Download Homework 1 Solutions - UCSD Math Department

Homework 1 Solutions
Problem 4.1.6
Part A
We can use induction to show that
n
X
i2 = n(n + 1)(2n + 1)/6.
i=0
First we verify that the above identity is true for n = 0 and n = 1.
For the inductive step, assume it holds true for some n ≥ 1 and we prove it for n + 1. To this
end, we compute:
!
n+1
n
X
X
i2 =
i2 + (n + 1)2
i=0
i=0
=
=
=
=
n(n + 1)(2n + 1)
+ (n + 1)2
6
(n + 1)[n(2n + 1) + 6n + 6]
6
(n + 1)(2n2 + 7n + 6)
6
(n + 1)(n + 2)(2n + 3)
6
so the result is true by induction.
Part B
We only consider the first integral, namely for the function x2 1[0,a) , the remaining ones being
similar. We consider N > 0, and we fix an integer ` such that
`
`+1
≤a< N .
2N
2
We calculate the upper Riemann sum with dyadic intervals of side
1
1
.
2N
We have
UN (x2 1[0,a) (x)) =
≤
X
sup
i i+1
i∈Z [ 2N , 2N )
x2 1[0,a) (x) ·
1
2N
`
1 X
sup x2 .
2N
i i+1
[
,
)
i=0
2N
2N
In the sum above, the index i goes only up to ` because for i ≥ ` + 1, the dyadic interval does not
intersect the support [0, a) of the function since a < `+1
. Also, note the less or equal sign that
2N
) is not entirely contained
opens the second line. This is because the last dyadic interval [ 2`N , `+1
2N
in the support [0, a), but rather extends beyond that support. By allowing a larger interval, the
supremum will possibly get bigger, hence the ≤ sign.
We continue the calculation:
UN
` 1 X i+1 2
= N
2
2N
i=0
1
(12 + 22 + . . . + (` + 1)2 )
23N
1 (` + 1)(` + 2)(2` + 3)
= 3N
2
6
` + 1 ` + 2 2` + 3 1
= N · N ·
·
2
2
2N
6
=
Using
`
2N
≤ a, we see that
UN ≤
1
`
+ N
N
2
2
`
2
`
3
1
1
3
1
·
+ N · 2· N + N · ≤a a+ N
2a + N · .
N
2
2
2
2
6
2
2
6
Making N → ∞, we obtain
lim UN ≤
N →∞
Thus the upper Riemann integral of
x2 1[0,a)
a3
.
3
is
U (x2 1[0,a) ) ≤
a3
.
3
Now, we repeat the above argument for the lower Riemann sums. We find
X
1
LN (x2 1[0,a) (x)) =
inf x2 1[0,a) (x) · N
i i+1
2
[ N, N)
i∈Z
=
1
2N
2
2
`
X
i=0
inf
i i+1
, N)
[ N
2
2
x2
` 1 X i 2
= N
2
2N
i=0
1
(12 + 22 + . . . + `2 )
23N
1 `(` + 1)(2` + 1)
= 3N
2
6
` ` + 1 2` + 1 1
·
= N · N ·
2
2
2N
6
=
Using
`+1
2N
LN
> a, we see that
`+1
1
`+1
`+1
1
1
1
1
1
≥
−
·
·
2
·
−
·
>
a
−
·
a
·
2a
−
· .
2N
2N
2N
2N
2N
6
2N
2N
6
Making N → ∞, we obtain
lim LN ≥
N →∞
Thus the lower Riemann integral of
x2 1
[0,a)
a3
.
3
is
L(x2 1[0,a) ) ≥
a3
.
3
Putting everything together, we obtain
a3
a3
≤ L(x2 1[0,a) ) ≤ U (x2 1[0,a) ) ≤ .
3
3
Therefore, all the inequalities above should be equalities, and as a result
Z
a3
x2 1[0,a) |dx| = .
3
R
The remaining there integrals are entirely similar, and are left to the reader.
Problem 4.1.4
We show that a set X ⊂ Rn has volume 0 if and only if for every > 0 there exists N such that:
X
voln (C) ≤ .
C∈DN (Rn )
C∩X6=∅
Fix any set X ⊂ Rn . We begin by finding the upper Riemann sum of the characteristic function
1X . This will be necessary in the proof below. We have
X UN (1X ) =
sup 1X voln C.
C∈DN
C
We can make this more explicit. Indeed,
• if C ∩ X = ∅ then the characteristic function will be 0 over all of C, so the sup will be 0, and
we can eliminate any such C’s from the sum.
• On the other hand, if C ∩ X 6= ∅, then the characteristic function will be 1 for at least one
value of x ∈ C, so the sup will be 1.
This means that the above sum can be rewritten as:
X UN (1X ) =
sup 1X voln C =
C∈DN
C
X
voln (C)
C∈DN (Rn )
C∩X6=∅
With this understood, we are now ready to prove Proposition 4.1.23. Assume first that X is a
set of volume zero
Z
voln (X) =
1X |dn x| = 0.
Rn
The above integral is then obtained as the limit of the upper Riemann sum
Z
0 = voln (X) =
1X = lim UN (1X ).
Rn
N →∞
Thus, using the definition of the limit for the sequence UN (1X ) which converges to 0 by assumption,
we see that for all > 0, there exists N such that
UN (1X ) < .
Using the above calculation of the upper Riemann sum, this translates into
X
voln (C) < .
C∈DN (Rn )
C∩X6=∅
The converse is similar. Indeed, assume that for all > 0, we can find N such that
X
C∈DN (Rn )
C∩X6=∅
voln (C) < .
This is simply the statement that
UN (1X ) < .
Since the upper Riemann sums form a decreasing sequence, we conclude that for all k ≥ N , we
have
Uk (1X ) ≤ UN (1X ) < .
This is exactly the -definition of the fact that
lim Uk (1X ) = 0.
k→∞
Therefore, the limit above, i.e. the upper Riemann integral of the function 1X , has to be ≤ 0:
U (1X ) ≤ 0.
Since the lower Riemann integral is clearly at least equal to 0 because 1X ≥ 0, we must have
0 ≤ L(1X ) ≤ U (1X ) ≤ 0.
This shows that we must have equality throughout, hence 1X is integrable and the integral is zero.
This means
voln (X) = 0,
as needed.
Problem 3
This was proved in class. Please consult your notes.
Problem 4
Consider the Dirichlet function f : [0, 1] → Rn given by:
(
1
if x rational, where x = p/q written in lowest order terms
g(x) = q
0 if x irrational
Part (i)
Calculate the lower Riemann sums and lower Riemann integral.
Note that at all rational numbers in [0, 1], g takes on positive values, and for any dyadic paving
of [0, 1], every subinterval will contain an irrational number, which means that g will always take
on a value of 0 on every subinterval of every paving. Thus the lower Riemann sum is always 0 for
any choice of N , which means that the lower Riemann integral is also 0.
Part (ii)
Explain that g is discontinuous at all rational numbers.
Consider any rational number
x=
p
∈ [0, 1] =⇒ g(x) = 1/q.
q
If we take any interval that contains x, say (x − n1 , x + n1 ) for n > 0, it also contain at least one
irrational number yn . In particular,
|yn − x| <
1
=⇒ lim yn = x.
n→∞
n
Nonetheless, g(yn ) = 0 since yn is irrational, hence
1
lim g(yn ) = 0 6= g(x) = .
n→∞
q
This shows g is not continuous at the rational numbers.
Part (iii)
Show that for any > 0, there are finitely many rational numbers x ∈ [0, 1] such that g(x) > .
Consider any > 0. Then the only values of x where g(x) > are rational values of 0 ≤ x =
p/q ≤ 1 where
1/q > q < 1/
Let E be 1/, rounded up to the nearest integer. Then there are at most E different values that
q can take on. For example, if we had = 1/4 then q could take on values of 1, 2, or 3, but not
values of 4 or higher. But we also require that 0 ≤ x ≤ 1, which means that p cannot be less than
0 or greater that q. Thus there are only a finite number of choices for integers p and q such that
0 ≤ p/q ≤ 1, gcd(p, q) = 1, and 1/q > for any given > 0.
Part (iv)
From part (iii), if we fix > 0, there exist only finitely many rational numbers where g(x) > .
Now let x be an arbitrary irrational number in the interval [0, 1]. Let δ be the distance between x
and the closest rational number to x that satisfies g(xq ) > , i.e.
δ=
min
y∈[0,1] and g(y)>
|x − y| .
This min is guaranteed to exist since the set of rational numbers y ∈ [0, 1] where g(y) > is finite.
Now, every y ∈ (x − δ, x + δ) will satisfy g(y) ≤ , which means that |g(x) − g(y)| ≤ , and thus
that g is continuous at x.
Part (v)
In Problem 3 we proved that the set of rational numbers has measure 0, and in Problem 4 we
proved that g is continuous except at the set of rational numbers between 0 and 1. Now we apply
Theorem 4.4.6, which states that a function is integrable if and only if it is continuous except on a
set of measure 0, to conclude that g is integrable.
In particular, g integrable means that the limits of its upper and lower N th sums agree and
equal the integral. And we computed the limit of the lower N th sum in part (i), which was 0. Thus
R1
0 g(x) dx = 0.
Problem 5
Show that the composition of two integrable functions f, g : [0, 1] → [0, 1] may not necessarily be
integrable. For instance, you may take:
(
0 if x = 0
f (x) =
1 if 0 < x ≤ 1
and g : [0, 1] → [0, 1] could be the Dirichlet function in Problem 4.
Part 1
Explain that f and g are integrable.
We proved that g was integrable in Problem 4. To show that f is integrable note that it is only
discontinuous at 0, so it is integrable by Theorem 4.4.6.
Part 2
Calculate h = f ◦ g. Where is h continuous? Is h integrable?
g is 0 at all irrational numbers, and greater than 0 at all rational numbers. Thus, f (g(x)) will
be 0 if x irrational, and 1 if x rational. This is the characteristic function over the set of rational
numbers. We have seen in class that this function is not integrable. In fact, the function h is
discontinuous everywhere.