Solutions to Self-Scheduled Part of Math 318
Exam #2
1. (9 points) For each of the following statements, say whether it is true
or false. Please write “True” or “False” and not just “T” or “F” (since
these letters are easily mistaken for each other). If the statement is
true give a brief explanation why (this does not need to be a rigorous
proof); if it is false, give a counterexample.
(a) Let A = {a1 , a2 , . . .} be aPcountable set and let p1 , p2 , . . . be
positive numbers such that ∞
n=1 pn = 1. Define the set function
µ on B ⊂ A by
X
µ(B) =
pi .
ai ∈B
Then µ is a measure on the collection of subsets of A.
Answer. True. We check each of the four conditions for a set
function to be a measure. If B ⊂ A then, since each of the pi is
positive,
∞
X
X
0 ≤ µ(B) =
pi ≤
pi = 1.
ai ∈B
i=1
If B = ∅, then the sum is over no elements, so m(∅) = 0.
If B ⊆ C, then ai ∈ B implies ai ∈ C, and so positivity of the pi
implies
X
X
µ(B) =
pi ≤
pi = µ(C).
ai ∈B
ai ∈C
S∞
Finally, if B = n=1 Bn where the Bn are pairwise disjoint, then
each ai ∈ B belongs to exactly one of the Bn , so
µ(B) =
X
ai ∈B
pi =
∞ X
X
n=1 ai ∈Bn
pi =
∞
X
µ(Bn ).
n=1
Hence, all conditions are satisfied and we conclude that µ is a
measure of the collection of subsets of A.
(b) The function f : [0, 1] → R given by
(
1/x if x ∈ C\{0}
f (x) =
0
else
is a measurable function (here C is the Cantor set).
Answer. True. To see this, let c ∈ R. If c < 0, then f (x) > c for
all x ∈ [0, 1], and so
{x ∈ [0, 1] : c < f (x)} = [0, 1]
is measurable.
If c ≥ 0, then f (x) > 0 only if x ∈ C, so
{x ∈ [0, 1] : c < f (x)} ⊂ C.
Since C has measure zero, the above set is measurable with measure zero.
Therefore, we see that {x ∈ [0, 1] : c < f (x)} is measurable for
all c ∈ R, and so f is a measurable function.
(c) If f is bounded and measurable on a bounded, measurable set A,
then f is Lebesgue integrable on any measurable subset of A.
Answer. True. Let B ⊂ A be measurable. If we can show that f
is bounded and measurable on B, then Corollary 23.2 will imply
that f is Lebesgue integrable on B.
Now, f is bounded on A, so it must also be bounded on B. To
see that f is measurable on B, let c ∈ R. Then
{x ∈ B : c < f (x)} = {x ∈ A : c < f (x)} ∩ B
is the intersection of two measurable sets (since f is measurable
on A and B is measurable) and hence measurable. Since the
choice of c ∈ R was arbitrary, we see that {x ∈ B : c < f (x)} is
measurable for all c ∈ R, and so we conclude that f is measurable
on B, completing the proof.
2. (4 points) Give examples of each of the following, or explain why no
such example exists (this explanation does not need to be a proof, but
you should give a cogent explanation).
(a) A measurable set A ⊂ [0, 1] such that A (the closure of A) is not
measurable.
Answer. There is no such example. The closure of any set is a
closed set and hence measurable since every closed set is measurable.
(b) A function f : [0, 1] → R such that f 2 is measurable but f is not.
(Here f 2 is the function that takes x to (f (x))2 for all x ∈ [0, 1].)
Example. Let V ⊂ [0, 1] be a non-measurable set (such as the
one we constructed in class) and define
(
1
if x ∈ V
f (x) :=
−1 else.
Then f 2 is the constant function 1, which is certainly measurable.
However,
{x ∈ [0, 1] : 0 < f (x)} = V,
which is not measurable, so f is not a measurable function.
3. (4 points) Suppose f and g are continuous functions on an interval
[a, b] such that f = g a.e. Show that, in fact, f = g.
Proof. Since f and g are continuous, the function f − g is also continuous. If there exists x0 ∈ [a, b] such that f (x0 ) 6= g(x0 ), then
y0 = (f − g)(x0 ) = f (x0 ) − g(x0 ) 6= 0.
Let =
|f (x0 )−g(x0 )|
.
2
Then, since f − g is continuous,
(f − g)−1 (V (y0 ))
is open; moreover, it is non-empty since it contains x0 . Therefore,
(f − g)−1 (V (y0 )) has positive measure. However, for any x ∈ (f −
g)−1 (V (y0 )), we know that
|(f − g)(x)| = |f (x) − g(x)| > .
Hence, (f − g)−1 (V (y0 )) ⊂ {x ∈ [a, b] : f (x) 6= g(x)}, which means it
must have measure zero.
From this contradiction, then, we conclude that there is no such x0
where f and g differ, and so f = g.