MATH 202A - Problem Set 12
Walid Krichene (23265217)
November 21, 2012
(12.1) For x ∈ R, let Px = {(p, q) ∈ Z × N|p ∧ q and |x − pq | ≤
R|Px is an infinite set}. Then we have E is a Borel set, and m(E) = 0.
proof
1
q r },
were r > 2. Let E = {x ∈
For all x ∈ R, let
1
p
Qx = {q ∈ N|∃p ∈ Z : p ∧ q = 1 and |x − | ≤ r }
q
q
be the projection of Px on the second coordinate. We first observe that Px is infinite if and only if Qx is
infinite. Indeed
• Qx is infinite ⇒ Px is infinite: this follows from the fact that the projection
Px → Qx
(p, q) 7→ q
is a surjection.
• Qx is finite ⇒ Px is finite: for all q ∈ Qx , there are at most three integers p ∈ Z such that (p, q) ∈ Px
(indeed, if |x − pq1 | ≤ q13 and |x − pq2 | ≤ q13 , then | pq1 − pq2 | ≤ q23 , i.e. |p1 − p2 | ≤ 2/q r−1 ≤ 2 (q r−1 ≥ 1),
therefore there may be at most three distinct such integers p).
Now for all q ∈ N and for all n ∈ N let
p
1
Xqn = {x ∈ [−n, n]|∃p ∈ Z : p ∧ q = 1 and |x − | ≤ r }
q
q
we have x ∈ Xqn ⇔ q ∈ Qx and x ∈ [−n, n], therefore
x ∈ E∩[−n, n] ⇔ x ∈ [−n, n] and Px is infinite ⇔ x ∈ [−n, n] and Qx is infinite ⇔ {q ∈ N|x ∈ Xqn } is infinite
Therefore E ∩ [−n, n] is the set of points x that belong to Xqn for infinitely many indices q ∈ N. We observe
that
p
1 p
1
Xqn = ∪p∈Z,p∧q=1 [−n, n] ∩
− r, + r
q
q q
q
therefore it is a Borel set, and we observe that each intersection [−n, n] ∩ [p/q − 1/q r , p/q + 1/q r ] is nonempty
if and only if p/q − 1/q r ≤ n and p/q + 1/q r ≥ −n, i.e.
p ∈ [−nq − 1/q r−1 , nq + 1/q r−1 ] ⊆ [−nq − 1, nq + 1]
therefore we can write
Xqn ⊆ ∪p∈[−nq−1,nq+1]
1
p
1 p
1
− r, + r
q
q q
q
thus the measure of Xqn is m(Xqn ) ≤ (2nq + 3) q2r ≤ (2nq + 3q) q2r =
X
m(Xqn ) ≤
q∈N
X 4n + 6
q∈N
q r−1
4n+6
q r−1
and
<∞
therefore by the Borel-Cantelli theorem, m(E ∩ [−n, n]) = 0. Finally, since the sequence (E ∩ [−n, n])n is
non-decreasing, and ∪∞
n=1 E ∩ [−n, n] = E, we have by σ-additivity of the measure
m(E) = lim m(E ∩ [−n, n]) = 0
n→∞
(12.2) Let E be the set of all x ∈ [0, 1) such that x has decimal expansion x = 0.a1 a2 . . . in which no two
consecutive digits equal 7. Then E is a Borel set, and m(E) = 0.
proof Let En be the set of points x ∈ [0, 1) that have decimal expansion x = 0.a1 a2 . . . an . . . such that
the (a1 , . . . , an ) does not contain two consecutive elements equal to 7. Let Fn = Enc . We have Fn ⊆ Fn+1
(if (a1 , . . . , an ) contains two consecutive elements equal to 7, then so does (a1 , . . . , an+1 )), thus En+1 ⊆ En
and we have
E = ∩∞
n=1 En
We additionally write En as the disjoint union of two sets: the set of points in E that have digital expansion
with the nth digit equals 7 (respectively, not equal to 7)
En = En7 ∪ En7̄
where En7 is the set of points x ∈ En that have decimal expansion x = 0.a1 . . . an . . . such that an = 7, and
En7̄ is the set of points x ∈ En that have decimal expansion x = 0.a1 . . . an . . . such that an 6= 7.
Then we have the following equivalences: for all n ≥ 1
7
x ∈ En+1
⇔ x ∈ En7̄ and an+1 = 7
7̄
x ∈ En+1
⇔ x ∈ En and an+1 6= 7
Now let
un = m(En7 )
vn = m(En7̄ )
Then we have m(En ) = un + vn , and we have
1
vn
10
9
=
(un + vn )
10
un+1 =
vn+1
with initial terms u1 =
1
10
9
and v1 = 10
. Then we can write
n−1 un
0
α
u1
=
vn
(1 − α) (1 − α)
v1
1
. Therefore we have limn→∞ un = limn→∞ vn = 0.
with α = 10
Finally, since En+1 ⊆ En and m(E0 ) = 1 is finite, we have by σ-additivity
m(E) = m(∩∞
n=1 En )
= lim m(En )
n→∞
= lim un + vn
n→∞
=0
2
(12.3) Let (X, A, µ) be a measure space, with µ(X) = 1. A family of measurable sets {An } is said to be
mutually independent if for all finite N , and all {Bk }k∈{1,...,N } with Bk ∈ {Ak , X \ Ak } satisfies
N
Y
µ(∩N
k=1 Bk ) =
µ(Bk )
k=1
If {An } is a family of mutually independent sets, and
to An for infinitely many indices n.
P
n
µ(An ) = ∞, then almost every x ∈ X belongs
proof Let E = {x ∈ X|x ∈ An for infinitely many n}. Then we have x ∈ E if and only if ∀N ∈ N, ∃n ≥ N
such that x ∈ An , therefore
E = ∩N ∈N ∪n≥N An
and it follows that E is measurable. We have E c = ∪N ∈N ∩n≥N Acn . Let N ∈ N, the sequence of sets
+k c
(∩N
n=N An )k∈N is decreasing, thus by σ-additivity of µ, and since the space has finite measure, we have
+k c
N +k c
µ(∩k∈N (∩N
n=N An )) = lim µ(∩n=N An )
k→∞
for each k ∈ N, by mutual independence, we have
N +k c
µ(∩n=N
An ) =
N
+k
Y
µ(Acn )
n=N
=
N
+k
Y
(µ(X) − µ(An ))
n=N
=
N
+k
Y
(1 − µ(An ))
n=N
Where the sequence (
consider two cases:
QN +k
n=N (1
− µ(An )))k converges to 0: to prove this, let Pk =
QN +k
n=N (1
− µ(An )). We
1. if there exists n0 ≥ N such that µ(An ) = 1, then for all k ≥ n0 − N , Pk = 0, and the result follows.
2. if µ(An ) < 1 for all n ≥ N , then consider
ln Pk =
N
+k
X
ln(1 − µ(An ))
n=N
≤
N
+k
X
−µ(An )
n=N
using the fact that for all x ∈ [0, 1), ln(1−x) ≤ −x. By assumption, we have limk→∞
therefore
lim ln Pk = −∞
PN +k
n=N
µ(A) = ∞,
k→∞
therefore (Pk ) converges to 0 (for all > 0, there exists K such that for all k ≥ K, ln Pk ≤ ln , thus
for all k ≥ K, Pk ≤ ).
Therefore
N +k c
c
µ(∩∞
n=N An ) = µ(∩k∈N (∩n=N An )) = lim
k→∞
3
N
+k
Y
n=N
(1 − µ(An )) = 0
To conclude, we have by σ-additivity of µ
c
µ(E c ) = µ(∪N ∈N (∩∞
n=N An ))
X
c
≤
µ(∩∞
n=N An ) = 0
N ∈N
therefore m(E c ) = 0, which proves that almost every x is in E.
(12.4) Egoroff’s theorem: Let (X, A, µ) be a finite measure space. Suppose that (fn ) is a sequence of
measurable functions fn : X → R, that (fn ) converges to f : X → R almost everywhere, where f is a
measurable function. Then for all > 0, there exists E ∈ A such that µ(E) < and (fn ) converges to f
uniformly on X \ E.
proof We first show that for all η > 0, there exists Fη ∈ A and Nη ∈ N such that µ(F ) ≤ η and for all
n ≥ Nη , for all x ∈ Fηc , |fn (x) − f (x)| ≤ η. Let η > 0 be fixed, and consider the sets
GN,η = {x ∈ X|∀n ≥ N, |fn (x) − f (x)| ≤ η}
FN,η = GcN,η
where ∪∞
N =1 GN,η = X since for all x ∈ X, limn fn (x) = f (x) thus there exists N such that for all n ≥ N
|fn (x) − f (x)| ≤ η, i.e. x ∈ Gn,η .
We have for all N ∈ N, GN,η is the inverse image of the interval [0, η] by the measurable function |fn − f |,
thus GN,η is measurable and so is FN,η . And we have GN,η ⊆ GN +1,η , thus FN +1,η ⊆ FN,η . Therefore
c
c
∞
(FN,η )N is a non-increasing sequence of subsets, with intersection ∩∞
N =1 FN,η = (∪N =1 GN,η ) = X = ∅. by
σ-additivity of m,
lim m(FN,η ) = m(∅) = 0
n→∞
thus there exists Nη ∈ N such that m(FNη ,η ) ≤ η, and on FNc η ,η = GNη ,η , we have |fn − f | ≤ η. Let Fη
denote FNη ,η (to simplify notation).
Now fix > 0, and consider the sequence ηn = /2n . Consider the union
F = ∪∞
n=1 Fηn
then we have
• F is measurable as the countable union of measurable sets.
• the measure of F is
m(F ) = m(∪∞
n=1 Fηn ) ≤
∞
X
m(Fηn )
n=1
≤
∞
X
ηn = n=1
∞
X
1/2n = n=1
• on the complement of F , fn converges to f uniformly: indeed, for all η > 0, since limn ηn = 0, there
exists n0 such that for all ηn0 ≤ η, and we have for all Fηn0 ⊆ F , i.e. F c ⊆ Fηn0 , therefore for all
x ∈ F c , x ∈ Fηn0 , thus for all n ≥ Nηn0 , |fn (x) − f (x)| ≤ η. This proves that (fn ) converges to f
uniformly on F c .
4
(12.5) Let (X, A, µ) be a measure space, such that µ(X) < ∞. Then the dominated convergence theorem
follows from Egoroff’s theorem:
Let E ∈ A, and let (fn ) be a sequence of measurable functions that converges pointwise to f ,R and such
that there exists an integrable g with |fn (x)| ≤ g(x) for all x ∈ X. Then f is integrable, and E fn dµ n
R
converges to E f dµ.
proof
Lemma 1 If f R: X → R is integrable, then for all > 0, there exists η > 0 such that if F ∈ A has measure
µ(F ) ≤ η, then F hdµ ≤ .
First, we have f is measurable as the pointwise limit of the sequence of measurable functions (fn ). Then
we have for all x, |f (x)| = limn→∞ |fn (x)| ≤ g(x), thus
Z
Z
|f |dµ ≤
gdµ < ∞
X
X
therefore f is integrable.
Since g is integrable,
by Lemma 1, there exists η > 0 such that for every measurable set F ∈ A with
R
measure m(F ) ≤ η, F gdµ ≤ /3.
Applying Egoroff’s theorem to the sequence of functions (fn ) that converges pointwise to f , there exists
a measurable set F such that µ(F ) ≤ η and (fn ) converges uniformly to f on F c . Consider the constant
(assuming that µ(X) > 0, otherwise the result is trivial), then there exists N ∈ N such that for
c = 3µ(X)
all n ≥ N , for all x ∈ F c , |fn (x) − f (x)| ≤ c. Thus for all n ≥ N
Z
Z
Z
|
fn dµ −
f dµ| = | (fn − f )dµ|
E
E
Z E
≤
|fn − f |dµ
ZE
Z
=
|fn − f |dµ +
|fn − f |dµ
E∩F c
E∩F
To bound the first term in the sum, we have
Z
Z
|fn | + |f |dµ ≤ 2
E∩F
gdµ ≤ 2
E∩F
3
since µ(E ∩ F ) ≤ µ(F ) ≤ η. To bound the second term, we have
Z
Z
|fn − f |dµ ≤
cdµ ≤ cµ(E ∩ F c ) ≤ cµ(X) =
3
c
c
E∩F
E∩F
Adding the two terms, we obtain
Z
|
Z
fn dµ −
E
which proves that (∈E fn dµ) converges to
R
E
f dµ| ≤ E
f dµ.
5
(12.6) Let (X, A, µ) be a measure space such that µ(X) is finite. Let fn : X → [0, ∞) be a measurable
function, and suppose that (fn ) converges to f almost everywhere. Then
R
R
R
R
R
1. if X f dµ < ∞ and limn→∞ X fn dµ = X f dµ, then limn→∞ E fn dµ = E f dµ for every measurable
set E.
R
2. this is not true in general if X f dµ is not finite.
3. this is not true in general if fn : X → R instead of R+ , and assumed to be integrable.
R
4. under the assumptions of 1, we additionally have X |fn − f |dµ → 0
proof
Let A = {x ∈ X|fn (x) → f (x)}. Then A is measurable, and by assumption, m(Ac ) = 0.
1. Let E ∈ A be a measurable set. Since (fn ) is a sequence of nonegative measurable functions, we have
by Fatou’s Lemma
Z
Z
lim inf fn dµ ≤ lim inf
fn
n
n
E
E
R
R
but E lim inf fn dµ = E f dµ since the difference
Z
Z
Z
(lim inf fn − f )dµ =
(lim inf fn − f )dµ +
(lim inf fn − f )dµ = 0 + 0
E
n
n
A∩E
Ac ∩E
n
where the first term is zero since for all x ∈ A, lim inf n fn (x) = limn fn (x) = f (x), and the second
term is zero since m(Ac ∩ E) ≤ m(Ac ) = 0. Therefore we have
Z
Z
f dµ ≤ lim inf
fn
n
E
E
Similarly, we have
Z
Z
f dµ ≤ lim inf
n
Ec
Now we have for all k ∈ N
Z
Z
Z
sup
fn dµ = sup
fn dµ −
fn dµ
n≥k E
n≥k
X
Ec
Z
Z
≤ sup
fn dµ + sup −
fn dµ
n≥k X
n≥k
Ec
Z
Z
= sup
fn dµ − inf
fn dµ
n≥k
X
n≥k
fn
(1)
Ec
since X = E ∪ E c disjointly
Ec
since both sequences converge, so does the sum of sequences, and we have
Z
Z
Z
lim sup
fn dµ ≤ lim sup
fn dµ − lim inf
fn dµ
n
c
n
n
E
ZX
ZE
= lim sup
fn dµ − lim inf
fn dµ
n
n
X
Ec
Z
Z
Z
Z
≤
f dµ − lim inf
fn dµ
since
fn dµ
→
f dµ
n
Ec
X
X
n
ZX
Z
≤
f dµ −
lim inf fn dµ
by (1)
n
X
Ec
Z
Z
Z
=
f dµ +
f dµ −
lim inf fn dµ
n
Ec
Ec
ZE
=
f dµ
E
6
Therefore we have
Z
Z
n
Z
fn dµ ≤
lim sup
E
n
E
which proves that all terms are equal, thus
E
Z
n→∞
R
X
fn dµ ≤ lim sup
n
E
fn dµ n converges, and
R
Z
lim
2. Counterexample with
Z
f dµ ≤ lim inf
fn dµ =
E
f dµ
E
f dµ = ∞: consider the sequence of functions
fn : R+ → R+


n(1 − nx)
x 7→ fn (x) = 0


1
if 0 ≤ x ≤ 1/n
if 1/n < x ≤ 1
otherwise
then we have
• (fn ) converges pointwise to
f : R+ → R+


+∞ if x = 0
x 7→ f (x) = 0
if 0 < x ≤ 1


1
otherwise
R
R
• for all n ∈ N, R+ fn dm = R+ f dm = ∞.
R
R
• however, [0,1] fn dm = 1/2 for all n, while [0,1] f dm = 0.
3. Counterexample with fn : X → R instead of R+ , and integrable.
fn : R → R
(
x/n2
x 7→ fn (x) =
0
if − n ≤ x ≤ n
if 1/n < x ≤ 1
then we have
• for all n, fn is integrable and
R
R
|fn |dm = 1
• (fn ) converges pointwise to the function f identically zero on R
R
R
• for all n ∈ N, R fn dm = R f dm = 0.
R
R
• however, R+ fn dm = 1/2 for all n, while R+ f dm = 0.
4. Let B be the measurable set of points where f is nonzero,
B = {x ∈ X|f (x) > 0}
We have
Z
Z
|f − fn |dµ =
X
Z
|f − fn |dµ
fn dµ +
Bc
7
B
fn dµ
E
In order to apply Egoroff’s theorem, we need a measure space with finite measure. Define the measure
ν by
ν : A → R+
Z
E 7→ ν(E) =
f dµ
E
since f is measurable and non-negative, ν is a measure, and we have
Z
ν(X) =
f dµ < ∞
X
We assume ν(X) > 0, otherwise f is zero almost everywhere, and the result becomes trivial. Now
consider the functions
gn : X → R+
(
fn (x)
if x ∈ B
x 7→ gn (x) = f (x)
0
otherwise
We have
• gn is a measurable function
• gn converges pointwise to 1B on A (indeed, if x ∈ A ∩ B c , then gn (x) = 1B
R (x) = 0, and if
x ∈ A ∩ B, then (fn (x)) → f (x) > 0, thus (gn (x)) → 1 = 1B (x)), and ν(Ac ) = Ac f dµ = 0 since
µ(Ac ) = 0. Therefore (gn ) → 1B almost everywhere.
We can now apply Egoroff’sR theorem on (gn ). Let > 0. By Egoroff’s theorem, there exists F ∈ A
such that ν(F ) ≤ /4 (i.e. F f dµ ≤ /4) and (gn ) converges uniformly to 1B on F c . Consider the
constant c = 4ν(X)
(note that this is well defined since we assumed ν(X) > 0). Then there exists N1
such that for all n ≥ N1 , for all x ∈ F c , |gn (X) − 1B (x)| ≤ c. In particular,
∀x ∈ B ∩ F c , |fn (x) − f (x)| ≤ cf (x)
We then have
Z
Z
Z
|f − fn |dµ =
X
fn +
ZB
c
≤
ZB∩F
fn +
Bc
Z
|f − fn |dµ +
|f − fn |dµ
B∩F c
Z
Z
f dµ +
fn dµ +
|f − fn |dµ
B∩F
B∩F
B∩F c
R
R
since
f dµ converges
to E f dµ for Rall E ∈ A (by 1), there
exists N2 such that for all n ≥ N2 ,
E n
R
R
R
f ≤ /4 + B c f dµ = /4 + 0 and B∩F fn dµ ≤ /4 + B∩F f dµ ≤ /4 + /4. Therefore we have
Bc n
for all n ≥ max(N1 , N2 ),
Z
Z
|f − fn |dµ ≤ 3/4 +
|f − fn |dµ
X
B∩F c
Z
≤ 3/4 + c
f dµ
c
ZB∩F
≤ 3/4 + c
f dµ
X
= 3/4 + /4
which proves
R
X
|f − fn |dµ converges to 0.
8
Rn
(1 − nx )n ex/2 dx = 2.
(12.7)
limn→∞
proof
For all n ∈ N, let
0
fn : R+ → R
x 7→ f (x) = (1 − x/n)n ex/2 1[0,n]
We seek to compute limn→∞
R
R+
fn (x)dx. We first observe that for all n ∈ N, fn (x) ≤ fn+1 (x).1 Let x ≥ 0.
Then limn→∞ fn (x) = e−x/2 , since for all n > x, fn (x) = (1 − x/n)n ex/2 = exp(n log(1 − x/n) + x/2) =
exp(n(−x/n + o(n)) + x/2) = exp(−x/2 + o(1)). Therefore for all n ∈ N, for all x ≥ 0
0 ≤ fn (x) ≤ e−x/2
where x 7→ e−x/2 is integrable on R+ . By the dominated convergence theorem, limn→∞
and
Z n
Z
x n x/2
(1 − ) e dx = lim
lim
fn (x)dx
n→∞ 0
n→∞ R
n
+
Z
e−x/2 dx
=
R+
=2
1f
n (x)
≤ fn+1 (x): this is true for x > n. To prove this for x ∈ [0, n), consider
h : [0, n) → R
x 7→ h(x)
= log(fn+1 (x)/fn (x))
x
x
= (n + 1) log(1 − ) − n log(1 − )
n
n
This is well defined since fn+1 (x)/fn (x) > 0. Then we have for all x ∈ [0, n)
h0 (x) = −
1
log(1 −
x
)
n+1
+
1
log(1 −
x
)
n
>0
therefore h is increasing, and for all x ∈ [0, n), h(x) ≥ h(0) = 0, therefore fn+1 (x)/fn (x) ≥ 1.
9
R
R+
fn (x)dx exists