18.014 Problem Set 3 Solutions
Sam Elder
October 8, 2015
Rb
[x] dx + a [−x] dx = a − b.
Rb
Solution. We rewrite the left side as a ([x] + [−x]) dx, then examine the integrand.
Problem 1 (1.15-3). Show that
Rb
a
Lemma. By the notation defined in Problem 4, [x] + [−x] = χZ (x) − 1.
Proof. We consider the two cases suggested by the characteristic function separately. If x ∈ Z, then [x] = x
and −x ∈ Z so [−x] = −x, hence [x] + [−x] = x + (−x) = 0. Meanwhile, χZ (x) − 1 = 1 − 1 = 0 so the
formula holds for x ∈ Z.
If x 6∈ Z, then let [x] = y ∈ Z, which satisfies y ≤ x < y + 1. Since x 6∈ Z, x 6= y and y < x < y + 1.
Multiplying by −1 and rearranging, −y−1 < −x < −y = (−y−1)+1. Therefore, by definition, [−x] = −y−1
and [x] + [−x] = y + (−y − 1) = −1. Meanwhile, χZ (x) − 1 = 0 − 1 = −1, so the formula holds for x 6∈ Z.
Now χZ (x) − 1 is a step function with steps at each integer. On the open intervals of the form (m, m + 1)
for m ∈ Z (as well as (a, m) and (m, b) at the ends), it takes on the value of −1. Since the sum of the widths
of these intervals is b − a, we have by definition
b
Z
(χZ (x) − 1) dx = (−1)(b − a) = a − b,
a
as desired.
Problem 2 (1.15-5a). Prove that
R2
0
[t2 ] dt = 5 −
√
2−
√
3.
Solution. We just need to write f (t) = [t2 ] as a step function on [0, 2]. Indeed, squaring gives
0≤t<1
√
1≤t< 2
√
√
2≤t< 3
√
3≤t<2
Therefore, we can write f (t) as a step function:


0




1
f (t) = 2



3



4
if
if
if
if
if
=⇒ 0 ≤ t2 < 1
=⇒ 1 ≤ t2 < 2
=⇒ 2 ≤ t2 < 3
=⇒ 3 ≤ t2 < 4.
0≤t<1
√
1≤t< 2
√
√
2≤t< 3
√
3≤t<2
t = 2.
1
By the definition of the integral of a step function, therefore,
Z 2
√
√
√
√
f (t) dt = 0(1 − 0) + 1( 2 − 1) + 2( 3 − 2) + 3(2 − 3)
0
√
√
√
√
= 2−1+2 3−2 2+6−3 3
√
√
= 5 − 2 − 3,
as desired.
Problem 3 (1.26-20). Compute the following integral:
Z −4
(x + 4)10 dx.
−2
[Hint: Theorem 1.18.]
Solution. As the hint suggests, we translate the integral by 4 using Theorem 1.18.
Z 0
Z −4
10
x10 dx.
(x + 4) dx =
−2
2
To write this in the form we know, we need to switch the bounds of integration using Theorem 1.19. Then
we can simply apply Theorem 1.15:
Z 0
Z 2
211
2048
x10 dx = −
x10 dx = −
= −
.
10 + 1
11
2
0
Problem 4. Compute the integral
Z
1
−1
1 + x + x2
dx.
1 + x2
Solution. Noticing the symmetric integral, we cleverly split the integral into its even and odd components:
Z 1
Z 1
Z 1
Z 1
1 + x + x2
x
1 + x2
x
dx
=
+
dx
=
dx.
1
dx
+
2
2
2
1
+
x
1
+
x
1
+
x
1
+
x2
−1
−1
−1
−1
The first integrand is constant, so has value 1(1 − (−1)) = 2. To handle the second integral, we will use the
following lemma:
Ra
Lemma. If f (x) is odd (f (−x) = −f (x)) and integrable on [−a, a], then −a f (x) dx = 0.
Proof. Indeed, we can rewrite using the reflection property:
Z a
Z a
Z
f (x) dx =
f (a + (−a) − x) dx =
−a
if f is odd. Therefore, 2
−a
Ra
−a
a
Z
−a
f (x) dx = 0 so
Ra
−a
a
f (−x) dx = −
f (x) dx
−a
f (x) dx = 0 as desired.
x
To apply this lemma, we need to demonstrate that f (x) = 1+x
2 is odd and integrable on [−1, 1]. Indeed,
−x
x
1+(−x)2 = − 1+x2 so f (−x) = −f (x) as desired. Moreover, on [−1, 1], it is increasing: If −1 ≤ x ≤ y ≤ 1,
|xy| = |x| · |y| ≤ 1 so xy ≤ 1 and therefore
0 ≤ (y − x)(1 − xy)
0 ≤ y − x − xy 2 + x2 y
x + xy 2 ≤ y + x2 y
x(1 + y 2 ) ≤ y(1 + x2 )
x
y
≤
,
1 + x2
1 + y2
2
where the last inequality Rrequires 1 + y 2 , 1 + x2 ≥ 1 > 0. Since our function is increasing, it is integrable.
1
x
Therefore by the lemma, −1 1+x
2 dx = 0 and we are left with
Z
1
−1
1 + x + x2
dx = 2 + 0 = 2 .
1 + x2
Problem 5. The characteristic function (also called the indicator function) of a subset A ⊆ R is the function
χA : R → R defined by
(
1 if x ∈ A
χA (x) :=
0 if x 6∈ A.
Prove that χQ is not integrable on the unit interval [0, 1]. (You may use the results of Exercises I 3.12-6 and
I 3.12-9 without proof.)
Solution. We will prove that χQ is not integrable on [0, 1] by showing the upper and lower integrals are not
equal.
Lemma. I(χQ ) = 1.
Proof. First, χQ (x) ≤ 1 for all x ∈ [0, 1], so the constant (step) function 1 is an upper bound. Its integral is
R1
1 dx = 1, so I(χQ ) ≤ 1.
0
Now let t(x) ≥ χQ (x) be a step function, and let [a, b] ⊆ [0, 1] be one of the steps. Then t(x) = c is
constant on (a, b). By Exercise I 3.12-6, there is some rational number r ∈ (a, b), so c = t(r) ≥ χQ (r) = 1.
R1
Therefore, t(x) = c ≥ 1 on this step, and hence on [0, 1]. Therefore by the Comparison Theorem, 0 t(x) dx ≥
R1
1 dx = 1.
0
Since this holds for any upper bound step function t(x), I(χQ ) ≥ 1. With both bounds, I(χQ ) = 1.
Lemma. I(χQ ) = 0.
Proof. This proof is very similar to the previous one. First, χQ (x) ≥ 0 for all x ∈ [0, 1], so the constant
R1
(step) function 0 is a lower bound. Its integral is 0 0 dx = 0, so I(χQ ) ≥ 0.
Now let s(x) ≤ χQ (x) be a step function, and let [a, b] ⊆ [0, 1] be one of the steps. Then s(x) = c is
constant on (a, b). By Exercise I 3.12-9, there is some irrational number r ∈ (a, b), so c = t(r) ≤ χQ (r) =
0. Therefore, s(x) = c ≤ 0 on this step, and hence on [0, 1]. Therefore by the Comparison Theorem,
R1
R1
s(x) dx ≤ 0 0 dx = 0.
0
Since this holds for any lower bound step function s(x), I(χQ ) ≤ 0. With both bounds, I(χQ ) = 0.
Having shown that I(χQ ) = 1 6= 0 = I(χQ ), we conclude from the definition of integrability that χQ is
not integrable on [0, 1].
Problem 6. Let S0 = [0, 1] be the closed unit interval and let Si+1 be defined by removing the open middle
third of each interval in Si . Thus S1 = [0, 13 ] ∪ [ 23 , 1], S2 = [0, 19 ] ∪ [ 92 , 13 ] ∪ [ 23 , 94 ] ∪ [ 89 , 1], and so on; in general
Si is the union of 2i closed intervals of length 3−i . The Cantor set C is defined as the intersection of all the
Si :
\
C := {Si |i ≥ 0}.
R1
Prove that 0 χC (x) dx = 0 where χC is the characteristic function (as defined in the previous problem) of
the Cantor set.
Solution. We must demonstrate that I(χC ) ≤ 0 and I(χC ) ≥ 0. Since I(χC ) ≥ I(χC ), this will be enough
R1
to show that both are 0 and hence 0 χC (x) dx = 0.
The second bound is easier. Since χC (x) ≥ 0, we can take the constant (step) function lower bound
R1
s(x) = 0. Since 0 s(x) dx = 0, I(χC ) ≥ 0.
3
For the upper bound, we construct a sequence of upper bound step functions ti (x) ≥ χC (x) with
t (x) dx decreasing to 0. We can take these step functions directly from the definition of C: Let
0 i
ti (x) = χSi (x). Since Si is a disjoint union of 2i intervals, χSi (x) is a step function with steps consisting of the 2i intervals
composing Si and the 2i − 1 “deleted middle third” intervals between pairs of Si .
T
Since C = {Si : i ≥ 0}, C ⊂ Si for all i. This means that χC (x) ≤ χSi (x): If x ∈ C, x ∈ Si so
χC (x) = 1 = χSi (x) and if x 6∈ C, χC (x) = 0 ≤ χSi (x). Therefore, ti (x) is indeed an upper bound step
function for χC (x).
R1
Finally, we compute 0 ti (x) dx. Since Si consists of 2i closed intervals of width 3−i , we can write
i
Si = ∪2k=1 [ai , ai + 3−i ]. Then we can write
R1
Z
i
1
ti (x) dx =
0
2 Z
X
k=1
2 Z
X
k=1
χSi (x) dx +
ai
i
=
ai +3−i
k=1
ai +3−i
1 dx +
ai
=
−i
(3 ) +
i
2X
−1
χSi (x) dx
0 dx
ai +3−i
0 = 2i · 3−i =
k=1
k=1
ai +3−i
i
2X
−1 Z ai+1
k=1
i
2
X
i
2X
−1 Z ai+1
i
2
.
3
i
Therefore I(χC ) ≤ 32 for all i ∈ Z+ . Finally, we need to show that the infimum of these numbers is 0:
n o
i
Lemma. inf 23 : i ≥ 0 = 0.
2 i
3
≥ 0, so 0 is a lower bound and therefore the infimum y ≥ 0
i
i+1
is also a lower bound: Otherwise there is some i such that 23 > 23 y, but then 23
> y,
exists. Then
making y not a lower bound. But if y > 0, then 23 y > y so y cannot be the smallest lower bound. Hence we
must have the infimum y = 0 as desired.
n o
R1
i
Therefore, I(χC ) ≤ inf 23 : i ≥ 0 = 0, so as we reasoned above, 0 χC (x) = 0 as desired.
Proof. As the product of positive numbers,
3
2y
4