Real Analysis HW 5 Solutions
Problem 7: Let f be an increasing real-valued function on [0, 1]. For a natural number
n, define Pn to be the partition of [0, 1] into n subintervals on length 1/n. Show that
U (f, Pn ) − L(f, Pn ) ≤ 1/n[f (1) − f (0)]. Use Problem 5 to show that f is Riemann integrable
over [0, 1].
Proof: Define the interval Ik,n = ((k − 1)/n, k/n) for all natural numbers k, n. From the
definition of the upper and lower Darboux sums, and the fact that f is increasing,
n
U (f, Pn ) =
and
n
1X
1X
sup f (x) =
f (k/n)
n k=1 x∈Ik,n
n k=1
n
L(f, Pn ) =
n
1X
1X
inf f (x) =
f ((k − 1)/n)
n k=1 x∈Ik,n
n k=1
Therefore
n
1X
1
U (f, Pn ) − L(f, Pn ) =
(f (k/n) − f ((k − 1)/n)) = (f (1) − f (0)),
n k=1
n
since the above summation is telescoping. Now using the result of Problem 5, Pn is a sequence
of partitions such that
lim [U (f, Pn ) − L(f, Pn )] = 0
n→∞
Therefore f is Riemann integrable over [0, 1].
Problem 11: Does the Bounded Convergence Theorem hold for the Riemann integral?
Solution: No. Consider the sequence of functions
fn =
n
X
χ{qk } ,
k=1
where {qk }∞
in [0, 1]. It is easy to see for any natural
k=1 is an enumeration of the rationals
R
number n, fn is Riemann integrable and [0,1] fn = 0, since for any partition Pk of size 1/k
of [0, 1], we have
n
U (fn , Pk ) ≤ , and L(fn , Pk ) = 0.
k
1
Therefore limk→∞ (U (fn , Pk ) − L(fn , Pk )) = 0. However, fn converges pointwise on [0, 1] to
f = χQ , which is not Riemann integrable on [0, 1].
Problem 15: Verify the assertions in the last Remark of this section.
Solution: Yup, the Remark is true
Problem 16:
R Let f be a nonnegative bounded measurable function on a set of finite measure
E. Assume E f = 0. Show that f = 0 a.e. on E.
Solution: By Chebychev’s inequality, we have
Z
m({E : f > 1/n}) ≤ n
f = 0,
∀n ≥ 1.
E
We conclude that m({E : f > 0}) = 0, since {E : f > 0} =
countable union of measure 0 sets. Therefore f = 0 a.e. on E.
S∞
n=1 {E
: f > 1/n} is a
Problem
19: For a number α, define f (x) = xα for 0 < x ≤ 1, and f (0) = 0. Compute
R1
f.
0
Solution: To do this, first consider the sequence of truncated functions {fn } given by
fn = f χ[1/n,1] . Clearly this sequence is increasing, non-negative, and converges to f pointwise.
Therefore by the monotone convergence theorem
Z
Z
f = lim
f.
n→∞
[0,1]
[1/n,1]
Since f is continuous and bounded on [1/n, 1] for every n ≥ 1, the Lebesgue and Riemann
integrals coincide,
(
Z
Z 1
1
(1 − n−α−1 ) if α =
6 −1
.
f = (R)
xα dx = 1+α
log n
if α = −1
[1/n,1]
1/n
Taking the limit we obtain
(
Z
f=
[0,1]
1
1+α
∞
if α > −1
.
if α ≤ −1
Problem 21: Let the function f be nonnegative and integrable over E andR > 0. Show
there is a simple function η on E that has finite support, 0 ≤ η ≤ f on E and E |f − η| < .
If E Ris a closed, bounded interval, show there is a step function h on E that has finite support
and E |f − h| < .
Solution: Let > 0. We know by the definition of the integral for non-negative measurable
functions that there exists a bounded, measurable function of finite support h ≤ f , such
that
Z
Z
h>
f − /2.
E
E
2
We also know that for any given bounded, measurable function h with finite support, there
exists a simple function η ≤ h with the same support as h such that
Z
Z
η>
h − /2.
E
E
Therefore we see that there is a simple function η with finite support such that η ≤ f , and
Z
(f − η) < E
In the case that E is a closed and bounded interval, let η =
such that
Z
(f − η) < /2.
Pn
k=1 ck χEk
be a simple function
E
k
For each k, since m(Ek ) < ∞, there exists a finite collection of disjoint open intervals {Iik }ni=1
such that
!
Z nk
nk
[
X
χIik = m Ek ∆ Iik < /2n.
χEk −
R
i=1
i=1
Moreover, since E is a closed bounded interval, we may assume that each Iik is contained
inside E, but may no longer be open. Define a function ψ by
ψ=
nk
n X
X
ck χIik .
k=1 i=1
Note that this function is a step function even though some of the intervals may overlap,
since any finite collection of intervals that cover E define a parition of E by ordering their
endpoints. It follows that
Z
Z
n
X
X
χIik | ≤ /2.
ck
|χEk −
|ψ − η| ≤
E
k=1
E
i
Therefore ψ is a step function such that,
Z
|ψ − f | < E
Problem 22: Let {fn } be a sequence of nonnegative measurable functions of R that converges pointwise on R to f integrable. Show that if
Z
Z
Z
Z
f = lim
fn , then
f = lim
fn
R
n→∞
E
R
for any measurable set E.
3
n→∞
E
R
R
R
Solution: Let E be a measurable set. Using the fact that R f = E f + R∼E f , Fatou’s
lemma on R ∼ E implies
Z
Z
Z
Z
f ≤ lim inf
fn =
f − lim sup fn .
n
R∼E
Therefore
R∼E
n
R
Z
E
Z
f ≥ lim sup
fn
n
E
However Fatou’s Lemma on E gives
Z
E
Z
f ≤ lim inf
n
E
fn .
E
These two inequalities conclude the proof.
Problem 25: Let {fn } be a sequence of nonnegative measurable functions on E that converge pointwise on E to f . Suppose fn ≤ f on E for each n. Show that
Z
Z
fn =
f.
lim
n→∞
Solution: Clearly for each n,
R
E
fn ≤
E
E
R
f , therefore
Z
Z
lim sup fn ≤
f.
E
n
E
E
However, since {fn } are nonnegative, Fatou’s Lemma implies
Z
Z
f ≤ lim inf
fn .
E
n
The conclusion follows from these two inequalities.
4
E