REAL ANALYSIS I HOMEWORK 4
CİHAN BAHRAN
The questions are from Stein and Shakarchi’s text, Chapter 2.
1.
Given a collection of sets E1 , E2 , . . . , En , construct another collection
S
S
∗
∗
∗
∗
∗
E1 , E2 , . . . , EN
, with N = 2n − 1, so that nk=1 Ek = N
j=1 Ej ; the collection {Ej }
S
∗
is disjoint; also Ek = Ej∗ ⊆Ek Ej , for every k.
Observe that it suffices to construct such a collection with N ≤ 2n − 1 because in case
N < 2n − 1 we can pad empty sets to get to exactly 2n − 1 sets and the conclusions
will still hold.
Let C = {E1 , . . . , En }. Let D be the collection of all subsets of X = nk=1 Ek which have
expressions of the type E10 ∩ · · · ∩ En0 where Ei0 is either Ei or Eic for each i = 1, . . . , n.
The function
S
f : 2C → D
E 7→
\
E∩
E∈E
\
Ec
E ∈E
/
is surjective by the definition of D, hence |D| ≤ |2C | = 2n . Note that if E = ∅, then
f (E) = E1c ∩ · · · ∩ Enc = X c = ∅. So we can restrict f as f : 2C r {∅} → Dr {∅}.
First, we show that
X=
[
f (E) =
E∈2Cr{∅}
[
D.
(?)
D∈Dr{∅}
The second equality directly follows from f being surjective. For the first equality, let
x ∈ X. If we let Ex = {E ∈ C : x ∈ E}, we have x ∈ f (Ex ) by the definition of f .
Second, we show disjointness. Say x ∈ f (E). Then for every E ∈ E, we have x ∈ E.
Hence E ⊆ Ex . Conversely whenever E ∈
/ E, we have x ∈
/ E. Thus E = Ex . From here
we deduce that
f (E) ∩ f (E 0 ) 6= ∅ ⇒ E = E 0 .
Third, for E ∈ C we claim that
E=
[
f (E) =
E∈2Cr{∅}
[
D.
D∈Dr{∅}
D⊆E
f (E)⊆E
Again the second equality is immediate. For the first equality the ⊇ part is trivial. Let
x ∈ E, so E ∈ Ex , equivalently E ∈ Ex , and x ∈ f (Ex ) so we are done.
Thus writing Dr {∅} = {E ∗ , . . . , E ∗ }, we have N ≤ 2n − 1 with the desired properties.
1
N
2. In analogy to Proposition 2.5, prove that if f is integrable on Rd and δ > 0,
then f (δx) converges to f (x) in the L1 -norm as δ → 1.
1
REAL ANALYSIS I HOMEWORK 4
2
Since the set of continuous functions with compact support is dense in L1 (Rd ), given
ε > 0 there exists such a function g such that ||f − g|| ≤ ε. Writing fδ for the function
x 7→ f (δx), we have
||fδ − gδ || = ||(f − g)δ || = δ −d ||f − g|| .
Since g is uniformly continuous on K := supp g, there exists λ > 0 such that |g(x) −
g(y)| ≤ ε/m(K) whenever |x − y| ≤ λ and x, y ∈ K. Since K is compact, there exists
M > 0 such that x ∈ K implies |x| ≤ M . Choosing δ close enough to 1, we can
guarantee that |1 − δ| ≤ λ/M . Therefore for every x ∈ K we get
|x − δx| = |1 − δ||x| ≤ λ .
Thus for every x ∈ K
|g(x) − g(δx)| ≤ ε/m(K) .
Thus ||g − gδ || =
Z
K
|g − gδ | ≤ ε. Hence
||f − fδ || ≤ ||f − g|| + ||g − gδ || + ||fδ − gδ ||
= (δ −d + 1)||f − g|| + ||g − gδ ||
≤ (δ −d + 1)ε + ε = ε(δ −d + 2)
and given δ even closer to 1 we can make sure that δ −d < 2. So given ε > 0, we can
pick η > 0 such that |1 − δ| < η implies
||f − fδ || ≤ 4ε .
This yields
lim ||f − fδ || = 0 .
δ→1−
5. Suppose F is a closed set in R, whose complement has finite measure, and let
δ(x) denote the distance from x to F , that is,
δ(x) = d(x, F ) = inf{|x − y| : y ∈ F } .
Consider
I(x) =
Z
R
δ(y)
dy .
|x − y|2
(a) Prove that δ is continuous, by showing that it satisfies the Lipschitz condition
|δ(x) − δ(y)| ≤ |x − y|.
(b) Show that I(x) = ∞ for each x ∈
/ F.
(c) Show that I(x) < ∞ for a.e. x ∈ F . This may be surprising in view of the
fact that the Lipschitz condition cancels only one power of |x − y| in the
integrand of I.
[Hint: For the last part, investigate
R
F
I(x)dx.]
(a) For every z ∈ F we have
|x − z| ≤ |x − y| + |y − z| .
REAL ANALYSIS I HOMEWORK 4
3
Hence
δ(x) = inf{z ∈ F : |x − z|} ≤ inf{z ∈ F : |x − y| + |y − z|}
≤ |x − y| + inf{z ∈ F : |y − z|} = |x − y| + δ(y)
so δ(x) − δ(y) ≤ |x − y|. Changing the roles of x and y, we get δ(y) − δ(x) ≤ |y − x| =
|x − y|. Thus
|δ(x) − δ(y)| ≤ |x − y| .
(b) Note that x ∈
/ F means that x has a ball around it disjoint from F since F is closed
and hence δ := d(x, F ) = δ(x) > 0. Thus |x − y| ≤ δ/2 implies
|δ(x) − δ(y)| ≤ δ/2 .
So δ(y) ≥ δ/2. Thus
Z x+δ/2
x−δ/2
δ(y)
dy ≥
|x − y|2
Z x+δ/2
x−δ/2
δ/2
dy =
|x − y|2
Z δ/2
−δ/2
δ/2
du = δ
u2
Z δ/2
0
du
= δ · ∞ = ∞.
u2
(c) Consider the function
f :F ×R→R
(x, y) 7→
δ(y)
.
|x − y|2
Note that f (x, x) = ∞ for every x ∈ F . f is continuous since δ is continuous. Also
clearly f is non-negative. Hence by Tonelli’s theorem, we have
Z
Z Z
δ(y)
I(x)dx =
dxdy
F
R F |x − y|2
ÇZ
å
Z
1
= δ(y)
dx dy
F |x − y|2
R
ÇZ
å
Z
1
=
δ(y)
dx dy .
F |x − y|2
Fc
Observe that for fixed y ∈ F c , for every x ∈ F we have δ(y) ≤ |x − y| and hence
F ⊆ {x ∈ R : |x − y| ≥ δ(y)}
= {x ∈ R : x − y ≥ δ(y)} ∪ {x ∈ R : x − y ≤ −δ(y)}
= [δ(y) + y, ∞) ∪ (−∞, y − δ(y)]
and so
Z
F
Z ∞
y−δ(y)
1
1
dy
+
dy
2
|x − y|2
y+δ(y) |x − y|
−∞
Z ∞
Z y−δ(y)
1
1
=
dy +
dy
2
(x − y)2
y+δ(y) (x − y)
−∞
Z ∞
Z −δ(y)
1
1
=
du
+
du
2
u2
δ(y) u
−∞
1
dx ≤
|x − y|2
=2
Z ∞
δ(y)
= 2 lim
L→∞
Z
L
−1 1
du
=
2
lim
L→∞ u u2
δ(y)
Ç
1
1
−
δ(y) L
å
=
2
.
δ(y)
REAL ANALYSIS I HOMEWORK 4
4
Thus
Z
I(x)dx ≤
Z
F
Fc
δ(y)
2
dy = 2m(F c ) < ∞
δ(y)
since m(F c ) < ∞ by assumption. So we showed that
finite almost everywhere for x ∈ F .
R
F
I(x)dx is finite hence I(x) is
7. Let Γ ⊆ Rd × R, Γ = {(x, y) ∈ Rd × R : y = f (x)}, and assume f is measurable
on Rd . Show that Γ is a measurable subset of Rd+1 , and m(Γ) = 0.
Consider the function
F : Rd × R → R
(x, y) 7→ f (x) .
Given α ∈ R, the set
F −1 ((α, ∞)) = f −1 ((α, ∞)) × R
is measurable in Rd+1 ; hence G is a measurable function. Also the projection
Rd × R → R
(x, y) 7→ y
is measurable (in fact continuous). Thus their difference
G : Rd × R → R
(x, y) 7→ f (x) − y
is a measurable function, thus G−1 ({0}) = Γ is a measurable subset of Rd+1 . And by
Corollary 2.3.3,
m(Γ) =
Z
Rd
m(Γx )dx =
Z
Rd
m({f (x)})dx =
Z
0dx = 0.
10. Suppose f ≥ 0, and let E2k = {x : f (x) > 2k } and Fk = {x : 2k < f (x) ≤
2k+1 }. If f is finite almost everywhere, then
∞
[
Fk = {f (x) > 0} ,
k=−∞
and the sets Fk are disjoint.
Prove that f is integrable if and only if
∞
X
2k m(Fk ) < ∞,
if and only if
k=−∞
∞
X
2k m(E2k ) < ∞ .
k=−∞
Use this result to verify the following assertions. Let

|x|−a
f (x) = 
0
if |x| ≤ 1,
otherwise,
and

|x|−b
g(x) = 
0
if |x| > 1,
otherwise.
Then f is integrable on Rd if and only if a < d; also g is integrable on Rd if and
only if b > d.
REAL ANALYSIS I HOMEWORK 4
5
Note that Fk ’s are disjoint so letting E = {x : f (x) > 0}, as f = 0 outside E by
non-negativity we have
Z
f=
Z
f=
∞ Z
X
f
k=−∞ Fk
E
and by the definition of Fk , we have
∞
X
2k m(Fk ) ≤
k=−∞
so
R
f is finite if and only if
∞ Z
X
f≤
−∞ Fk
∞
X
∞
X
∞
X
2k+1 m(Fk ) = 2 ·
k=−∞
2k m(Fk )
k=−∞
2k m(Fk ) is finite.
k=−∞
Observe that for every k, we have
E2k = E2k+1 t Fk
hence
∞
X
2k m(E2k ) =
k=−∞
=
Now since
∞
X
2k+1 m(E2k+1 ) =
k=−∞
∞
X
2k m(E2k+1 ) +
k=−∞
∞
X
∞
X
2k m(Fk )
k=−∞
∞
X
1
2k+1 m(E2k+1 ) +
2k m(Fk ) .
2 k=−∞
k=−∞
∞
X
2k m(E2k ), we obtain
k=−∞
∞
X
2k m(Fk ) =
k=−∞
∞
1 X
2k m(E2k )
2 k=−∞
so we get the second if and only if.
Consider the given f and E2k ’s defined by it. If a < 0 then f is clearly integrable. So
we may assume a > 0. Then we have
E2k = {x ∈ Rd : |x|−a > 2k , |x| ≤ 1}
= {x ∈ Rd : |x|a < 2−k , |x| ≤ 1}
= {x ∈ Rd : |x| < min{1, 2−k/a }}
and since 2−k/a ≤ 1 if and only if −k/a ≤ 0 if and only if k ≥ 0, we get

B
if k < 0
d (1)
E2k =  R −k/a
BRd (2
) if k ≥ 0
where BRd (r) denotes the ball centered at the origin of radius r in Rd . Thus
∞
X
k=−∞
2k m(E2k ) =
−1
X
k=−∞
2k m(BRd (1)) +
∞
X
2k m(B
−k/a
))
Rd (2
k=0
Observe that a ball of radius r in Rd contains a cube of side-length r and be contained
in a cube of side-length 2r in Rd . Thus
rd ≤ m(BRd (r)) ≤ (2r)d = 2d rd
REAL ANALYSIS I HOMEWORK 4
therefore
−1
X
2k +
k=−∞
so as
∞
X
2k (2−k/a )d
≤
k=0
−1
X
k=∞
X
Ñ
−1
X
2k m(Ek ) ≤ 2d
k=−∞
6
2k +
k=−∞
2k = 1, it is enough to determine when
k=−∞
∞
X
2k (2−k/a )d
é
k=0
∞
X
2k (2−k/a )d
converges for the
k=0
integrability of f . And
∞
X
2k (2−k/a )d
=
k=0
converges iff 2
1−d/a
∞
X
2k−kd/a
=
k=0
∞
X
2k(1−d/a)
=
k=0
∞
X
(2(1−d/a) )k
k=0
< 1 iff 1 − d/a < 0 iff d/a > 1 iff a < d.
Now consider the given g and E2k ’s defined by it. We may define g to be 1 when |x| ≤ 1
which does not affect the integrability of g. If b < 0 then g is clearly not integrable. So
we may assume b > 0. In this case we have
E2k = {x ∈ Rd : |x|−b > 2k , |x| > 1} ∪ {x ∈ Rd : 1 > 2k , |x| ≤ 1}
= {x ∈ Rd : |x|b < 2−k , |x| > 1} ∪ {x ∈ Rd : 1 > 2k , |x| ≤ 1}
= {x ∈ Rd : 1 < |x| < 2−k/b } ∪ {x ∈ Rd : 1 > 2k , |x| ≤ 1} .
Note that since 2−k/b > 1 iff −k/b > 0 iff k < 0; so E2k = ∅ if k ≥ 0. And for k < 0 we
have
E2k = {x ∈ Rd : 1 < |x| < 2−k/b } ∪ {x ∈ R : |x| ≤ 1}
= {x ∈ Rd : |x| < 2−k/b } .
Because E2k = BRd (2−k/b ), we have
−1
X
2k (2−k/b )d ≤
k=−∞
Since
−1
X
k=−∞
2k (2−k/b )d =
−1
X
2k m(Ek ) ≤ 2d
k=−∞
−1
X
2k(1−d/b) =
k=−∞
−1
X
2k (2−k/b )d .
k=−∞
∞
X
2(d/b−1)k
converges iff 2(d/b−1) < 1 iff b > d,
k=1
the series
∞
X
2k m(E2k ) =
k=−∞
−1
X
2k m(E2k )
k=−∞
also converges iff b > d as desired.
13. Give an example of two measurable sets A and B such that A + B is not
measurable.
[Hint: In R2 take A = {0} × [0, 1] and B = N × {0}.]
Let A, B be as in the hint where N ⊆ R is a nonmeasurable set. Then
A + B = {(0, y) + (x, 0) : y ∈ [0, 1], x ∈ N } = N × [0, 1]
By Lemma 3.5, both A and B are null sets, hence are measurable. But A + B cannot
be measurable, because since m([0, 1]) = 1 > 0, the measurability of A + B would imply
the measurability of N by Proposition 3.4.
REAL ANALYSIS I HOMEWORK 4
7
15. Consider the function defined over R by

x−1/2
f (x) = 
0
if 0 < x < 1,
otherwise.
For a fixed enumeration {rn }∞
n=1 of the rationals Q, let
∞
X
2−n f (x − r
F (x) =
n ).
n=1
Prove that F is integrable, hence the series defining F converges for almost every
x ∈ R. However, observe that the series is unbounded on every interval, and in
f that agrees with F a.e. is unbounded in any interval.
fact, any function F
Note that
Z
f χ(1/n,1) =
Z 1
√ 1
dx = 2 x
x
1/n
f=
Z
= 2 − 2 1/n
1/n
so by monotone convergence theorem
Z
»
−1/2
»
f χ(0,1) = n→∞
lim (2 − 2 1/n) = 2 .
Note that by the translation invariance of the Lebesgue integral, we have
Z
f (x)dx =
Z
f (x − r)dx
for any r ∈ R. Thus
Z X
N
2−n f (x − r
n )dx =
Z
N
X
2−n
f (x)dx =
n=1
n=1
N
X
2−n+1
=
n=1
N
−1
X
2−n
n=0
and again by the monotone convergence theorem
Z
F =
∞
X
2−n
= 2.
n=0
Let I be a interval. Let rN be a rational number in I. Then for every M > 1, whenever
x ∈ (rN , rN + 2−2N /M 2 ) we have 0 < x − rN < 2−2N /M 2 < 1 so since f is decreasing
f (x − rN ) > f (2−2N /M 2 ) = 2N M .
But (rN , rN +2−2N /M 2 ) intersects I in a nonempty interval which has positive measure.
So on a set of positive measure contained in I, we have
F (x) =
∞
X
2−n f (x − r
n)
≥ 2−N f (x − rN ) > M .
n=1
f
F
f also has to be larger than M in a set of positive
So if
is a.e. equal to F then F
measure zero contained in I. Since M was arbitrary, we get the desired conclusion.
18. Let f be a measurable finite-valued function on [0, 1], and suppose that |f (x) −
f (y)| is integrable on [0, 1] × [0, 1]. Show that f (x) is integrable on [0, 1].
We are assuming that the function
F : [0, 1] × [0, 1] → R
(x, y) 7→ |f (x) − f (y)|
REAL ANALYSIS I HOMEWORK 4
8
is integrable. So by a part of Fubini’s theorem, for almost every y ∈ [0, 1] the function
[0, 1] → R
x 7→ |f (x) − f (y)|
is in L1 ([0, 1]). So pick and fix such a y. Then the function
g : [0, 1] → R
x 7→ f (x) − f (y)
is in L1 ([0, 1]) , because |g| is. Now since f (y) is finite, the constant function
c : [0, 1] → R
x 7→ f (y)
is also in L1 ([0, 1]) (its integral is f (y)). Thus g + c = f ∈ L1 ([0, 1]).
19. Suppose f is integrable on Rd . For each α > 0, let Eα = {x : |f (x)| > α}.
Prove that
Z
Rd
Z ∞
|f (x)|dx =
0
m(Eα )dα.
Consider the function
F : (0, ∞) × Rd → R

1
(α, x) 7→ 
if |f (x)| > α,
0 otherwise.
Since F is non-negative and measurable (since F = χ|f (x)|>α and f is measurable). So
by Tonelli’s theorem we have
Z ∞Z
d
Z ∞0ÅZ R
0
Rd
F (α, x)dxdα =
ã
χEα (x)dx dα =
Z ∞
0
m(Eα )dα =
Z ∞
Z
d
F (α, x)dαdx
0
ZR ÅZ
∞
d
ZR
Rd
0
ã
χ(0,|f (x)|) (α)dα dx
|f (x)|dx .