Homework 6 Solutions - Math 321, Spring 2015
1. Given a nonconstant non-decreasing function α : [a, b] → R, let Rα [a, b] denote the collection
of all bounded functions on [a, b] which are Riemann-Stieltjes integrable with respect to α.
Is Rα [a, b] a vector space, a lattice, an algebra?
Solution. They are all a vector space, a lattice, an algebra. (See also Theorem 6.12-6.13 in
the textbook.) To show that f, g ∈ Rα [a, b] =⇒ f + g, cf, |f | ∈ Rα [a, b], we use (verify it !)
Uα (f + g, P ) − Lα (f + g, P ) ≤ (Uα (f, P ) − Lα (f, P )) + (Uα (g, P ) − Lα (g, P ))
Uα (cf, P ) − Lα (cf, P ) = cUα (f, P ) − cLα (g, P )
Uα (|f |, P ) − Lα (|g|, P ) ≤ Uα (f, P ) − Lα (g, P )
To show that f g ∈ Rα [a, b]. First we prove in the special case that f, g are non-negative
then general case follows by decomposing f = f + − f − , g = g + − g − and we know that
f + , f − , g + , g − ∈ Rα [a, b] since it is a lattice.
Now assume that f, g are non-negative, we use
|Uα (f g, P ) − Lα (f g, P )| ≤ |Uα (f, P )Uα (g, P ) − Lα (f, P )Lα (g, P )|
≤ |Uα (f, P )||Uα (g, P ) − Lα (g, P )| + |Lα (g, P )||Uα (f, P ) − Lα (f, P )|
Here |Uα (f, P )|, |Lα (g, P )| are bounded by an absolute constant.
2. This problem focuses on computing the Riemann-Stieltjes integral for specific choices of
integrators.
(a) Let x0 = a < x1 < x2 < · · · < xn = b be a fixed collection of points in [a, b], and let
α be an increasing step function on [a, b] that is constant on each of the open intervals
(xi−1 , xi ) and has jumps of size αi = α(xi +) − α(xi −) at each of the points xi . For i = 0
and n, we make the obvious adjustments
α0 = α(a+) − α(a−),
αn = α(b) − α(b−).
If f ∈ B[a, b] is continuous at each of the points xi , show that f ∈ Rα [a, b] and
Z b
n
X
f dα =
f (xi )αi .
a
i=0
Solution. Intuitively, the point xi will contribute the value of the integral if α jumps at
xi which αi measures how large does α jump. . Given > 0, for each xi there is a
δi such that |x − xi | < δ =⇒ |f (x) − f (xi )| < . Choose δ = min{δ1 , . . . , δn } then
|f (x) − f (xi )| < . (or one can argue that f is uniformly continuous). Also assume
δ < 21 min1≤i≤n |xi − xi+1 |.
Refine the partition P as follows:
For each x0 , x1 , . . . , xn−1 choose a point yi ∈ (xi , xi + δ).
For each x1 , x2 , . . . , xn choose a point zi ∈ (xi − δ, xi ). Then we have a new partition P ∗
{x0 , y0 , z1 , x1 , y1 , z2 , x2 , y2 , . . . , zn−1 , xn−1 , yn−1 , zn , xn }
1
2
Now using definition of δ and the fact that α is constant along (xi , xi+1 )
n
X
∗
Lα (f, P ) >
(f (xi ) − )αi
i=1
Uα (f, P ∗ ) <
n
X
(f (xi ) + )αi
i=1
We have
Z b
Z b
n
n
X
X
∗
∗
f dα ≤
f dα < Uα (f, P ) <
(f (xi ) + )αi
(f (xi ) − )αi < Lα (f, P ) <
a
a
i=1
i=1
Since > 0 is arbitrary (small), let → 0, we are done.
Rn
(b) If f is continuous on [1, n], compute 1 f (x)d[x], where [x] is the greatest integer in x.
Rt
What is the value of 1 f (x)d[x] if t is not an integer?
Solution. Apply part a) with x0 = 1, x1 = 2, . . . , xn−1 = n since α0 = 0, α1 = α2 = · · · =
αn−1 = 1, we have
Z n
n
n−1
X
X
f (k)
f (n)αi =
f d[x] =
1
i=0
k=2
i.e. on [1, N ], [x] jumps at 2, 3, . . . , N . In the same way,
Z t
[t]
X
f (k)
f d[x] =
1
k=2
In we want the sum from 1 to n we could take e.g.
Z t
[t]
X
f (k)
f d[x] =
0.99
k=1
where 0.99 could be replaced by any numbers in [0, 1).
3. Determine, with adequate justification, whether each of the following statements is true or
false.
(a) An equicontinuous, pointwise bounded subset of C[a, b] is compact.
Solution. The statement is easily seen to be false; the class of constant functions { n1 :
n ≥ 1} provides a counterexample. The collection is equicontinuous but not closed (the
constant function zero is a limit point outside the set), hence not compact.
Remark: Note however that an equicontinuous, pointwise bounded and closed subset of
C(X) is in fact compact. Just follow the proof of the Arzela-Ascoli theorem.
(b) The function χQ is Riemann integrable on [0, 1].
Solution. The function χQ is not Riemann integrable on [0, 1] or on any interval [a, b]
for that matter. Since Q is dense in R, it follows that U (χQ , P ) = 1 and L(χQ ) = 0
for any partition P of [a, b]. By Riemann’s condition on integrability, χQ fails to be
integrable.
3
(c) The function χ∆ is Riemann integrable on [0, 1], where ∆ denotes the Cantor middle-third
set. (We have already run into this set in Homework 2, Problem 5).
Solution. The statement is true. We will again use Riemann’s condition for integrability.
Let us recall from construction of the Cantor set that ∆ = ∩∞
n=1 ∆n , where ∆n is the
set obtained at the nth stage of the construction. In particular, ∆n is the union of 2n
chosen intervals each of length 3−n . Let Pn denote the partition on [0, 1] generated by
the intervals chosen at the nth step of the iteration. Then
Mi = 1, mi = 0 if I = [xi−1 , xi ] is a chosen interval,
Mi = mi = 0 if I is not chosen.
This implies that
U (χ∆ , Pn ) − L(χ∆ , Pn ) =
X
(Mi − mi )∆xi = (1)(2n )(3−n ) =
i
2 n
→ 0 as n → ∞.
3
(d)
T
α {Rα [a, b]
: α increasing} = C[a, b].
Solution. The statement is true. We have already seen in class that C[a, b] ⊆ Rα [a, b] for
any increasing α. Conversely, suppose that f ∈ Rα [a, b] for every increasing α. We aim
to show that f in continuous on [a, b] and argue by contradiction. Assume if possible
that f is discontinuous at the point c (say from the right without loss of generality), and
define α to be the increasing function χ(c,b] . Let P = {a = x0 < x1 < · · · < xn = b}
denote any partition of [a, b] in which xi−1 = c for some i. (Without loss of generality
one can always ensure this after passing to a refinement - why?)
U (f, P ) − L(f, P ) ≥ (Mi − mi )∆αi ≥ |f (c+) − f (c)| > 0.
Thus Riemann’s condition does not hold.
(e) If f is a monotone function and α is both continuous and non-decreasing, then f ∈
Rα [a, b].
Solution. The statement is true. Let us fix > 0 and aim to find a partition P of [a, b] for
which Uα (f, P ) − Lα (f, P ) < . Since α is continuous on [a, b], it is uniformly continuous;
hence there exists δ > 0 such that α(x) − α(y) < for all a ≤ y < x ≤ b, 0 < x − y < δ.
Choosing P = {a = x0 < x1 < x2 < · · · < xn = b} to be a partition of [a, b] into
subintervals of length < δ, and recalling that f is monotone, we find that
n
X
U (f, P ) − L(f, P ) =
|f (xi ) − f (xi−1 )|∆αi
i=0
n
X
<
|f (xi ) − f (xi−1 )| = |f (b) − f (a)|,
i=0
where the last step uses the monotonicity of f . Riemann’s condition on integrability
proves the statement.
(f) There exists a non-decreasing function α : [a, b] → R and a function f ∈ Rα [a, b] such
that f and α share a common-sided discontinuity.
4
Solution. The statement is false. If a nondecreasing α : [a, b] → R and a function
f : [a, b] → R share a common-sided discontinuity, then f 6∈ Rα [a, b].
Suppose that c is a point of common-sided discontinuity, say from the right. Choose a
partition P of [a, b] in which c is one of the partitioning points. Then
U (f, P ) − L(f, P ) ≥ |f (c+) − f (c)||α(c+) − α(c)|,
which violates Riemann’s condition for integrability.
(g) If f ∈ Rα [a, b] with m ≤ f ≤ M and if ϕ is continuous on [m, M ], then ϕ◦f ∈ Rα [a, b].
(1)
Solution. The statement is true.
Fix > 0. Since ϕ is uniformly continuous on [m, M ], let us choose δ > 0 such that
whenever |u − v| < δ. Since f is assumed to be Riemann|ϕ(u) − ϕ(v)| < 2(α(b)−α(a))
Stieltjes integrable on on [a, b], we know from Riemann’s condition that there exists a
partition P = {a = x0 < x1 < · · · < xn = b} of [a, b] such that
n
X
δ
,
Uα (f, P ) − Lα (f, P ) =
(Mi − mi )∆αi <
4||ϕ
◦
f
||
∞
i=0
Mi (f ) = sup f (x), mi (f ) = inf f (x), Ii = [xi−1 , xi ].
x∈Ii
x∈Ii
We will decompose the set of indices i ∈ {0, 1, · · · , n − 1} into two disjoint subsets A
and B: i ∈ A if Mi − mi ≤ δ and i ∈ B otherwise. Then it follows from (1) above that
X
X
X
δ
(Mi − mi )∆αi ≤
∆αi ≤
δ
, i.e.,
∆αi <
.
4||ϕ ◦ f ||∞
4||ϕ ◦ f ||∞
i∈B
i∈B
i∈B
Now, for the same partition P as above we can estimate
Uα (ϕ ◦ f, P ) − Lα (ϕ ◦ f, P )
hX X i
+
Mi (ϕ ◦ f ) − mi (ϕ ◦ f ) ∆αi
=
i∈A
i∈B
X
X
≤
∆αi + 2||ϕ ◦ f ||∞
∆αi
2(α(b) − α(a)) i∈A
i∈B
≤
(α(b) − α(a)) + 2||ϕ ◦ f ||∞
2(α(b) − α(a))
4||ϕ ◦ f ||∞
≤ + = ,
2 2
verifying Riemann’s condition.