Math 2210 Real Analysis Problem Set 3
Solutions
I. Minevich (small corrections by R. Kenyon)
November 27, 2009
p. 48 # 3. If {fn } is a sequence of measurable functions on X, then
{x : lim fn (x) exists} is a measurable set.
Solution.
{x : lim fn (x) exists} = {x : ∀ > 0 ∃N ∀n, m > N |fn (x) − fm (x)| < since the real (or complex) numbers are complete, so a sequence (namely
(fn (x))∞
n=1 ) converges iff it is Cauchy. Thus
{x : lim fn (x) exists} =
∞ [
∞ \
∞ \
∞
\
k=1 N =1 m=N n=N
{x : fn (x)−fm (x) <
1
1
}∩{x : fm (x)−fn (x) < }.
k
k
Since fm and fn are measurable, so is ±(fm − fn ), so this is a countable
intersection of a countable union of a countable intersection of measurable
sets, which is measurable because the measurable sets form a σ-algebra.
p. 48 # 5. If X = A ∪ B where A, B ∈ M, a function f on X is measurable
iff f is measurable on A and on B.
Solution.
If f is measurable on A and on B and f : (X, M) → (Y, N ) then ∀E ∈
N , f −1 (E) = (f −1 (E) ∩ A) ∪ (f −1 (E) ∩ B) = (f |A )−1 (E) ∪ (f |B )−1 (E): a
union of two measurable sets, hence measurable.
If f is measurable on A ∪ B then (f |A )−1 (E) = f −1 (E) ∩ A: an intersection of two measurable sets, hence measurable; similarly for f |B .
1
p. 48 # 9. Let f : [0, 1] → [0, 1] be the Cantor function (§1.5), and
let g(x) = f (x) + x.
a. g is a bijection from [0, 1] to [0, 2], and h = g −1 is continuous from [0, 2]
to [0, 1].
b. If C is the Cantor set, m(g(C)) = 1.
c. By Exercise 29 of Chapter 1, g(C) contains a Lebesgue nonmeasurable
set A. Let B = g −1 (A). Then B is Lebesgue measurable but not Borel.
d. There exist a Lebesgue measurable function F and a continuous function G on R such that F ◦ G is not Lebesgue measurable.
Solution.
a. Since f (x) is monotone increasing, if y > x then f (y) ≥ f (x), so
g(y) = f (y) + y > f (x) + x = g(x), which shows g is injective. It is surjective
by the intermediate value theorem because it is continuous (being the sum
of two continuous functions), g(0) = 0, and g(1) = 2. Since g is monotone
and bijective, g((a, b)) = (g(a), g(b)), so g is also an open map, which implies
h is continuous.
b. Since g is surjective, g([0, 1] \ C) t g(C) = [0, 2], so we just have to
show m(g([0, 1] \ C)) = 1 (we are using the fact that C is measurable). But
[0, 1] \ C is a disjoint union of open intervals (a, b) on which f (x) is constant, so g([0, 1] \ C) is a disjoint union of open intervals (g(a), g(a) + b − a),
hence m(g([0, 1] \ C)) = the length of the union of these intervals, which is
1. Therefore m(g(C)) = 2 − 1 = 1.
c. Since A ⊂ g(C), we have B ⊂ C, so m(B) ≤ m(C) = 0, and any
null set is measurable. Now if B is Borel then the continuity of h ensures
h(B) = A is Borel: a contradiction since A is not even in the Lebesgue σalgebra.
d. Let F = χB , so F (x) = 1 if x ∈ B and F (x) = 0 if x ∈
/ B, and G = h.
−1
Then F is Lebesgue measurable since F (a, ∞) = ∅, B, or R, all of which
are Lebesgue measurable. And (F ◦ G)−1 (1/2, ∞) = G−1 ◦ F −1 (1/2, ∞) =
2
G−1 (B) = g(B) = A, which is not Lebesgue measurable, so F ◦ G is not
Lebesgue measurable.
p. 59 # 22. Let µ be counting measure on N. Interpret Fatou’s Lemma
and the monotone and dominated convergence theorems as statements about
infinite series.
Solution.
R
P
P
k
k
By setting fk = ∞
a
χ
,
we
have
fk = ∞
{n}
n=1 n
n=1 an . In each
P statement
k ∞
we consider a sequence of fk ’s, so we have a sequence of series { ∞
n=1 an }k=1 .
Fatou’s lemma states
∞
X
n=1
lim inf
k
akn
≤ lim inf
k
∞
X
akn .
n=1
The monotone convergence theorem says that if ∀n ∀k akn ≤ ak+1
n , and
k
∀n an = limk→∞ an , then
∞
X
n=1
an = lim
k→∞
∞
X
akn .
n=1
P
k
The dominated convergence theorem says that if ∞
n=1 an is an absolutely
convergent
akn → an and (b) there exists a
P∞ series for all k such that (a)P∀n
∞
series n=1 bn with bn ≥ 0 for all n and n=1 bn < ∞ such that ∀n ∀k |akn | ≤
bn , then
∞
∞
∞
X
X
X
|an | < ∞ and
an = lim
akn .
n=1
n=1
k→∞
n=1
p. 59 # 25. Let f (x) = x−1/2 if 0 < x < 1, f (x)P= 0 otherwise. let {rn } be
−n
an enumeration of the rationals, and set g(x) = ∞
f (x − rn ).
1 2
a. g ∈ L1 (m), and in particular g < ∞ a.e.
b. g is discontinuous at every point and unbounded on every interval, and
it remains so after any modification on a Lebesgue null set.
c. g 2 < ∞ a.e., but g 2 is not integrable on any interval.
3
Solution.
a. By the monotone convergence theorem,
Z ∞
Z ∞
Z
∞
X
−n
|g(x)|dx =
g(x)dx =
2
−∞
−∞
rn
n=1
1
x
−1/2
dx =
0
n=1
f (x − rn )dx.
−∞
n=1
But this equals
Z
Z rn +1
∞
∞
X
X
−1/2
−n
−n
(x−rn )
dx =
2
2
∞
∞
X
n=1
21−n x1/2 |10
=
∞
X
21−n = 2.
n=1
if ∃E ∈ M, m(E) = > 0, such that g = ∞ on E then
RIn particular,
R
g ≥ E g ≥ · ∞ = ∞ > 2: a contradiction.
R
b. g is unbounded on any interval because any interval (a, b) contains a
rational rn for some n, and for any large M ∈ R and any x ∈ (rn , b) ∩ (rn , r +
1/(22n M 2 )) we have f (x − r)2−n > M , and g(x) ≥ f (x − rn )2−n since g(x)
is a sum of nonnegative terms, hence we can pick x ∈ (a, b) so that g(x) is as
large as we want. Any modification on a null set will still leave uncountably
many such x ∈ (rn , b) ∩ (rn , r + 1/(22n M 2 )), so g will remain unbounded.
Now suppose g is continuous at a point x0 . Then there exists δ such that
|x − x0 | < δ ⇒ |g(x) − g(x0 )| < 1. But g is unbounded on (x0 − δ, x0 + δ): a
contradiction.
c. Because g < ∞ a.e., g 2 < ∞ a.e. since g(x) = ∞ ⇔ g(x)2 = ∞. Again
using the monotone convergence theorem,
!2
Z
Z b X
Z bX
∞
∞
2
−n
2−2n (f (x − rn ))2 dx
|g| =
2 f (x − rn ) dx ≥
(a,b)
a
a n=1
n=1
=
∞
X
f (x − rn )2 dx
2
(1)
a
n=1
Let rk ∈ (a, b). Then
Z b
Z
−2k
2
−2k
(1) ≥ 2
f (x−rk ) dx = 2
a
b
Z
−2n
min{b,rk +1}
rk
1
dx = 2−2k (ln(min{b, 1})−ln(0)) = ∞.
x − rk
p. 63 # 32. Suppose µ(X) < ∞. If f and g are complex-valued
measurable functions on X, define
Z
|f − g|
ρ(f, g) =
dµ.
1 + |f − g|
4
Then ρ is a metric on the space of measurable functions if we identify functions that are equal a.e., and fn → f with respect to this metric iff fn → f
in measure.
Solution.
To prove ρ is a metric, first we check that ρ(f, g) = 0 ⇔ f = g. Obviously,
ρ(f, f ) = 0. If ρ(f, g) = 0 but f 6= g then ∃E ∈ M|m(E) > 0 and ∀x ∈
R |f −g|
|f −g|
>
0
on
E,
so
ρ(f,
g)
≥
> 0.
Ef (x) 6= g(x), which implies 1+|f
−g|
E 1+|f −g|
Second, obviously ρ(f, g) = ρ(g, f ). Third, we check the triangle inequality:
|f − h|
1
1
=1−
≤1−
1 + |f − h|
1 + |f − h|
1 + |f − g| + |g − h|
=
|f − g|
|g − h|
|f − g|
|g − h|
+
≤
+
1 + |f − g| + |g − h| 1 + |f − g| + |g − h|
1 + |f − g| 1 + |g − h|
Now suppose fn → f in this metric but not in measure. Then ∃ > 0, δ >
0 ∀N ∃n > N µ({x : |fn (x) − f (x)| ≥ }) ≥ δ. This implies
Z
|fn − f |
≥
δ > 0,
ρ(fn , f ) ≥
1+
{x:|fn (x)−f (x)|≥} 1 + |fn − f |
a contradiction to convergence in the metric.
Finally, suppose
R f|fnn −f→| f in measure but not in the metric. Then ∃ >
0 ∀N ∃n > N 1+|f
≥ . Without loss of generality we may take <
n −f |
|fn −f |
2µ(X). Letting M = µ(X), g = 1+|f
, and E = {x : g(x) ≥ /(2M )}, we
n −f |
have µ(E) ≥ /2: otherwise, since g ≤ 1 we have
Z
Z
Z
≤
g=
g+
g < µ(E) +
µ(X \ E) < +
M = ,
2M
2 2M
X
E
X\E
a contradiction. But the fact that µ(E) ≥ /2 means exactly that
µ({x : |fn (x) − f (x)| ≥
/(2M )
> 0}) ≥ ,
1 − /(2M )
2
a contradiction to convergence in measure.
5
p. 63 # 33. If fn ≥ 0 and fn → f in measure, then
R
f ≤ lim inf
R
fn .
Solution.
By theorem 2.30, there is a subsequence fnk that converges to f a.e., so
lim inf fn (x) ≤ f (x) a.e. If there exists a measurable
S∞ set E, µ(E)
S∞ > 0,
such that ∀x ∈ E lim inf fn (x) < f (x) then E = k=1 Ek = k=1 {x :
lim inf fn (x) ≤ f (x) − k1 }, so ∃k|µ(Ek ) = δ > 0. But fn → f in measure,
so ∃N ∀n > N µ({x : |fn (x) − f (x)| ≥ k1 })
R < δ:
R a contradiction, hence
R
lim inf fn (x) = f (x) a.e. By Fatou’s lemma, f = lim inf fn ≤ lim inf fn .
p. 63 # 34. Suppose |fn | ≤ g ∈ L1 and fn → f in measure.
R
R
a. f = lim fn .
b. fn → f in L1 .
Solution.
a. By theorem 2.30 there exists a subsequence fnk → f a.e., so by Proposition
2.11 f is measurable (up to a null set). Looking at the imaginary and real
parts of fn and f separately, we can assume the fn and g are all real-valued,
so g +fn ≥ 0 a.e. and g −fn ≥ 0 a.e. By definition of convergence in measure,
g + fn → g + f and g − fn → g − f in measure, so by exercise 33,
Z
Z
Z
Z
Z
Z
g + f = g + f ≤ lim inf (g + fn ) = g + lim inf fn
so
R
R
f ≤ lim inf fn , and
Z
Z
Z
Z
Z
Z
g − f = g − f ≤ lim inf (g − fn ) = g − lim sup fn
so
R
R
f ≥ lim sup fn . Together,
Z
Z
Z
Z
lim inf fn ≥ f ≥ lim sup fn ≥ lim inf fn
andR we have equality everywhere:
lim fn .
R
f = lim inf
R
R
fn = lim sup fn =
b. Since |fn − f | → 0 in measure and |fn − f | ≤R 2g ∈ L1R, we can apply part (a) to the sequence (|fn − f |)∞
0 = lim |fn − f |.
n=1 to get 0 =
p. 63 # 38. Suppose fn → f in measure and gn → g in measure.
6
a. fn + gn → f + g in measure..
b. fn gn → f g in measure if µ(X) < ∞, but not necessarily if µ(X) = ∞.
Solution.
a. Given > 0, δ > 0, ∃N1 , N2 ∀n > N1 µ({x : |fn (x) − f (x)| ≥ /2}) < δ/2
and ∀n > N2 µ({x : |gn (x) − g(x)| ≥ /2}) < δ/2. Letting N = max{N1 , N2 }
we have ∀n > N µ({x : |fn (x) + gn (x) − (f (x) + g(x))| ≥ }) < δ since
≤ |fn (x) + gn (x) − (f (x) + g(x))| ≤ |fn (x) − f (x)| + |gn (x) − g(x)| implies
|fn (x) − f (x)| ≥ /2 or |gn (x) − g(x)| ≥ /2.
b. Note first that if fn → f in measure then f must be finite a.e.; similarly for g. Let An = {x : f (x) > n} and Bn = {x : g(x) > n}. Since
∞
µ(X) < ∞, and X = ∪∞
n=1 An = ∪n=1 Bn , given > 0, there exists N1 such
that for n > N1 , µ(An ), µ(Bn ) < .
There exists N2 such that for n ≥ N2 , µ{x : |fn (x) − f (x)| > /n} < and µ{x : |gn (x) − g(x)| > /n} < . Let Cn = {x : |fn (x) − f (x)| < /n}
and Dn = {x : |gn (x) − g(x)| < /n}. Now
|fn (x)gn (x) − f (x)g(x)| ≤ |fn (x)||gn (x) − g(x)| + |g(x)||fn (x) − f (x)|
and if n ≥ max{N1 , N2 } then the right-hand side at most 2 on An ∩ Bn ∩
Cn ∩ Dn , a set whose complement has small measure.
For a counterexample when µ(X) = ∞, let f (x) = g(x) = x and
fn (x) = gn (x) = x +
1
χ[n,n+1) .
n
Clearly fn → f and gn → g in measure. However for x ∈ [n, n + 1),
|fn (x)gn (x) − f (x)g(x)| = (x + 1/n)2 − x2 = 2x/n + 1/n2 > 2. So fn gn 9 f g
in measure.
p. 63 # 42. Let µ be counting measure on N. Then fn → f in measure iff fn → f uniformly.
Solution.
If fn → f in measure then, given > 0, ∃N ∀n > N µ({x : |fn (x) − f (x)| ≥
}) < 1/2, i.e. {x : |fn (x) − f (x)| ≥ } = ∅, so ∀x|fn (x) − f (x)| < , which
is uniform convergence.
7
If fn → f uniformly then, given δ > 0 and > 0, ∃N ∀n > N ∀x|fn (x) −
f (x)| < , so µ({x : |fn (x) − f (x)| ≥ }) = 0 < δ, which is convergence in
measure.
p. R77 # 55.
the existance and equality
R1
R 1 E = [0, 1] × [0, 1].
R 1 Investigate
R 1 Let
2
of E f dm , 0 0 f (x, y)dxdy, and 0 0 f (x, y)dydx for the following f .
a. f (x, y) = (x2 − y 2 )(x2 + y 2 )−2 .
b. f (x, y) = (1 − xy)−a (a > 0).
c. f (x, y) = (x − 1/2)−3 if 0 < y < |x − 1/2|, f (x, y) = 0 otherwise.
Solution.
The strategy is to first integrate |f |. If the integral exists and is finite, then
f ∈ L1 so by Fubini’s theorem, all three integrals are equal.
a.
1
1
1
1
|x2 − y 2 |
|f |dm =
|f (x, y)|dxdy =
dxdy
2
2 2
0
0
0
0 (x + y )
y
1 !
Z 1 Z y 2
Z 1 2
Z 1
y − x2
x − y2
x x −
dy
=
dx +
dx dy =
2
2 2
2
2 2
x2 + y 2 0 x2 + y 2 y
0
0 (x + y )
y (x + y )
0
Z 1
1
1
=
−
dy = ln y|10 − tan−1 y|10 = ∞
2
y 1+y
0
R
so f dm2 does not exist. But
Z
1
Z
0
1
Z
0
and
Z 1Z
0
0
Z
2
x2 − y 2
dxdy =
(x2 + y 2 )2
1
Z
1
Z
x2 − y 2
dydx =
(x2 + y 2 )2
Z
0
Z
1
0
Z
1
Z 1
−1
−x π
−1 1
dy
=
dy
=
−
tan
y|
=
−
0
2
x2 + y 2 0
4
0 1+y
1
Z 1
y 1
π
−1
1
dx
=
dx
=
tan
x|
=
0
2
x2 + y 2 0
4
0 1+x
b. Case 1. a < 2, a 6= 1:
Z
2
Z
1
Z
|f |dm =
1
−a
Z
(1−xy) dxdy =
0
0
0
8
1
1
Z 1
(1 − xy)1−a 1
(1 − y)1−a 1
dy =
−
dy
y(a − 1) 0
a−1 0
y
y
Z
1/2
=
0
(1 − y)1−a 1
−
y
y
Z
1
dy +
1/2
(1 − y)1−a 1
−
y
y
dy
Note that the function is bounded in the first integral since
(1 − y)1−a 1
−
= lim (1 − a)(1x )−a = 1 − a
lim
y→0
y→0
y
y
And the denominator in the second integral is ≥ 1/2 so the integral is
Z 1
(1/2)2−a
≤2
((1 − y)1−a − 1)dy =
−1
2−a
1/2
Thus, by Fubini’s theorem, all three integrals are equal.
Case 2. a = 1:
Z
Z
Z 1Z 1
−1
2
(1−xy) dxdy =
|f |dm =
0
1
Z 1
ln(1 − xy) ln(1 − y)
dy = −
dy
−y
y
0
0
1
0
0
1/2
Z
=−
0
ln(1 − y)
dy −
y
Z
1
1/2
ln(1 − y)
dy < ∞
y
because
ln(1 − y)
−1
= lim
= −1
y→0
y→0 1 − y
y
so the function is bounded from 0 to 1/2, which makes the first integral finite,
and
Z 1
Z 1
ln(2) 1
ln(1 − y)
−
dy ≤ −2
ln(1 − y)dy = 2(
+ ) < ∞.
y
2
2
1/2
1/2
lim
Again, by Fubini’s theorem, all three integrals are equal.
Case 3. a ≥ R2: From case 1, we can see that the second integral will be
1
infinite since 1/2 (1 − y)−b = ∞ when b > 1. Since the function is symmetric
in x and y, integration with respect to dxdy is the same as with respect to
dydx, so none of the integrals exist.
c.
Z
2
Z
1/2
Z
|f |dm =
0
0
1/2−x
1
−x
2
−3
Z
1
Z
dydx +
1/2
9
0
x−1/2
−3
1
x−
dydx
2
Z
1/2
=
0
R
1
−x
2
−2
−2
−1 12 −1 1
1
1
1
x−
dx =
−x
− x−
= ∞+∞ = ∞
1
2
2
2
1/2
Z
dx+
1
0
2
2
so f dm does not exist. Also,
−3
−3
Z 1 Z x−1/2 Z 1Z 1
Z 1/2 Z 1/2−x 1
1
x−
dydx+
dydx = ∞−∞
f (x, y)dydx =
x−
2
2
1/2 0
0
0
0
0
R1R1
so 0 0 f (x, y)dydx does not exist. And
−3
−3
Z 1Z 1
Z 1/2 Z 1/2−y Z 1/2 Z 1 1
1
x−
f (x, y)dxdy =
dxdy+
x−
dxdy
2
2
0
0
0
0
0
y−1/2
"Z
#
Z 1/2 1/2 1
1
1
1
1
−
−
=−
dy +
dy = −∞
2 0
y2 4
4 (y − 1)2
0
p. 77 # 59. Let f (x) = x−1 sin x.
R∞
a. Show that 0 |f (x)|dx = ∞.
Rb
b. Show that limb→∞ 0 f (x)dx = π/2 by integrating e−xy sin x with respect to x and y. (In view of part (a), some care is needed in passing
to the limit as b → ∞.)
Solution.
a.
Z ∞
∞ Z (k+1)π
∞ X
X
sin x |
sin
x|
1
1
dx ≥
√
dx
≥
x x
4 2 k=1 k +
0
k=1 kπ
1
4
1
+
k+
3
4
=∞
since this is the harmonic series. In more detail, by trapezoidal rule,
Z (k+1)π
| sin x|
π 1
1
1
1
3
dx ≥ ·
| sin((k + )π)| +
| sin((k + )π)|
x
2 2 (k + 14 )π
4
4
(k + 43 )π
kπ
1
1
1
= √
+
.
k + 43
4 2 k + 41
b. On the one hand,
Z ∞Z ∞
Z
−xy
e
sin xdydx =
0
0
0
∞
∞
Z ∞
−1 −xy
sin x
e
sin x dx =
dx.
x
x
0
0
10
On the other hand,
Z
Z ∞Z ∞
−xy
e
sin xdxdy =
0
0
0
∞
∞
−e−xy cos x − ye−xy sin x dy
y2 + 1
0
Noting that | cos x| ≤ 1 and | sin x| ≤ 1, the limits at ∞ are 0:
Z ∞
1
π
dy = tan−1 y|∞
=
0 =
2
y +1
2
0
11