#### You've reached the end of your free preview.

Want to read all 8 pages?

**Unformatted text preview: **Real Analysis Homework IX solutions
November 22nd, 2019 Exercise 5.1.1
Let f : [0, 1] → R be defined by f (x) = x3 and let P = {0, 0.1, 0.4, 1}. Compute L(P, f ) and
U (P, f ).
We must compute the data mi , Mi , and ∆xi . Since f is monotone increasing , mi is
achieved at the left endpoint of [xi−1 , xi ] and Mi is achieved at the right endpoint . Hence,
m1 = 0
m2 = 0.001
m3 = 0.064 M1 = 0.001
M2 = 0.064
M3 = 1 ∆x1 = 0.1 − 0 = 0.1
∆x2 = 0.4 − 0.1 − 0.3
∆x3 = 1 − 0.4 = 0.6 Finally, we compute the upper and lower sums:
L(P, f ) =
U (P, f ) = 3
X
i=1
3
X m1 ∆xi = 0.0387
M1 ∆xi = 0.6193 i=1 Exercise 5.1.2
R1
Let f : [0, 1] → R be defined by f (x) = x. Show that f ∈ R[0, 1] and compute 0 f using
the definition of the Riemann integral (but feel free to use any of the propositions from this
section).
R1
R1
R1
Proof. We will show that f ∈ R[0, 1] with 0 f = 0 f = 0 f = 12 . Fix n ∈ N, and let Pn
consist of n + 1 evenly spaced points in [0, 1]:
1 2
n−1
Pn = 0, , · · · ,
,1
n n
n 1 We will compute L(Pn , f ) and U (Pn , f ). Then,
i−1
i
i−i
mi = inf x :
≤x≤
=
n
n
n
i−1
i
i
Mi = sup x :
≤x≤
=
n
n
n
and so
L(Pn , f ) = n
X mi ∆xi = i=1 U (Pn , f ) = n
X
i=1 n
1 X
n(n − 1)
(i − 1) =
2
n i=1
2n2 n
1 X
n(n + 1)
Mi ∆xi = 2
i=
n i=1
2n2 Pn
n(n+1)
(using the result that
). Thus, by proposition 5.1.8 and the definitions of
k=1 k =
2
the upper and lower Darboux sums:
1 Z
L(Pn , f ) ≤ f≤ f ≤ U (Pn , f ) 0 n(n − 1)
≤
2n2 1 Z Z 0 1 Z
f≤ 1 f≤ 0 0 n(n + 1)
2n2 Taking n → ∞ and using the fact that limits preserve inequalities, we find that
Z 1 Z
f= 0 Hence, f ∈ R[0, 1] and R1
0 1 f=
0 1
2 f = 12 . Exercise 5.1.3
Let f : [a, b] → R be a bounded function, and suppose there exists a sequence of partitions
{Pk }∞
k=1 such that
lim U (Pk , f ) − L(Pk , f ) = 0
k→∞ Show that f is Riemann integrable with
Z b
f = lim U (Pk , f ) = lim L(Pk , f ).
a k→∞ k→∞ Proof. We first show f is Riemann integrable. Proposition 5.1.13 says that it is enough to
show that for every > 0 there exists a k such that U (Pk , f ) − L(Pk , f ) < . But, since
limk→∞ U (Pk , f ) − L(Pk , f ) = 0, there exists an N such that for every n ≥ N ,
2 |U (Pk , f ) − L(Pk , f )| < . In particular, for k = N , U (Pk , f ) − L(Pk , f ) < and so f is
Riemann integrable.
Rb
We now need to show the second assertion. Notice that since a f is both an upper
bound of all of the lower sums and a lower bound to the upper sums, we have
Z b
0≤
f − L(Pk , f ) ≤ U (Pk , f ) − L(Pk , f )
a Rb
The Squeeze theorem, implies that limk→∞ a f − L(Pk , f ) = 0, so,
Z b
f = lim L(Pk , f )
a k→∞ It follows by limit arithmetic that limk→∞ U (Pk , f ) exists as well, and
Z b
f = lim L(Pk , f ) = lim U (Pk , f )
a k→∞ k→∞ Exercise 5.1.4
Finish the proof of 5.1.7. That is, prove that if P˜ is a refinement of P , then U (P˜ , f ) ≤
U (P, f ).
The proof can be obtained in much that same ways at the argument for lower sums (up to
changing infs to sups and reversing inequalities). However, it is less work to use the following
argument to get it ‘for free’ from the argument for lower sums:
Proof. It has already been shown that L(P˜ , f ) ≥ L(P, f ). We claim that U (Q, f ) =
−L(Q, −f ) for any function f and partition Q. Assuming this is true, then we have that
L(P˜ , −f ) ≥ L(P, −f )
from the book, so multiplying by −1 gives the result.
We now prove the claim. Let Q = {x0 , x1 , · · · , xn } be a partition. Then,
mi (−f ) = inf{−f (x) : x ∈ [xi−1 , xi ]} = − sup{f (x) : x ∈ [xi−1 , xi ]} = −Mi (f )
To see why, notice if y = sup{f (x) : x ∈ [xi−1 , xi ]}, then −y ≤ f (x) for all x ∈ [xi−1 , xi ], then
certainly −y is a lower bound of {−f (x) : x ∈ [xi−1 , xi ]}. Moreover, if z is another lower
bound for {−f (x) : x ∈ [xi−1 , xi ]}, then −z is an upper bound for {f (x) : x ∈ [xi−1 , xi ]}, so
−z ≥ y by the definition of sup. Thus, −y ≥ z for all lower bounds, so −y = inf{−f (x) :
x ∈ [xi−1 , xi ]}. Hence,
−L(Q, −f ) = − n
X
i=1 mi (−f )∆xi = n
X −mi (−f )∆xi = i=1 n
X
i=1 3 Mi (f )∆xi = U (Q, f ) Exercise 5.1.5
Suppose f : [−1, 1] → R is defined as
(
1 x>0
f (x) =
0 x≤0
Prove that f ∈ R[−1, 1] and compute R1
−1 f using the definition of the integral. Proof. For each > 0, let P be the partition P = { −1, 0, , 1 }. Then,
U (P , f ) = 0(1) + 1() + 1(1 − ) = 1
L(P , f ) = 0(1) + 0() + 1(1 − ) = 1 − .
Thus, Proposition 5.1.8 and the definition of the lower and upper Darboux integrals imply
that
Z b
Z b
1 − = L(P , f ) ≤
f≤
f ≤ U (P , f ) = 1.
a a Taking → 0, the Squeeze theorem implies that
Rb
Riemann integrable with a f = 1. Rb
a f = Rb
a f = 1 so, by definition, f is Exercise 5.1.6
Let c ∈ (a, b) and d ∈ R. Define f : [a, b] → R by
(
d x=c
f (x) =
0 else
Prove that f ∈ R[a, b] and compute Rb
a f using the definition of the integral. Proof. We will prove that f ∈ R[a, b] with
Pn = {a, c − Rb
a f = 0. Let Pn be a partition given by c−a
b−c
,c +
, b}
n+1
n+1 4 For this partition,
mi,n = inf{f (x) : x ∈ [xi , xi+1 ] = 0
c−a
M1,n = sup{f (x) : x ∈ [a, c −=0
n+1
c−a
b−c
M2,n = sup{f (x) : x ∈ [c −
,c +=d
n+1
n+1
b−c
M3,n = sup{f (x) : x ∈ [c +
, b] = 0
n+1
c−a
n
∆x1 = c −
−a=
(c − a)
n+1
n+1
b−c
c−a
b−a
∆x2 = c +
− c−
=
n+1
n+1
n+1
b−c
n
∆x1 = b − c −
=
(b − c)
n+1
n+1
In particular, using Proposition 5.1.8 and the definition of the Darboux sums,
Z b 0 = L(Pn , f ) ≤ f≤ f ≤ U (Pn , f ) =
a a Taking n → ∞, we find that Rb
a f= b Z d(b − a)
n+1 f = 0 , so f ∈ R[a, b] with
a Rb Rb
a f = 0. Exercise 5.1.7
Suppose f : [a, b] → R is Riemann integrable. Let > 0 be given. Then show that there
exists a partition P = {x0 , x1 , · · · , xn } such that for any choice of numbers {c1 , c2 , · · · , cn }
with ck ∈ [xk−1 , xk ] for all k = 1, 2, · · · , n, then Z n b X f
−
f
(c
)∆x k
k < a k=1 Proof. To begin, we find a partition ‘fine enough’ that L(P, f ) and U (P, f ) are good apRb
Rb
proximations of a f . Let > 0 be given. Since f is Riemann integrable, a f is the sup
Rb
of the set of lower sums. Thus, there exists a partition PL such that a f − L(PL , f ) < .
Rb
Similarly, a f is the inf of the set of upper sums, so there exists a partition PU such that
Rb
U (PU , f ) − a f < . Let P be the common refinement of PL and PU , P = PL ∪ PU . Then,
Z b f − L(P, f ) < and a Z
U (P, f ) − f <
a 5 b by proposition 5.1.7. Write P as an ordered list of points P = {x0 , x1 , · · · , xn }, and suppose
we have points {c1 , c2 , · · · , ck } with each ck ∈ [xk−1 , xk ]. Then,
mk = inf x∈[xk−1 ,xk ] f (x) ≤ f (ck ) ≤ sup f (x) = Mk . x∈[xk−1 ,xk ] Multiplying by ∆xk and summing gives
Z b f − < L(P, f ) =
a ∞
X
k=1 so mk ∆xk ≤ ∞
X f (ck )∆xk ≤ k=1 ∞
X Z
Mk ∆xk = U (P, f ) < k=1 b f + .
a Z n b X f−
f (ck )∆xk < . a k=1 Exercise 5.1.8
Let f : [a, b] → R be a Riemann integrable
a−β b−β function. Let α > 0 and β ∈ R. Define
g(x) = f (αx + β) on the interval I = α , α . Show that g is Riemann integrable on I.
Proof. We will make use of proposition 5.1.13. Let > 0 be arbitrary. Since f is Riemann
integrable, there exists a partition P = {x0 , x1 , · · · , xn } such that
U (P, f ) − L(P, f ) < α
Consider the partition Q = {y0 , y1 , · · · , yn } = x0α−β , x1α−β , · · · , xnα−β . Then, Q is a partition of I. Moreover, we have that we can express the ‘partition data’ for (Q, g) in terms of
the data for (P, f ):
mi (g) = inf{g(y) : y ∈ [yi−1 , yi ]}
xi−1 − β xi − β
= inf f (αy + β) : y ∈
,
α
α
= inf{f (x) : x ∈ [xi−1 , xi ]}
= mi (f )
Mi (g) = sup{g(y) : y ∈ [yi−1 , yi ]}
xi−1 − β xi − β
= sup f (αy + β) : y ∈
,
α
α
= sup{f (x) : x ∈ [xi−1 , xi ]}
= Mi (f )
xi − β xi−1 − β
xi − xi−1
1
∆yi = yi − yi−1 =
−
=
= ∆xi
α
α
α
α 6 Thus,
L(Q, g) =
U (Q, g) = n
X n
X mi (g)∆yi = i=1 i=1 n
X n
X Mi (g)∆yi = i=1 i=1 so
U (Q, g) − L(Q, g) =
and g is Riemann integrable on I. 1
1
mi (f ) ∆xi = L(P, f )
α
α
1
1
Mi (f ) ∆xi = U (P, f )
α
α 1
(U (P, f ) − L(P, f )) <
α Exercise 5.1.9
Suppose f, g : [0, 1] → R are such that for all x ∈ (0, 1], we
R 1 haveR f1 (x) = g(x). Suppose f is
Riemann integrable. Prove g is Riemann integrable and 0 f = 0 g.
Proof. We will show
Z 1 1 Z
g≥ 0 and f, 1 Z g≤ 0 0 from which it will follow that
Z 1 1 Z
g= 0 Z
g= 0 Z 1 f
0 1 f
0 R1
R1
so g is Riemann integrable with 0 g = 0 f .
Since g(x) = f (x) on (0, 1], we have
max |g(x)| ≤ max |f (x)| + |g(0)| ≤ B x∈[0,1] x∈[0,1] for some B ∈ R. Let > 0 be arbitrary, and let P be a partition such that
Z b
Z b
f − L(P, f ) <
and
U (P, f ) −
f<
2
a
a
(This can be done as in problem 5.1.7 by choosing different partitions PL and PU for the
lower and upper sum and taking the common refinement P ). Now, let P˜ = P ∪ 4B
. Then,
Z b f − L(P˜ , g) = a Z b
˜
˜
˜
f − L(P , f ) + L(P , f ) − L(P , g) a Since P˜ is a refinement of P , the first term in parentheses is less than /2. The second term
can be rewritten as
L(P˜ , f ) − L(P˜ , g) = n
X
(mk (f ) − mk (g))∆xk = (m1 (f ) − m1 (g))
≤ 2B
≤
4B
4B
2
k=1 7 since g(x) = f (x) in [xk−1 , xk ] for k ≥ 2, so mk (g) = mk (f ) for k ≥ 2. Thus,
Z b f − L(P˜ , f ) < a Since > 0 was arbitrary, this shows that
Z b
Z b
g = sup{L(P, g) : P is a partition} ≥
f.
a a Similar reasoning with the upper sums shows that
Z b g = inf{U (P, g) : P is a partition} ≤ a Z b f
a Rb
Rb
Rb
Rb
Rb
Since Proposition 5.1.8 states that a g ≤ a g, this shows that a g = a g = a f so g is
Rb
Rb
Riemann integrable with a g = a f . Exercise 5.1.10
Let f : [0, 1] → R be a bounded function. Let Pn = {x0 , x1 , · · · , xn } be the uniform partition;
that is, xj = nj . Is {L(Pn , f )}∞
n=1 always monotone?
No. Consider
(
1 13 ≤ x ≤ 23
f (x) =
0 else
Then, whenever Pn contains the points 13 and 23 (which happens when n is a multiple of 3),
R1
L(Pn , f ) = 13 = 0 f . However, for n not a multiple of three, L(P, f ) < 13 , since over the
subintervals containing 13 and 23 , mi = 0. In particular,
L(P1 , f ) = 0
L(P2 , f ) = 0
1
L(P3 , f ) =
3
L(P4 , f ) = 0
so L(Pn , f ) cannot be monotone. 8 ...

View
Full Document

- Fall '19
- Riemann integral, Riemann sum, Riemann, f using