Problem 1
Part a
Suppose E ⊂ N , where N is the set described in section 1.1, and assume that E
is Lebesgue measurable. Suppose m(E) > 0. Then at least one of E ∩ [0, 21 ] and
E ∩ [ 21 , 1] has positive Lebesgue measure. Call this set E ∗ . Then, consider the
family of translations of E ∗ by n1 for n ≥ 3 to the right if E ∗ is the left part of E
and to the left if E ∗ is the right part of E. These translations are disjoint subsets
of [0, 1]. Therefore, the union of these sets must have measure no more than 1,
and because they’re disjoint and the Lebesgue
invariant,
P∞ measure is translation
P∞
∗
it follows that the union has measure j=1 µ(E ∗ ± j) =
j=1 µ(E ) = ∞,
which is a contradiction.
Part b
Suppose m(E) > 0, and decompose E as suggested in the problem:
[
E=
E ∩ Nr
r∈R
Note that E ∩ Nr is a subset of Nr , and that the E ∩ Nr ’s are disjoint. By part
a, either one of the E ∩ Nr ’s is non-measurable, in whichP
case we’re done, or all
∞
of the E ∩ Nr ’s have measure zero. But then m(E) = r=1 m(E ∩ Nr ) = 0,
which contradicts the assumption that m(E) > 0.
Problem 2
Suppose that f is a measurable function from X into R by the definition on
page 45. In other words, suppose we know that f −1 (O) is measurable for every
open set O ⊂ R. We need to show:
f −1 (∞) is measurable: {∞} This is the intersection
T of the countable family
of open sets (a, ∞], where a ∈ Z. So f −1 (∞) = a∈Z f −1 ((a, ∞])), and all
of the f −1 ((a, ∞]) sets are measurable because f is measurable. So f −1 (∞) is
measurable. The same argument works for f −1 (−∞).
Therefore, the set Y is measurable, where Y = f −1 (R). Note that every
measurable subset of X that is entirely contained in Y is also a measurable
subset of Y . Furthermore, if O is an open set in the topology of R, then it is
open in the topology of R. Therefore, for any open set O ⊂ R, we have that
f −1 (O) is a measurable subset of X by measurability of f on R. Hence, f −1 (O)
is a measurable subset of Y .
Conversely, suppose f satisfies the conditions given in the problem. I seek
to show that f −1 (B) is measurable for every Borel set in R. But every element
of this σ-algebra can be expressed as B1 ∪ B2 , where B1 ⊂ R is a Borel set, and
B2 ⊂ {−∞, ∞}. By assumption, f −1 (B1 ) is a measurable subset of Y (hence of
X) and f −1 (B2 ) is measurable because it is either empty, f −1 (∞), f −1 (−∞),
1
or the union on f −1 (∞) and f −1 (−∞). This proves that f −1 (B1 ) and f −1 (B2 )
are measurable, so f −1 (B) is measurable.
Problem 3
The problem doesn’t really specify what the codomain of f is here. I will assume
it is the real numbers.
Note that limn→∞ fn (x) exists if and only if fn (x) is a Cauchy sequence.
This means that for any ǫ > 0, there exists an N0 such that for all m, n > N0
we have that fn (x) and fm (x) differ by less than ǫ.
Note that we only need to check this for ǫj = 1j . Let’s unravel this condition
and turn quantifiers into unions and intersections. The set of points where fn (x)
and fm (x) differ by less than 1j is the same as the set of points for which there
1
of q.
exists a rational number q such that fn (x) and fm (x) are both within 2j
(just take a rational number arbitrarily close to the midpoint of fn and fm ).
Taking the union over all q’s, therefore, will give the set of points where fn (x)
and fm (x) differ by no more than 1j .
Then, we need this to hold for every n, m > N , so we want to take an
intersection over all such n, m. We want to take the union in N , because we
just need the condition to hold for some N depending on j. Finally, we take
the intersection over all j to arrive at our crazy set
∞ [
∞ \
∞
∞ [
\
\
j=1 N =1 n=N m=N q∈Q
{fn (x) ∈ (q −
1
1
1
1
, q + )} ∩ {fm (x) ∈ (q − , q + )}
2j
2j
2j
2j
These sets are measurable by the assumption that the fn ’s are measurable.
Problem 4
Part a
We first need to show that g(x) is a bijection from [0, 1] to [0, 2]. Note that the
Cantor function f (x) is nondecreasing and x is strictly increasing. Therefore
g(x) is strictly increasing and therefore injective. To check that it’s surjective,
we only need to verify that g(0) = 0 and g(1) = 2, but that’s trivial because
f (0) = 0 and f (1) = 1 by definition.
We need to check the continuity of g −1 . To see this, we use the fact that
for any x2 > x1 , we have that g(x2 ) − g(x1 ) = f (x2 ) − f (x1 ) + x2 − x1 . But
f (x2 ) − f (x1 ) is nonnegative, so g(x2 ) − g(x1 ) is at least x2 − x1 .
Therefore, for any y1 and y2 in [0, 2], we have that g −1 (y2 ) − g −1 (y1 ) is no
more than g(g −1 (y2 )) − g(g −1 (y1 )) = y2 − y1 . So g −1 is continuous (in fact, it
is Lipschitz continuous with Lipschitz constant 1).
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Part b
Let C be the Cantor set, and let C c be the complement of the Cantor set in
[0, 1]. Consider the set g(C c ). Note that C c (and hence, by continuity, f (C c ))
are countable unions of intervals (aj , bj ). Furthermore, the function f is constant
on each of these intervals (aj , bj ). Therefore, g(bj ) − g(aj ) = bj − aj for each
such interval. Therefore, the length of g((aj , bj )) is the same as the length of
(aj , bj ). Thus, g(C c ) has measure exactly equal to 1, and therefore g(C) has
measure 1 as well.
Part c
Consider the sets A and B described in this problem statement. Then B =
g −1 (A) is a subset of the Cantor set C, and therefore a subset of a set of
Lebesgue measure zero. By completeness of the Lebesgue measure, it follows
that B is Lebesgue measurable (and has measure zero).
To see that B is not a Borel set, notice that g and g −1 are continuous, so by
Corollary 2.2 in Folland, B is a Borel set if and only if g(B) = (g −1 )−1 (B) = A is
a Borel set. But A is not Lebesgue measurable so it’s not a Borel set; therefore,
B is also not a Borel set.
Part d
We’re looking for functions F and G with F Lebesgue measurable and G continuous such that F ◦ G is not Lebesgue measurable. This means we need to
find a set S that is Borel measurable so that G−1 (F −1 (S)) is not even Lebesgue
measurable. If we take G = g, then G−1 is a continuous function that maps the
Lebesgue measurable set B to the non-measurable set A. So we’re done if we
can show that there is a Lebesgue measurable function F and a Borel set S on
the reals such that F −1 (S) is equal to B. But this is easy: just take F to be the
indicator function of the set B and S = {1}. Then F is Lebesgue measurable:
f −1 of any set containing 1 but not 0 is B, f −1 of any set containing 0 but
not 1 is B c , f −1 of any set containing both 0 and 1 is R, and f −1 of any set
not containing either 0 or 1 is empty. Then G−1 (F −1 (S)) is A, which is not a
Lebesgue measurable set.
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