24 1 Lebesgue Measure
1.4 Properties of Lebesgue Measure
Now we will prove that Lebesgue measure is a measure on Rd (in the sense of
Definition 1.1) with respect to the Lebesgue σ-algebra LRd .
Theorem 1.39 (Countable Additivity of Lebesgue Measure). Suppose
that E1 , E2 , . . . are disjoint Lebesgue measurable subsets of Rd . Then
∞ X∞
S
Ek =
|Ek |.
k=1 k=1
Proof. Step 1. Assume that each set Ek is bounded. By subadditivity,
∞
X∞
S
Ek ≤ |Ek |,
k=1 k=1
so our goal is to prove the opposite inequality.
Fix ε > 0. Then, since Ek is measurable, Theorem 1.37 implies that there
exists a closed set Fk ⊆ Ek such that
ε
|Ek \Fk | < .
2k
Since we have assumed that each Ek is bounded, the sets Fk are disjoint and
compact. Therefore, by Exercise 1.28(b) combined with monotonicity, for each
integer N ∈ N we have
N
X
N
N
∞
S S S
|Fk | = Fk ≤ Ek ≤ Ek .
k=1 k=1 k=1 k=1
Taking the limit as N → ∞, we obtain
∞
X N
X
∞
S
|Fk | = lim |Fk | ≤ Ek .
N →∞ k=1
k=1 k=1
Finally,
∞
X ∞
X
|Ek | = |Fk ∪ (Ek \Fk )|
k=1 k=1
∞
X
≤ |Fk | + |Ek \Fk |
k=1
∞
X ε
≤ |Fk | + k
2
k=1
c 2011 by Christopher Heil
1.4 Properties of Lebesgue Measure 25
∞
X
= |Fk | + ε
k=1
∞
S
≤ Ek + ε,
k=1
and since ε is arbitrary this shows that
∞
X
∞
S
|Ek | ≤ Ek .
k=1 k=1
Step 2. Now let the Ek be arbitrary disjoint measurable subsets of Rd . For
each j, k ∈ N, set
Ekj = {x ∈ Ek : j − 1 ≤ |x| < j}.
Then {Ekj }k,j is a countable collection of disjoint measurable sets, and for each
fixed k ∈ N we have j Ekj = Ek . Since each set Ekj is bounded, by applying
S
Step 1 twice we obtain
∞
∞ S ∞
X∞ X ∞ X∞ ∞ ∞
X
j j
S S S j
Ek
=
Ek = |E k | =
E k
=
|Ek |. ⊓⊔
k=1 k=1 j=1 k=1 j=1 k=1 j=1 k=1
Corollary 1.40. Lebesgue measure | · | is a measure on Rd with respect to the
σ-algebra LRd (in the sense of Definition 1.1). ♦
Now we will derive some of the important properties of Lebesgue measure.
We begin by improving on what we know about monotonicity, which tells us
that if A is a measurable set that is contained in another measurable set B,
then |A| ≤ |B|. However, we can write B is the disjoint union of A and B\A,
and the set B\A = B ∩ AC is measurable, so by additivity we know a little
more, namely that
|B| = |A| + |B\A|.
This equation is true whenever A and B are measurable sets such that A ⊆ B.
We are tempted to conclude that |B\A| = |B| − |A|, but of course this has no
meaning if |A| and |B| are both infinite. On the other hand, the next lemma
shows that this equality does hold when |A| is finite (even if |B| is infinite).
Lemma 1.41. If A ⊆ B are Lebesgue measurable sets and |A| < ∞ then
|B\A| = |B| − |A|,
in the sense that if |B| < ∞ then both sides above are finite and equal, while
if |B| = ∞ then both sides are ∞. ♦
Now we consider a nested increasing sequence of sets E1 ⊆ E2 ⊆ · · · . Is
it true that the measures of the sets Ek will increase to the measure of Ek ?
S
26 1 Lebesgue Measure
Theorem 1.42 (Continuity from Below). If Ek ⊆ Rd are measurable and
E1 ⊆ E2 ⊆ · · · , then
S∞
Ek = lim |Ek |.
k=1 k→∞
Proof. If we set E0 = ∅, then
∞
S ∞
S
Ek = (Ej \Ej−1 ),
k=1 j=1
and the sets on the right-hand side above are measurable and disjoint. There-
fore, by countable additivity,
S∞ S∞
E k
=
(Ej \E )
j−1
k=1 j=1
∞
X
= |Ej \Ej−1 |
j=1
N
X
= lim |Ej \Ej−1 |
N →∞
j=1
SN
= lim (Ej \Ej−1 )
N →∞ j=1
= lim |EN |.
N →∞
Here is an alternative proof. If |Ek | = ∞ for some k then there is nothing
to prove, so we may assume that |Ek | is finite for every k. In this case, we can
turn the computation above into a telescoping sum:
∞ ∞
S S
Ek
=
(Ej \E )
j−1
k=1 j=1
∞
X
= |Ej \Ej−1 |
j=1
∞
X
= |Ej | − |Ej−1 |
j=1
= lim |EN | − |E0 |
N →∞
= lim |EN |. ⊓
⊔
N →∞
However, the analogue of Theorem 1.42 for nested decreasing sets does not
always hold!
1.4 Properties of Lebesgue Measure 27
Example 1.43. Let Bn (0) = {x ∈ Rd : |x| < n} be the open ball in Rd
centered at the origin with radius n, and set En = Bn (0)C = Rd \Bn (0). Then
E1 ⊇ E2 ⊇ · · · and E = ∩En = ∅, so
∞
T
Ek = 0 yet lim |Ek | = ∞. ♦
k=1 k→∞
The problem in this example is that nested sets having infinite measure
can decrease to a set that has finite measure. The next exercise shows that
“continuity from above” holds as long as the sets in the sequence have finite
measure from some point onward.
Exercise 1.44 (Continuity from Above). Let Ek be measurable subsets
of Rd for k ∈ N. Show that if E1 ⊇ E2 ⊇ · · · and |Ek | < ∞ for some k, then
T∞
E k = lim |Ek |. ♦
k=1 k→∞
Additional Problems
P if {Qk } is a countable collection of nonoverlapping boxes,
1.21. Show that
then Qk = |Qk |.
S
1.22. Show that if A and B are any measurable subsets of Rd , then
|A ∪ B| + |A ∩ B| = |A| + |B|.
1.23. Find the Lebesgue measures of the following subsets of R2 .
(a) A = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ x}.
(b) B = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ x2 }.
1.24. Show that the Lebesgue measure of an arbitrary rectangle or parallelo-
gram in R2 coincides with its volume in the usual sense.
1.25. Modify the Cantor middle-thirds set construction by removing stage n
intervals of relative length θn from Fn to form Fn+1 . Show that the generalized
Cantor set P = ∩Fn is perfect and has measure |P | = 1 − δ where δ =
Q ∞
k=1 (1 − θk ). Conclude that if θk → 0 quickly enough, then P has positive
measure.
1.26. Show that continuity from below holds for Lebesgue exterior measure,
i.e., if E1 ⊆ E2 ⊆ · · · is any nested increasing sequence of subsets of Rd , then
| Ek |e = limk→∞ |Ek |e (in contrast, Problem 1.31 shows that continuity from
S
above can fail for exterior measure, even if |Ek |e < ∞ for every k).
