0 in each [xi−1 , xi ] except at most one point. Since P is arbitrary, this
implies that the lower Riemann-Stieltjes integral is
L(f, α) = sup L(P, f, α) = 0.
P
Because we have already known that f ∈ R(α), we get
Z
f dα = L(f, α) = 0.
b
Z
6.2 Suppose f ≥ 0, f is continuous on [a, b], and
f dx = 0. Prove that f (x) = 0 for
a
all x ∈ [a, b].
Rudin’s Ex. 2
∗
∗
Proof Suppose f (x ) > 0 for some x ∈ [a, b]. Since f is continuous on [a, b], for
= f (x∗ )/2 > 0, there exist δ > 0 such that |x − x∗ | < δ and x ∈ [a, b] imply
|f (x) − f (x∗ )| <
f (x∗ )
.
2
Thus, we know that there is an interval whose length is at least δ, say [γ, γ+δ] ⊂ [a, b],
on which
f (x∗ )
f (x∗ )
=
.
f (x) > f (x∗ ) −
2
2
By Theorem 6.12, we have
Z
b
Z
f dx =
a
γ
Z
γ+δ
f dx +
a
Z
b
f dx ≥ 0 +
f dx +
γ
γ+δ
f (x∗ )
δ + 0 > 0,
2
b
Z
f dx = 0. Therefore, for every x ∈ [a, b], we
which contradicts to the hypothesis
a
have f (x) = 0.
1
6.3 Define three functions β1 , β2 , β3 as follows: βj (x) = 0 if x < 0, βj (x) = 1 if x > 0
for j = 1, 2, 3; and β1 (0) = 0, β2 (0) = 1, β3 (0) = 12 . Let f be a bounded function on
[−1, 1].
(a) Prove that f ∈ R(β1 ) if and only if f (0+) = f (0) and that then
Z
f dβ1 = f (0).
(b) State and prove a similar result for β2 .
(c) Prove that f ∈ R(β3 ) if and only if f is continuous at 0.
(d) If f is continuous at 0 prove that
Z
Z
Z
f dβ1 = f dβ2 = f dβ3 = f (0).
Proof For each j, j = 1, 2, 3, and any partition P of [−1, 1], by the definitions, it is
easy to to see that
L(P, f, βj ) ≤ f (0) ≤ U (P, f, βj ),
which gives
Z
Z
f dβj ≤ f (0) ≤
f dβj .
Z
Hence, we have
f dβj = f (0) whenever the integrals exist.
Let > 0 be given.
(a) If f ∈ R(β1 ) on [−1, 1], by Theorem 6.12, we know that f ∈ R(β1 ) on [0, 1]. By
Theorem 6.6, there exists a partition P of [0, 1] such that
U[0,1] (P, f, β1 ) − L[0,1] (P, f, β1 ) < .
More explicitly, if we write
P : 0 < δ < x2 < · · · < xn = 1,
then for any 0 ≤ s < δ, by the definitions of the upper and lower Riemann-Stieltjes
sums, we have
L[0,1] (P, f, β1 ) ≤ f (s) ≤ U[0,1] (P, f, β1 ),
L[0,1] (P, f, β1 ) ≤ f (0) ≤ U[0,1] (P, f, β1 ),
which imply
|f (s) − f (0)| ≤ U[0,1] (P, f, β1 ) − L[0,1] (P, f, β1 ) < This means that f (0+) = lim+ f (x) = f (0).
x→0
Conversely, suppose f (0+) = f (0). Hence, there is δ > 0 such that 0 ≤ s < δ implies
|f (s) − f (0)| < /4.
2
Rudin’ Ex. 3
We fixed a point s∗ , with 0 < s∗ < δ. Put
M = sup f (x),
x∈[0,s∗ ]
m=
inf
x∈[0,s∗ ]
f (x).
Then there exist s1 , s2 ∈ [0, s∗ ] such that
M − /4 < f (s1 ),
m + /4 > f (s2 ).
Hence,
M − m < f (s1 ) − f (s2 ) + /2 ≤ |f (s1 ) − f (s2 )| + /2
≤ |f (s1 ) − f (0)| + |f (s2 ) − f (0)| + /2 < .
Choose a partition P of [−1, 1],
P : −1 = x0 < · · · < xm = 0 < xm+1 = s∗ < · · · < xn = 1,
since
U (P, f, β1 ) = M,
L(P, f, β1 ) = m,
we have
U (P, f, β1 ) − L(P, f, β1 ) = M − m < .
By Theorem 6.6, we know that f ∈ R(β1 ) on [−1, 1].
(b) Similarly to part (a), f ∈ R(β2 ) if and only if f (0−) = f (0). The proof is nearly
the same as part (a).
(c) We need to modify the proof of part (a) in the current case.
If f ∈ R(β3 ) on [−1, 1], by Theorem 6.6, there exists a partition P of [−1, 1] such
that
U (P, f, β3 ) − L(P, f, β3 ) < /2.
Without loss of generality, we can assume
P : − 1 = x0 < · · · < xk−1 = −δ < xk = 0 < xk+1 = δ < · · · < xn = 1,
otherwise a refinement can be chosen as in this form which still satisfies the last
inequality. For any two numbers s ∈ (−δ, 0] and t ∈ [0, δ), by the definitions of the
upper and lower Riemann-Stieltjes sums, we have
1
f (s) +
2
1
L(P, f, β3 ) ≤ f (0) +
2
L(P, f, β3 ) ≤
1
f (0) ≤ U (P, f, β3 ),
2
1
f (t) ≤ U (P, f, β3 ).
2
We also have
L(P, f, β3 ) ≤ f (0) ≤ U (P, f, β3 ).
3
These inequalities give that
for −δ < s ≤ 0, |f (s) − f (0)| ≤ 2[U (P, f, β3 ) − L(P, f, β3 )] < ;
for 0 ≤ t ≤ δ,
|f (t) − f (0)| ≤ 2[U (P, f, β3 ) − L(P, f, β3 )] < .
Hence, we have f (0−) = f (0) = f (0+), so that f is continuous at 0.
Conversely, suppose f (0−) = f (0) = f (0+). We similarly construct two partitions
P1 and P2 of [0, 1] and [−1, 0], respectively, such that
U[−1,0] (P1 , f, β3 )−L[−1,0] (P1 , f, β3 ) < /2,
U[0,1] (P2 , f, β3 )−L[0,1] (P2 , f, β3 ) < /2,
By combining them to obtain a partition P of [−1, 1], we have
U[−1,1] (P, f, β3 ) − L[−1,1] (P, f, β3 )
= U[−1,0] (P1 , f, β3 ) − L[−1,0] (P1 , f, β3 ) + U[0,1] (P2 , f, β3 ) − L[0,1] (P2 , f, β3 )
< /2 + /2 = .
It follows from Theorem 6.6 that f ∈ R(β3 ) on [−1, 1].
(d) The result follows from part (a), (b), (c) and the discussion at the beginning.
6.4 If f (x) = 0 for all irrational x, f (x) = 1 for all rational x, prove that f ∈
/ R on [a, b]
for any a < b.
Proof By the fact that rational and irrational numbers are both sense in every [a, b]
for any a < b, we know that for every partition of [a, b],
U (P, f ) =
n
X
sup f (x) ∆xi = b − a,
i=1 [xi−1 ,xi ]
Z
b
i=1
Z
f dx = 0 < b − a =
Hence
a
U (P, f ) =
n
X
inf
[xi−1 ,xi ]
b
f dx. By the definition, f ∈
/ R.
a
4
f (x) ∆xi = 0.
Rudin’ Ex. 4