Math 318 HW #7 Solutions
1. Exercise 20.12. Let C be the Cantor set. Let D ⊂ [0, 1] be a nowhere dense measurable set
with m(D) > 0. Then there is a non-measurable set B ⊂ D. At each stage of the construction
of C and of D, a certain finite number of open intervals of [0, 1] are deleted (put into [0, 1]\C
or [0, 1]\D). Let g map the intervals put into [0, 1]\D at the nth stage linearly onto the
intervals put into [0, 1]\C at the nth stage, for n = 1, 2, . . .. Thus g is monotone and defined
at every element of [0, 1]\D, mapping onto [0, 1]\C.
(a) Using the fact that g is increasing and that [0, 1]\D is dense in [0, 1], prove that for every
x0 ∈ D,
lim g(x) = sup{g(x) : x ∈ [0, 1]\D and x < x0 }
x→x−
0
and
lim g(x) = inf{g(x) : x ∈ [0, 1]\D and x > x0 }.
x→x+
0
Proof. Since [0, 1]\D is dense, we know that every x0 ∈ D is a limit point of [0, 1]\D, so
limx→x− g(x) makes sense. Now,
0
{g(x) : x ∈ [0, 1]\D and x < x0 }
is nonempty (unless x0 = 0, in which case limx→x− g(x) is vacuous) and, since g is
0
increasing, bounded above by g(x1 ) for any x1 > x0 such that x1 ∈ [0, 1]\D. Therefore,
sup{g(x) : x ∈ [0, 1]\D and x < x0 } = M exists. But then, for any > 0, there exists
x0 ∈ [0, 1]\D with x0 < x0 such that g(x0 ) > M − . Letting δ = x0 − x0 , we know that
x0 < x < x0 implies that g(x0 ) ≤ g(x) ≤ M , and so
g(x0 ) − M ≤ g(x) − M ≤ 0.
Therefore, since g(x0 ) − M > −, the above implies that
− < g(x) − M ≤ 0,
so we can conclude that |x − x0 | < δ and x < x0 implies that
|g(x) − M | < .
Since the choice of > 0 was arbitrary, we conclude that
lim g(x) = M = sup{g(x) : x ∈ [0, 1]\D and x < x0 },
x→x−
0
as desired.
An equivalent argument shows that limx→x+ g(x) = inf{g(x) : x ∈ [0, 1]\D and x >
0
x0 }.
(b) Use (a) to show that for x0 ∈ D, limx→x− g(x) ≤ limx→x+ g(x).
0
1
0
Proof. Since g is monotone increasing, every element of the set {g(x) : x ∈ [0, 1]\D and x >
x0 } is an upper bound on the set {g(x) : x ∈ [0, 1]\D and x < x0 }. Therefore, by definition of the supremum,
lim g(x) = sup{g(x) : x ∈ [0, 1]\D and x < x0 }
x→x−
0
is less than or equal to every element of {g(x) : x ∈ [0, 1]\D and x > x0 }. In other
words, limx→x− g(x) is a lower bound on the set {g(x) : x ∈ [0, 1]\D and x > x0 }, and
0
so, by the definition of the infimum, is less than or equal to
lim g(x) ≤ inf{g(x) : x ∈ [0, 1]\D and x > x0 } = lim g(x),
x→x−
0
x→x+
0
as desired.
(c) Show that if a = limx→x− g(x) < limx→x+ g(x) = b for some x0 ∈ D, then C would
0
0
contain the interval (a, b), contradicting the fact that C is nowhere dense. Therefore
limx→x− g(x) = limx→x+ g(x) = limx→x0 g(x), and the function
0
0
(
g(x)
f (x) =
limx0 →x g(x0 )
if x ∈ [0, 1]\D
if x ∈ D
is continuous and monotone.
Proof. Suppose y ∈ (a, b). Then g(x) = y implies that x ∈ [0, 1]\D and that both
x ≥ x0 and x ≤ x0 . These three conditions cannot all be satisfied simultaneously (since
x0 ∈
/ [0, 1]\D), so it must not be the case that g(x) = y for any x ∈ [0, 1]\D. But then
y is not in the image of g, which is [0, 1]\C, so y ∈ C.
But the above argument was independent of y, so it holds for all y ∈ (a, b), so the whole
interval (a, b) ⊂ C, which is clearly impossible since m((a, b)) = b−a > 0 = m(C). From
the contradiction, then, we conclude that
lim g(x) = lim g(x),
x→x−
0
x→x+
0
as desired.
(d) Let B0 = f (B). Show that B0 is measurable but f −1 (B0 ) = B is not, even though f is
a measurable function.
Since f −1 (B0 ) is not measurable, it is not a Borel set. By exercise 16.15, B0 is not a
Borel set, even though it is measurable.
Proof. Since B0 ⊂ C and C has measure zero, we know that B0 is measurable with
measure zero.
Now, B is non-measurable by construction, but we do need to make sure that f −1 (B0 ) =
B. Of course, this is obvious since f is monotone and, hence, one-to-one.
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(e) Using the fact that χB = χB0 ◦ f , show that composition in the opposite order in
Corollary 17.7 is false.
Answer. We check that, for any x ∈ [0, 1],
(
1 if f (x) ∈ B0
(χB0 ◦ f )(x) = χB0 (f (x)) =
= χf −1 (B0 ) (x) = χB (x).
0 else
Now, f is certainly continuous and, since B0 is measurable, χB0 is a measurable function.
On the other hand, B is not measurable, so χB is not a measurable function. Therefore,
the composition χB0 ◦ f provides a counterexample to the statement “If f : A → R is
continuous and g : f (A) → R is measurable, then g ◦ f is measurable”.
2. Exercise 20.16.
(a) Prove or disprove: if f = g a.e., and g is continuous, then f is continuous a.e.
Counterexample. Let f be the Dirichlet function on [0, 1], i.e.,
(
1 if x ∈ Q
f (x) =
0 else
for all x ∈ [0, 1], and let g(x) = 0 for all x ∈ [0, 1]. Then g is obviously continuous, and
{x ∈ [0, 1] : f (x) 6= g(x)} = Q ∩ [0, 1]
has measure zero, so f = g a.e.
However, as we saw last semester, f is nowhere-continuous, so it is definitely not continuous a.e.
(b) Prove or disprove the converse of (a).
Counterexample. The converse of (a) is also false. Suppose
(
0 if x < 0
f (x) =
1 else.
Then f is continuous except at x = 0; certainly {0} has measure zero, so f is continuous
a.e.
Now, suppose there were a continuous function g such that f = g a.e., meaning that
A = {x ∈ R : f (x) 6= g(x)} has measure zero. Then certainly there exists x0 < 0 such
that x0 ∈
/ A, meaning that
g(x0 ) = f (x0 ) = 0.
Likewise, there exists x1 > 0 such that x1 ∈
/ A, so
g(x1 ) = f (x1 ) = 1.
Thus, the Intermediate Value Theorem implies that there exists c ∈ (x0 , x1 ) such that
g(c) = 1/2. Therefore, g −1 ((0, 1)) is open (since g is continuous) and non-empty (since
c is an element), so m(g −1 ((0, 1))) > 0. But it must be the case that g −1 ((0, 1)) ⊂ A,
which has measure zero.
From this contradiction, then, we conclude that there cannot be a continuous function
g such that f = g a.e., so f provides a counterexample to the converse of (a).
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3. We say that a sequence (fn ) defined on a measurable set A ⊆ R converges in measure to a
function f : A → R if
lim m{x ∈ A : |fn (x) − f (x)| ≥ δ} = 0
n→∞
for all δ > 0.
(a) Show that if (fn ) converges to f almost everywhere on A, then it converges to f in
measure (cf. Exercise 20.26).
Proof. Let δ > 0. Since (fn ) → f almost everywhere, the set
B := {x ∈ A : lim fn (x) 6= f (x)}
n→∞
has measure zero.
For each k = 1, 2, . . ., define
Bk :=
∞
[
{x ∈ A : |fn (x) − f (x)| ≥ δ}.
n=k
Then certainly B1 ⊃ B2 ⊃ B3 ⊃ · · · and so, by Corollary 10.10,
!
∞
\
∗
Bk = lim m∗ (Bk )
m
k→∞
k=1
(1)
Now, for each x ∈
/ B, there exists N ∈ N such that n ≥ N implies |fn (x) − f (x)| < δ, so
∞
\
Bk ⊆ B,
k=1
meaning that
T∞
K=1 Bk
has measure zero. By (1), then,
lim m∗ (Bk ) = 0.
k→∞
But now {x ∈ A : |fn (x) − f (x)| ≥ δ} ⊂ Bn for each n, so
lim m∗ {x ∈ A : |fn (x) − f (x)| ≥ δ} = 0.
n→∞
Since fn and f are both measurable, the function |fn − f | is measurable, meaning that
the set {x ∈ A : |fn (x) − f (x)| ≥ δ} is measurable by Proposition 17.2(2), so the above
implies that
lim m{x ∈ A : |fn (x) − f (x)| ≥ δ} = 0,
n→∞
as desired.
(b) Suppose (fn ) converges in measure to f on A. Show that (fn ) converges in measure to
a function g if and only if f = g (a.e.).
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Proof. (⇒) Suppose (fn ) converges in measure to both f and g. Let
B := {x ∈ A : f (x) 6= g(x)};
our goal is to show that m(B) = 0.
Let x0 ∈ B. Then |f (x0 ) − g(x0 )| > 0, so, if k ∈ N such that |f (x0 ) − g(x0 )| ≥ 1/k, we
have
1
≤ |f (x0 )−g(x0 )| = |f (x0 )−fn (x0 )+fn (x0 )−g(x0 )| ≤ |f (x0 )−fn (x0 )|+|fn (x0 )−g(x0 )|.
k
This implies that
1
1
x0 ∈ x ∈ A : |fn (x) − f (x)| ≥
or |fn (x) − g(x)| ≥
2k
2k
1
1
= x ∈ A : |fn (x) − f (x)| ≥
∪ x ∈ A : |fn (x) − g(x)| ≥
2k
2k
for all n ∈ N.
Therefore, if we let Bk = {x ∈ A : |f (x) − g(x)| ≥ 1/k}, we have that
1
1
∪ x ∈ A : |fn (x) − g(x)| ≥
Bk ⊂ x ∈ A : |fn (x) − f (x)| ≥
2k
2k
so, by monotonicity of the Lebesgue measure,
1
1
m(Bk ) ≤ m
x ∈ A : |fn (x) − f (x)| ≥
∪ x ∈ A : |fn (x) − g(x)| ≥
2k
2k
1
1
= m x ∈ A : |fn (x) − f (x)| ≥
+ m x ∈ A : |fn (x) − g(x)| ≥
.
2k
2k
Therefore, by the Order Limit Theorem,
1
1
+ lim m x ∈ A : |fn (x) − g(x)| ≥
m(Bk ) ≤ lim m x ∈ A : |fn (x) − f (x)| ≥
n→∞
n→∞
2k
2k
= 0,
so we can conclude that m(Bk ) = 0 for all k.
S
But now B = ∞
k=1 Bk and B1 ⊂ B2 ⊂ · · · , so Corollary 10.9 implies that
!
∞
[
m(B) = m
Bk = lim m(Bk ) = lim 0 = 0.
k→∞
k=1
k→∞
Thus, we conclude that f = g a.e.
(⇐) On the other hand, suppose f = g a.e. Then
B = {x ∈ A : f (x) 6= g(x)}
has measure zero. Let δ > 0. Then
{x ∈ A : |fn (x)−g(x)| ≥ δ} ⊂ {x ∈ A : |fn (x)−f (x)| ≥ δ}∪{x ∈ B : |fn (x)−g(x)| ≥ δ}.
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The latter is a subset of B and, hence, has measure zero, so this implies that
m ({x ∈ A : |fn (x) − g(x)| ≥ δ}) ≤ m ({x ∈ A : |fn (x) − f (x)| ≥ δ})
for all n, so indeed
lim m ({x ∈ A : |fn (x) − g(x)| ≥ δ}) = 0
n→∞
for all δ > 0, and so (fn ) converges in measure to g.
4. Exercise 20.36. Prove that if f is monotone on a bounded measurable set A, then f is
measurable on A.
Proof. Let c ∈ R. Consider the set {x ∈ A : c ≤ f (x)}. This is a subset of A, and hence
is bounded below, so it has an infimum, m. Essentially by Problem 3 from last semester’s
take-home final, limx→m+ f (x) = y0 exists and, by construction, y0 ≥ c. Hence, for all x ≥ m
such that x ∈ A, we have that f (x) ≥ y0 ≥ c. On the other hand, since f is increasing,
f (x) ≥ c implies that x ≥ m. Therefore,
{x ∈ A : c ≤ f (x)} = [m, +∞) ∩ A,
which is measurable since it is the intersection of two measurable sets.
Therefore, using Proposition 17.2(2), we see that f is a measurable function, as desired.
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