Departments of Mathematics
Fall 2014
Montana State University
Prof. Kevin Wildrick
Measure Theory
Problem Set 1Solutions
Due Tuesday, September 8th , 11:00 am.
Note: There are many ways to solve these problems. The way that I have done it below,
problems 1 and 4b are solved independently, while solutions to the rest of the problems require
the solutions to problems 1 and 4b.
1. Let {Ej }j∈N be a countable sequence of subsets of Rn . Show that


[
X
m∗ 
Ej  ≤
m∗ (Ej ).
j∈N
j∈N
Solution: Let > 0. For each j ∈ N, the definition of m∗ (Ej ) implies that we may find a
countable collection {Ij,k }k∈N of cells such that
S
• Ej ⊆ k∈N Ij,k ,
P
• m∗ (Ej ) ≥ k∈N vol(Ij,k ) − /2j+1 .
Then,
[
Ej ⊆
j∈N
[ [
Ij,k .
j∈N k∈N
Hence, by definition

m∗ 

[
Ej  ≤
j∈N
XX
vol(Ij,k )
j∈N k∈N
≤
X
m∗ (Ej ) +
j∈N
2j+1


X
≤
m∗ (Ej ) + .
j∈N
2. (?) Let E be a countable set in Rn . Show that m(E) = 0. Give an example of a measurable
uncountable set E ⊆ R with m(E) = 0.
Solution We first show that it E = {p} for some point p ∈ Rn , then E is measurable
and has zero measure. To this end, note that in this case E is a closed set and hence
measurable. Moreover, for any > 0 we may find a cell that contains p and has volume
no greater than . For example, the cube centered at p with sidelengths 1/n will work.
This cell covers E, and so by definition m(E) ≤ . Since was arbitrary, this implies that
m(E) = 0.
Now, note that if E = {pi }i∈N is an arbitrary countable subset of Rn , then it is the
countable union of the singleton sets defined by its elements:
[
E=
{pi }.
i∈N
Hence, by the previous argument, it is the countable union of measurable sets and hence
measurable, and moreover by the subadditivity of m, we see that
X
m(E) ≤
m({pi }) = 0.
i∈N
We now construct an uncountable set with with zero measure. Let S denote the collection
of all finite sequences with entries in {0, 1}. Denote the length of a sequence σ ∈ S by |σ|.
We declare that the empty sequence has length 0.
Define I∅ = [0, 1]. Now, suppose that Iσ has been defined for all sequences σ ∈ S with
|σ| = k ∈ N. We then define Iσ,0 and Iσ,1 to be the left-most and right-most closed
subintervals of Iσ with length 1/3 of that of Iσ .
For each k ∈ N, let
[
Ck =
Iσ .
σ∈S,|σ|=k
Since Ck is a finite union of closed sets, it is closed and hence measurable. The Cantor set
C is defined by
\
C=
Ck .
k∈N
Since C is the intersection of closed sets, it is closed and hence measurable. Note also that
C ⊆ [0, 1]. Hence,
1 = m([0, 1]) = m([0, 1] ∩ C) + m([0, 1]\C) = m(C) + m([0, 1]\C).
(1)
The complement [0, 1]\C is the countable union of disjoint open intervals given by
[
[
[0, 1]\C =
Iσ \(Iσ,0 ∪ Iσ,1 ).
k∈N σ∈S,|σ|=k
For a fixed k ∈ N, there are 2k sequences σ ∈ S with length k; for each such σ the open
interval Iσ \(Iσ,0 ∪ Iσ,1 ) has length 3−(k+1) . Using exercise 3, we therefore see that
m([0, 1]\C) =
X 2k
= 1.
3k+1
k∈N
This, along with (1), shows that m(C) = 0.
To see that C is uncountable, we argue by contradiction. By construction, there is a
surjection (in fact it is a bijection) from C to the set S∞ of countably infinte sequences of
zeros and ones. Hence it suffices to show that S∞ is not countable. Suppose that S∞ is
countable, and let {σl }l∈N be an enumeration of it. Denote by σl (k) the kth entry of σl .
Now, define τ ∈ σ∞ by
τ (k) = 1 − σk (k).
Then τ 6= σl for any l ∈ N, a contradiction.
As a bonus, we prove that C contains no interval. First, note that any interval contains
some interval of the form
Iˆσ := Iσ \(Iσ,0 ∪ Iσ,1 )
where σ ∈ S. Hence, if C contains any interval, it will contain some Iˆσ . However, by
construction C is precisely the complement of the collection of all such intervals. This is a
contradiction.
S
3. (?) Show that if E = k∈N Ik is a countable union of cells in Rn with disjoint interiors,
then E is measurable and
X
m(E) =
vol Ik .
k∈N
Solution Since each cell is measurable, and the countable union of measurable sets is
measurable, we see that E is measurable.
Since the measure of a cell is its volume, the sub-additivity of m implies that
X
X
m(E) ≤
m(Ik ) =
vol Ik .
k∈N
k∈N
Denote the interior of a cell I by I o . Then
E⊇
[
Iko .
k∈N
Hence, by monotonicity of m,
!
m(E) ≥ m
[
Iko
.
k∈N
Note that k∈N Iko is a countable union of disjoint measurable sets. Hence, by exercise 4b,
we have that
!
X
X
[
X
o
vol(Ik ).
vol(Iko ) =
m
Ik =
m(Iko ) =
S
k∈N
k∈N
k∈N
k∈N
This completes the proof.
4. Let {Ei }i∈N be a countable sequence of measurable subsets of Rn .
(a) (?) Show that
!
m
[
Ei
i∈N
= lim m
n→∞
n
[
!
Ei
.
i=0
(b) Show that if the sequence is pair-wise disjoint, then
!
∞
[
X
m
Ei =
m(Ei ).
i∈N
i=0
(c) Show that if the sequence is increasing, i.e., E0 ⊆ E1 ⊆ E2 ⊆ . . . , then
!
[
m
Ei = lim m(Ei ).
i∈N
i→∞
(d) Show that if the sequence is decreasing, i.e., E0 ⊇ E1 ⊇ E2 ⊇ . . . , and there is a
number i0 ∈ N such that m(Ei ) < ∞, then
!
\
m
Ei = lim m(Ei ).
i∈N
i→∞
Show that this can fail if m(Ei ) = ∞ for each i ∈ N.
f0 = E0 , and define inductively
Solution (a) Let E
n−1
[
fn = En \
E
!
.
Ei
i=0
fn is measurable, and E
fn ∩ E
g
Then for each n ∈ N, the set E
m = ∅ if n 6= m. Moreover, for
each n ∈ N
n
n
[
[
fi =
E
Ei ,
i=1
i=1
and
[
fi =
E
[
Ei .
i∈N
i∈N
By exercise 4b,
!
m
[
=
fi
E
m
n
[
fi ),
m(E
i=0
i∈N
and
∞
X
!
=
fi
E
i=0
n
X
fi ).
m(E
i=0
Thus
!
[
m
Ei
!
[
=m
i∈N
fi
E
i∈N
=
∞
X
fi )
m(E
i=0
= lim
n→∞
n
X
fi )
m(E
i=0
= lim m
n→∞
= lim m
n→∞
n
[
i=0
n
[
!
fi
E
!
Ei
.
i=0
(b) We first claim that for any n ∈ N,
m
n
[
!
Ei
=
i=0
n
X
m(Ei ).
i=0
Clearly this is true for n = 0. Suppose that it is true for some given n ∈ N. Write
!
n+1
n
[
[
Ei =
Ei ∪ En+1 .
i=0
Since
i=0
Sn
i=0 Ei
is measurable, disjointness yields
!
!
!!
n+1
n+1
n
[
[
[
Ei
+m
m
Ei = m
Ei ∩
i=0
i=0
i=0
=m
n
[
!
Ei
i=0
=
n+1
X
i=0
m(Ei ).
+ m(En+1 ).
n+1
[
i=0
!
Ei
\
n
[
i=0
!!
Ei
This proves the claim by induction.
Applying the claim and monotonicity,
∞
X
n
X
m(Ei ) = lim
n→∞
i=0
m(Ei ) = lim m
n→∞
i=0
≤ lim m
n→∞
!
Ei
i=0
∞
[
!
Ei
i=0
∞
[
=m
n
[
!
Ei
.
[
fi .
E
i=0
Sub-additivity of the measure implies that
!
∞
∞
[
X
m
Ei ≤
m(Ei ),
i=0
i=0
completing the proof.
(c) Since the sequence is increasing,
n
[
Ei = En .
i=0
Hence, this follows immediately from (a).
(d) Consider, for each i ∈ N, the set
fi := Ei \Ei .
E
0
Then

!
Ei0 \
\
Ei
= Ei0 \ 

\
Ei  =
i≥i0
i∈N
i≥i0
fi }i≥i is increasing, so by (c)
Moreover, the sequence {E
0


!!
\
[
fi  = lim m(E
fi ).
m Ei0 \
Ei
= m
E
i→∞
i≥i0
i∈N
Consider any i ≥ i0 . Since Ei is measurable and the sequence {Ei } is decreasing,
fi ).
m(Ei0 ) = m(Ei0 ∩ Ei ) + m(Ei0 \Ei ) = m(Ei ) + m(E
Since m(Ei0 ) < ∞, all quantities above are also finite, and so
!!
\
m Ei0 \
Ei
= lim m(Ei0 ) − m(Ei ) = m(Ei0 ) − lim m(Ei ).
i∈N
Similarly, since
T
i∈N Ei
i→∞
i→∞
is measurable,
!!
m(Ei0 ) = m Ei0 \
\
Ei
!
+m
i∈N
Again, since m(Ei0 ) < ∞, we may write
!!
\
m Ei0 \
Ei
= m(Ei0 ) − m
i∈N
Together with (2), this yields the desired equality.
\
Ei
.
i∈N
!
\
i∈N
Ei
.
(2)
5. (?) Find a measurable subset A of R with the property that m(A) > 0, but A contains no
cells of positive length. This indicates that an analogously defined “inner measure” would
not coincide with the outer measure m∗ .
Solution Consider the set A = R\Q of irrational numbers. Since Q is countable, it is
measurable and has measure 0. Hence, A is also measurable. As a result,
m(R) = m(R ∩ A) + m(R\A) = m(A) + m(Q) = m(A).
However, the rational numbers are dense in R, and so A cannot contain any interval.