PROBLEM SET FOR LECTURES 23 AND 24
P∞
1) Show that if aj ≥ 0 for j ∈ N and j=1 aj < M for some M > 0, then given
P∞
ε > 0 there exists some n ∈ N such that j=n aj < ε.
2) Show replacing “open” with “open or closed” in the definition of measure zero
doesn’t change anything (i.e. a set is of measure zero with respect to one definition if and only if it is of measure zero with respect to the other).
3) Observe if E is a set of measure zero then any F ⊆ E is also of measure zero.
Prove the following converse: if E ⊆ R is such that every proper subset F ⊂ E
is of measure zero, then E is of measure zero.
4) Show that the collection of sets of measure zero is closed under reflections,
translations and dilations. In other words, given a set E ⊂ R of measure zero
and a ∈ R, λ > 0 show −E := {−x : x ∈ E}, E + a := {x + a : x ∈ E} and
{λx : x ∈ E} are all of measure zero.
5) Let a < b and
S suppose I is a collection of open intervals which cover [a, b]; that
is, [a, b] ⊆ I∈I I. Show there exists some finite sub-collection J ⊆ I which
also covers [a, b] using the following methods.
a) Consider the set
X := {z ∈ [a, b] : there exists some finite J ⊆ I which covers [a, z]}
and show sup X = b. From this deduced the desired result.
b) Give an alternative proof using the nested interval theorem.
The property we have proved here is known as compactness.
6) By using the result of Q 5), give alternative proofs of the following familiar
theorems.
a) If f : [a, b] → R is continuous, then it is bounded.
b) If f : [a, b] → R is continuous, then it is absolutely continuous.
Thus, we see that the fact [a, b] is compact underlies many of the key properties
of continuous functions on closed, bounded intervals we proved in the previous
quarter.
7) a) Recall that a function f : [a, b] → R is Lipschitz continuous if there exists
some M > 0 such that
|f (x) − f (y)| ≤ M |x − y|
for all x, y ∈ [a, b].
Show that if E ⊆ [0, 1] is of measure zero, then f (E) is of measure zero.
b) A function f : [a, b] → R is absolutely continuous if for every > 0, there
exists some δ > 0 such that whenever (a1 , b1 ), . . . , (an , bn ) are disjoint subintervals in [a, b] such that
n
X
bj − aj < δ
j=1
it follows
n
X
|f (bj ) − f (aj )| < .
j=1
Show that if f : [a, b] → R is Lipschitz, then√it is absolutely continuous.
c) Show that f : [0, 1] → R given by f (x) = x is absolutely continuous but
not Lipschitz. (Hint: the problem of showing the function is absolutely
1
2
PROBLEM SET FOR LECTURES 23 AND 24
continuous is significantly simplified once one realises that without loss of
generality one may choose the intervals to all lie next to one another).
d) If f : [a, b] → R is integrable, show that F : [a, b] → R given by
Z x
f
F (x) :=
a
is absolutely continuous.
e) Show if f : [a, b] → R is absolutely continuous and E ⊆ R is of measure zero,
then f (E) is of measure zero.
f) ** (This problem is stated to give context to the other parts of the question).
Find a counter-example to the following statement: “If f : [0, 1] → [0, 1] is
continuous and E ⊆ [0, 1] is of measure zero, then f (E) is of measure zero.”
8) a) Formulate a definition of measure zero for subsets of R2 using: i) open balls
and ii) open rectangles.
b) Show that it doesn’t matter which definition you choose (i.e. a set is of
measure zero with respect to one definition if and only if it is of measure zero
with respect to the other).
c) Show any line in R2 is of measure zero.
d) Let f : [a, b] → R be differentiable with continuous derivative. Show that the
graph
Γ := {(x, f (x)) : x ∈ [a, b]} ⊆ R2
is of measure zero.
9) a) * (For those of you who know a bit about convergence of series.) There is
an analogue for bases different from 10 of the usual decimal expansion of a
number. If b is an integer larger than 1 and 0 < x <P
1, show that there exist
∞
integral coefficients ck , 0 ≤ ck < b, such that x = k=1 ck b−k . Show that
−k
this expansion is unique unless x = cb for some integer 0 ≤ c < b, in which
there are precisely two such expansions.
b) When b = 3 in ii) the expansion is called the triadic or ternary expansion of
x. Show that the Cantor set C consists of all x such that x has some triadic
expansion for which every ck is either 0 or 2.
c) Use part b) to give a different proof of the fact the Cantor set is uncountable.
d) * (This problem uses the definition of a perfect set, which is given in the
notes. It is intended for enthusiasts.) Construct a subset of [0, 1] in the
same manner as the Cantor set by removing from each remaining interval a
subinterval of relative length θ, 0 < θ < 1. Show that the resulting set is
perfect and has measure zero.
e) Construct a Cantor-type subset of [0, 1] by removing from each interval remaining at the kth stage a subinterval of relative length θk ,P0 < θk < 1.
∞
Show that the resulting set has measure zero if and onlyPif k=1 θk = ∞
n
(that is, for every M > 0 there exists an n ∈ N such that k=1 θk > M ).
10) Let x ∈ R and N ∈ N. We consider the problem of approximating x by a
rational. In particular, we’d like to show that we can always find a rational
with small denominator which is close to x.
(a) (Trivial result) Let q ∈ N. As a warm up, show that there exists some
p ∈ Z such that
x − p < 1 .
q
q
(b) The previous trivial result can be significantly strengthened. Let N ∈ N be
given and now suppose x ∈ R \ Q. Show that there exists some p ∈ Z and
1 ≤ q ≤ N such that
x − p < 1 .
q
qN
PROBLEM SET FOR LECTURES 23 AND 24
3
To do this, first show use a simple proof by contradiction argument to show
there exists some 1 ≤ j ≤ N such that the interval [(j −1)/N, j/N ) contains
at least two elements of the set {kx − bkxc : 0 ≤ k ≤ N }.
(c) Conclude that for all x ∈ R \ Q there exists a infinitely many rationals p/q
such that
x − p < 1 .
q
q2
11) ** For almost all numbers one cannot improve the result of part c) of the previous
question. In particular, given ρ > 0 let Aρ denote the set of all x ∈ R for which
there exist infinitely many rationals p/q such that
x − p < 1 .
q
q 2+ρ
Show Aρ has measure zero as follows:
a) Show that the problem can be reduced to showing Aρ ∩ [−N, N ] has measure
zero for all N ∈ N.
b) Fix N ∈ N. For each q ∈ N and −N q ≤ p ≤ N q define an open interval Ip,q
with µ(Ip,q ) = 2q −(2+ρ) so that for each q0 ∈ N the collection
{Ip,q : q > q0 and − N q ≤ p ≤ N q}
covers Aρ ∩ [−N, N ].
c) If the intervals in part b) are chosen correctly, then one may use Q1) of this
sheet to conclude the proof.
Jonathan Hickman, Department of mathematics, University of Chicago, 5734 S. University Avenue, Eckhart hall Room 414, Chicago, Illinois, 60637.
E-mail address: [email protected]