Solutions to Homework #11
Problems:
1. Let A1 , A2 , B1 and B2 be subsets of the same set. Prove that
(a) (A1 ∪ A2 )4(B1 ∪ B2 ) ⊆ (A1 4B1 ) ∪ (A2 4B2 )
(b) (A1 ∩ A2 )4(B1 ∩ B2 ) ⊆ (A1 4B1 ) ∪ (A2 4B2 )
Recall that both formulas were used in verification of properties of Lebesgue
measure.
Solution: Let X denote the “big” set (which contains A1 , A2 , B1 and
B2 ). For simplicity of notation, we set Y c = X \ Y for every subset Y of X.
Note that for any subsets Y and Z of X we have (Y ∩ Z)c = Y c ∪ Z c ,
(Y ∪ Z)c = Y c ∩ Z c , Y \ Z = Y ∩ Z c and Y 4Z = (Y ∩ Z c ) ∪ (Y c ∩ Z).
Therefore,
(A1 ∪ A2 )4(B1 ∪ B2 ) = ((A1 ∪ A2 ) ∩ (B1 ∪ B2 )c ) ∪ ((A1 ∪ A2 )c ∩ (B1 ∪ B2 ))
= ((A1 ∪ A2 ) ∩ (B1c ∩ B2c )) ∪ ((Ac1 ∩ Ac2 ) ∩ (B1 ∪ B2 ))
= (A1 ∩ B1c ∩ B2c ) ∪ (A2 ∩ B1c ∩ B2c ) ∪ (Ac1 ∩ Ac2 ∩ B1 ) ∪ (Ac1 ∩ Ac2 ∩ B2 )
⊆ (A1 ∩ B1c ) ∪ (A2 ∩ B2c ) ∪ (Ac1 ∩ B1 ) ∪ (Ac2 ∩ B2 )
= ((A1 ∩B1c )∪(Ac1 ∩B1 ))∪((A2 ∩B2c )∪(Ac2 ∩B2 )) = (A1 4B1 )∪(A2 4B2 ),
which proves (a).
For (b) observe that for any Y, Z we have
(Y c 4Z c ) = (Y c ∩ (Z c )c ) ∪ ((Y c )c ∩ Z c ) = (Y c ∩ Z) ∪ (Y ∩ Z c ) = Y 4Z.
Hence (A1 ∩A2 )4(B1 ∩B2 ) = (A1 ∩A2 )c 4(B1 ∩B2 )c = (Ac1 ∪Ac2 )4(B1c ∪B2c ).
By (a), (Ac1 ∪ Ac2 )4(B1c ∪ B2c ) ⊆ (Ac1 4B1c ) ∪ (Ac2 4B2 )c = (A1 4B1 ) ∪
(A2 4B2 ). Combining the two inclusions, we obtain the assertion of (b).
2. Problem 1 from Kolmogorov-Fomin (p. 268). Note that this is a
problem about subsets of R2 (not R).
Solution: (a) We are asked to prove that every open subset of E = [0, 1]2
is measurable. Note that since E is NOT open in R2 , an open subset of E
is not the same as an open subset of R2 contained in E (for instance, E
satisfies the former condition but not the latter). However, we do know that
every open subset of E is of the form A ∩ E where A is open in R2 . Since
E is measurable and intersection of two measurable sets is measurable, it
suffices to show that every open subset of R2 is measurable.
1
2
So let A ⊆ R2 be an open set. We claim that A is a countable union
of rectangles (one can require that those rectangles are squares, but this
is not necessary for this problem). Since rectangles are elementary (hence
measurable) and countable unions of measurable sets are measurable (KF,
Theorem 9 on p.264), this would imply that A is measurable.
Take any point P ∈ A. Since A is open, there exists ε = ε(P ) > 0
such that Nε (P ), the open disk of radius ε centered at P (in the Euclidean
metric), is contained in A. Let δ = √ε2 . Clearly, if P = (a, b), then the open
square (a − δ, a + δ) × (b − δ, b + δ) is contained in Nε (P ) and hence also
in A. Since Q is dense in R, we can find rational numbers q1 , q2 , q3 , q4 such
that a − δ < q1 < a < q2 < a + δ and b − δ < q3 < b < q4 < b + δ, and let
RP be the open rectangle (q1 , q2 ) × (q3 , q4 ).
We claim that A = ∪P ∈A RP . Indeed, by construction each RP is contained in A, so ∪P ∈A RP ⊆ A. On the other hand, since P ∈ RP , we have
∪P ∈A RP ⊇ A. Note that each RP has rational coordinates (of its vertices),
and clearly there are countably many of rectangles with rational coordinates
(there is a natural bijection between the set of such rectangles and a subset
of Q4 = Q × Q × Q × Q and Q × Q × Q × Q is countable being a Cartesian
product of finitely many countable sets).
Thus, the collection {RP : P ∈ A} contains only countably many distinct
subsets, so A is a countable union of open rectangles.
(b) This follows from (a) since the complement of a closed set is open,
and complements of measurable sets are measurable (Corollary on p. 263 of
KF).
(c) follows from (a), (b) and Theorem 9 (p. 264) and Corollary (p. 263)
of KF.
3.
(a) Let A be a countable subset of [0, 1]. Prove that A has measure zero
(that is, A is measurable and µ(A) = 0).
(b) Prove that the Cantor set C has measure 0.
Solution: (a) First note that any point (that is, one element set) has measure zero since we can think of {a} as a closed interval [a, a]. If A is countable, then it is a countable disjoint union of its points (A = ta∈A {a}), so by
P
countable additivity of µ (Theorem 24.1(d)) we have µ(A) = a∈A µ({a}) =
0.
(b) We present two solutions. In both solutions we set E = [0, 1].
3
For the first solution, observe that proving that C has measure 0 is equivalent to proving that its complement E \ C has measure 1. Since by construction, E \ C is a (countable) disjoint union of intervals, it is measurable,
and its measure is equal to the sum of lengths of those intervals. By definition, E \ C is a disjoint union of 1 interval of length 31 , 2 intervals of length
4
2
n
9 , 4 intervals of length 27 etc. (for each n ∈ Z≥0 we have 2 intervals of
1
). Therefore,
length 3n+1
∞
∞ X
2n
1 X 2 n 1
1
µ(E \ C) =
= ·
= ·
= 1.
n+1
3
3
3
3 1 − 2/3
n=0
n=0
For the second solution, define the sets E0 = [0, 1] ⊃ E1 ⊃ E2 ⊃ . . .
as on page 41 of Rudin (so that C = ∩En ). By construction, each En is
a finite union of intervals of the same length, and En+1 is obtained from
En by removing the middle third of each interval of En . Thus, each En
is measurable and µ(En+1 ) = 23 µ(En ). Since µ(E0 ) = 1, we conclude that
µ(En ) = ( 23 )n .
The Cantor set C is measurable as a countable intersection of measurable
sets; moreover, C ⊂ En for each n, so µ(C) ≤ µ(En ) = ( 32 )n for each n.
Since ( 32 )n → 0 as n → ∞, we conclude that µ(C) = 0.
4. Let A be a subset of [0, 1], and suppose that for every ε > 0 there exists
a measurable set B such that µ∗ (A4B) < ε. Prove that A is measurable.
Solution: Take any ε > 0. By assumption there exists a measurable
set B such that µ∗ (A4B) < 2ε . Since B is measurable, there exists an
elementary set C such that µ∗ (B4C) < 2ε .
By formula (25) on page 306 in Rudin we have A4C ⊆ (A4B) ∪ (B4C).
Since µ∗ (X ∪ Y ) ≤ µ∗ (X) + µ∗ (Y ) for any two sets X and Y , we have
µ∗ (A4C) ≤ µ∗ (A4B) + µ∗ (B4C) < 2 · 2ε = ε. Since ε > 0 was arbitrary,
A is measurable by definition.
5. The main goal of this problem is to justify the definition of the Cantor
staircase function given in class.
Let a ≤ b be real numbers, let S be a subset of [a, b] which contains both
a and b, and let f : S → R be an increasing function (note: f is only defined
on S).
(a) Prove that f can be extended to an increasing function from [a, b]
to R, that is, there exists an increasing function F : [a, b] → R such
that F (x) = f (x) for all x ∈ S.
4
(b) Now assume that f (S) is a dense subset of [f (a), f (b)]. Prove that
there exists unique function F satisfying the conclusion of (a). Then
prove that this F is continuous.
Solution: (a) For each x ∈ [a, b] let Ax = {f (s) : s ∈ S and s ≤ x}, and
define F (x) = sup Ax . Note that Ax is always non-empty (since f (a) ∈ Sx )
and Ax is bounded from above since f (s) ≤ f (b) for all s ∈ S (since f is
increasing). Therefore, F (x) is indeed well defined.
Clearly, if x < y, then Ax ⊆ Ay , whence F (x) ≤ F (y), so F is increasing
(note that this has nothing to do with f being increasing). Finally, if x ∈ S,
then f (x) is the maximum element of Ax (this time because f is increasing),
so F (x) = f (x) for all x ∈ S.
(b) The existence of F is proved in (a). Suppose that there exist two
distinct functions F and F 0 satisfying the conditions of (a). Then there exists
x ∈ [a, b] with F (x) 6= F 0 (x); WOLOG assume that F (x) < F 0 (x). Note that
since both F and F 0 are increasing, we have f (a) = F (a) ≤ F (x) ≤ F (b) =
f (b), so F (x) ∈ [f (a), f (b)] and similarly F 0 (x) ∈ [f (a), f (b)]. Therefore,
(F (x), F 0 (x)) is a non-empty open subset of [f (a), f (b)].
Since f (S) is dense in [f (a), f (b)], it must intersect (F (x), F 0 (x)) nontrivially, so there exists s ∈ S such that F (x) < f (s) < F 0 (x). Since s ∈ S,
we have f (s) = F (s) = F 0 (s), so F (x) < F (s) and F 0 (s) < F 0 (x). Since F
and F 0 are both increasing, the first inequality implies that x < s, and the
second inequality implies that s < x, which is impossible.
Finally, since f (S) is dense in [f (a), f (b)] = [F (a), F (b)] and F ([a, b]) contains F (S) = f (S), it follows that F ([a, b]) is dense in [F (a), F (b)]. Therefore, F is continuous by Problem 4 in HW#8.
Cantor staircase function. We now use Problem 5 to justify the definition of the Cantor staircase function given in class. Recall that in class
we defined the function f : [0, 1] \ C → 1 (where C is the Cantor set) by
f (x) = 21 if 13 < x < 23 , f (x) = 41 if 19 < x < 29 , f (x) = 34 if 79 < x < 89 ,
1
2
7
8
f (x) = 18 if 27
< x < 27
, f (x) = 83 if 27
< x < 27
etc.
Note that f is only defined on [0, 1] \ C, the complement of C. It is
intuitively clear that f is increasing on [0, 1] \ C, but it takes a bit of work
to prove this formally (we skip this step here).
In class we claimed that f can be extended to a unique continuous increasing function F : [0, 1] → [0, 1] (and called this F the Cantor staircase
5
function). We can now justify this claim by applying the result of Problem 5(b). Let S = ([0, 1] \ C) ∪ {0, 1} and extend f to S by setting f (0) = 0
and f (1) = 1. To apply Problem 5(b) we only need to show that f (S) is
dense in [0, 1].
Let B = f (S). It is clear from definition that B consists of all rational
numbers in [0, 1] which have the form 2mn for some m, n ∈ Z≥0 . To show
that B is dense in [0, 1], it suffices to show that B is an ε-net in [0, 1] for
each ε > 0. Take any x ∈ [0, 1] and any ε > 0. Choose n ∈ N such that
2n ε > 1. Let m = [2n x], the greatest integer ≤ x. Then clearly 0 ≤ m ≤ 2n
and |2n x − m| < 1. Diving both inequalities by 2n , we get that 0 ≤ 2mn ≤ 1
(so 2mn ∈ B) and |x − 2mn | < 21n < ε. Therefore, B is indeed an ε-net in [0, 1].
Finally, for the graph of F see
http://en.wikipedia.org/wiki/Cantor_function
6 (bonus) Let F : [0, 1] → [0, 1] be the Cantor staircase function and
C the Cantor set. The argument from Lecture 24 shows that µ(F (C)) = 1
(while µ(C) = 0). Modify the construction of F to show that for every ε > 0
there exists a strictly increasing continuous function Fε : [0, 1] → [0, 1] such
that µ(Fε (C)) > 1 − ε.
Solution: First we introduce some terminology. Let I ⊆ [0, 1] be an
interval and ε > 0 a real number s.t. ε ≤ length(I). Define the ε-center
of I, denoted Cε (I) to be the open interval of length ε which has the same
center as I; more explicitly if I = (a, b), (a, b], [a, b) or [a, b], we set Cε (I) =
a+b+ε
( a+b−ε
2 ,
2 ).
Next we define a family of subsets of [0, 1] which we will call Cantor-type
sets. There will be exactly one such set corresponding to every sequence of
P∞ n−1
εn ≤ 1. THE Cantor set C will be
positive reals {εn }∞
n=1 with
n=1 2
the Cantor-type set corresponding to the sequence εn = ( 13 )n .
P∞ n−1
So let {εn }∞
εn ≤ 1 and εn > 0 for all n.
n=1 be a sequence with
n=1 2
Start with [0, 1] and remove its ε1 -center, call this set D1 . Clearly D1 is a
disjoint union of two intervals of the same length. Remove the ε2 -center of
each of those intervals and call the new set D2 . Continuing indefinitely, we
obtain a descending chain D1 ⊇ D2 ⊇ . . . of subsets of [0, 1]. The intersection
of all those sets ∩∞
n=1 Dn will be called the Cantor-type set associated to the
sequence {εn }.
We need to make sure that each step in the above procedure is well defined
in the sense that we are not going “out of bounds” by trying to remove
6
P
n−1 ε ≤ 1 comes into play.
too much. This is where the condition ∞
n
n=1 2
Indeed, suppose we have constructed Dn for some n. It is clear that Dn is a
disjoint union of 2n intervals of the same length. By construction [0, 1]\Dn is
P
a union of intervalsPof total length nk=1 2k−1 εk , so each of the 2n intervals in
Dn has length
1−
n
k−1 ε
k
k=1 2
2n
. This means that we can remove εn+1 -center
1−
Pn
2k−1 ε
k
k=1
from each of those intervals precisely when εn+1 ≤
. This
2n
Pn+1 k−1
inequality is equivalent to k=1 2 εk ≤ 1, and the latter holds by our
hypothesis.
Lemma: Let D be a Cantor type set. Then [0, 1] \ D is dense in [0, 1].
Proof: Suppose, by way of contradiction, that [0, 1] \ D is not dense in
[0, 1]. Then there is a non-empty open interval (a, b) ⊂ [0, 1] such that
(a, b) ∩ ([0, 1] \ D) = ∅ and therefore (a, b) ⊆ D. In particular, (a, b) ⊆ Dn
for each n (where Dn are defined as above). But by construction Dn is a
union of intervals of length less than 21n , so (a, b) ⊆ Dn implies b − a < 21n .
Since a < b, this cannot be true for all n, so we reached a contradiction. We can now construct a function with required properties. So take any
P
n−1 ε = ε < 1, so
ε > 0 (WOLOG ε < 1) and define εn = 4εn . Then ∞
n
n=1 2
2
the sequence εn satisfies the necessary condition.
Let C be the Cantor set and let D be the Cantor-type set corresponding
to {εn }. Note that each of the sets [0, 1] \ C and [0, 1] \ D is a countable
disjoint union of open intervals, and there is a natural bijection between
those sets of intervals which preserves the relative position of those intervals.
Let f : [0, 1] \ C → [0, 1] be the unique function which sends each interval
in [0, 1] \ C to the corresponding interval in [0, 1] \ D and is linear on each
interval.
Clearly, f is strictly increasing on [0, 1] \ C and f ([0, 1] \ C) = [0, 1] \ D.
By the above Lemma the set [0, 1] \ D is dense in [0, 1], hence by Problem 5
there exists a unique continuous increasing function F : [0, 1] → [0, 1] such
that F (x) = f (x) for all x ∈ [0, 1] \ C.
Since F ([0, 1]) = [0, 1] and F ([0, 1] \ C) = [0, 1] \ D, we must have F (C) ⊇
P
n−1 ε = 1− ε > 1−ε.
D, so µ(F (C)) ≥ µ(D) = 1−µ([0, 1]\D) = 1− ∞
n
n=1 2
2
It remains to show that F is strictly increasing (so far we only know that
F is increasing on [0, 1] and strictly increasing on [0, 1] \ C). Suppose that
F is not strictly increasing on [0, 1]. Then (since F is increasing) there exist
a < b in [0, 1] such that F (a) = F (b), so F must be constant on [a, b]. Since
[0, 1] \ C is dense in [0, 1] (again by the above lemma), [a, b] must contain
7
infinitely many points from [0, 1] \ C; in particular, there exist two distinct
points u, v ∈ [a, b] ∩ ([0, 1] \ C). But then F (u) = F (v), which contradicts
the fact that F is strictly increasing on [0, 1] \ C.
7. Problem 7 from Kolmogorov-Fomin (p. 268). Note that the hint given
in KF is essentially a sketch of the solutions. The things you need to justify
are
(a) C = ∪∞
n=−∞ Φn
(b) Φn ∩ Φm = ∅ if n 6= m
(c) Assume that Φ0 is measurable. Then each Φn is measurable and
µ(Φn ) = µ(Φ0 ) for all n ∈ Z
(d) the conclusion of (c) contradicts (33) in KF.
Remark: The Lebesgue measure on the circle C can be defined in exactly
the same way as on [0, 1] with the exception that we call a subset of C
elementary if it is a finite union of arcs.
Solution: We start by introducing some additional notations and slightly
rephrase the definition of the sets Φn .
Let R : C → C be the counterclockwise rotation by the angle 2πα (where
α is a fixed irrational number). Define the relation ∼ on C by x ∼ y ⇐⇒
there exists n ∈ Z such that y = Rn (x) (where Rn = R applied n times
= counterclockwise rotation by 2πnα). It is straightforward to check that
∼ is an equivalence relation and that every equivalence class with respect
to ∼ is countable (the latter holds since α is irrational; if α were rational,
equivalence classes would be finite).
Now let Φ0 be any subset of C which contains precisely one element from
each equivalence class, and for each n ∈ Z define Φn = Rn Φ0 = {Rn (x) :
x ∈ Φ0 }. We now proceed with verifying properties (a)-(d) above.
(a) By construction each Φn lies in C, so ∪∞
n=−∞ Φn ⊆ C. To prove the
reverse inclusion, take any x ∈ C. By definition of Φ0 there exists y ∈ Φ0
such that y ∼ x, so x = Rn (y) for some n ∈ Z. But then x ∈ Rn Φ0 = Φn as
desired.
(b) Suppose, by way of contradiction, that there exists some x ∈ C which
lies in Φn ∩ Φm for some n 6= m. Then there exist y, z ∈ Φ0 such that
x = Rn (y) = Rm (z). Hence y = Rn−m (z), so y ∼ z. Since both y and z lie
in Φ0 and since Φ0 contains only one element from each equivalence class,
we must have y = z. Hence y = Rn−m (y), which means that Rn−m must be
the trivial rotation. Since Rn−m is the rotation by the angle 2πα(n − m),
8
it follows that α(n − m) ∈ Z, which contradicts the assumption that α is
irrational.
Next we prove a general lemma:
Lemma: Let f : C → C be a rotation and A any subset of C. Then
(i) µ∗ (f (A)) = µ∗ (A)
(ii) If A is measurable, then f (A) is also measurable and therefore µ(f (A)) =
µ(A) by (a)
Proof of Lemma: (i) By definition,
(∞
)
X
∗
∞
µ (A) = inf
length(In ) where {In }n=1 is a countable cover of A by arcs .
n=1
∞
Note that if {In }∞
n=1 is any such cover of A, then {f (In )}n=1 is a countable
cover of f (A) by arcs. Moreover length(f (In )) = length(In ) since f is
P∞
P
a rotation, so ∞
n=1 length(In ), and it follows that
n=1 length(f (In )) =
∗
∗
µ (f (A)) ≤ µ (A).
Since the last inequality holds for any subset A and any rotation f and
since f −1 is also a rotation, it remains true if we replace A by f (A) and f by
f −1 . This way we get µ∗ (f −1 (f (A))) ≤ µ∗ (f (A)), that is, µ∗ (A) ≤ µ∗ (f (A)).
Thus, µ∗ (f (A)) ≤ µ∗ (A) and µ∗ (A) ≤ µ∗ (f (A)), so µ∗ (A) = µ∗ (f (A)).
(ii) Let ε > 0. Since A is measurable, there exists an elementary set B
such that µ∗ (A4B) < ε. By (i) we have µ∗ (f (A4B)) < ε. Since f is a
bijection, it is clear that f (A4B) = f (A)4f (B), so µ∗ (f (A)4f (B)) < ε.
Finally note that since B is a finite disjoint union of arcs, f (B) is also a
finite disjoint union of arcs, so f (B) is elementary and hence the previous
inequality implies that f (A) is measurable. (c) and (d) Assume that Φ0 is measurable. Then by the Lemma each Φn
is measurable and µ(Φn ) = µ(Φ0 ) for all n.
Since C = tn∈Z Φn , by countable additivity of µ we have
n=∞
n=∞
X
X
1 = µ(C) =
µ(Φn ) =
µ(Φ0 ).
n=−∞
This is impossible since if µ(Φ0 ) = 0, then
n=−∞
n=∞
P
n=−∞
then
n=∞
P
n=−∞
µ(Φ0 ) diverges.
µ(Φ0 ) = 0 and if µ(Φ0 ) > 0,