# §2 Outer Lebesgue measure by etssetcf

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```									6                                                Measure and Integration

§2 Outer Lebesgue measure
Deﬁnition (Outer Lebesgue measure)

Let E be any subset of R. We call the nonnegative number
∞                                           ∞
λ∗ (E) = inf          |Ij | : Ij are open intervals and            Ij ⊃ E
i=1                                          j=1

the outer Lebesgue measure of E.

Proposition 2.1 (Basic properties of the outer Lebesgue measure).

1. λ∗ (∅) = 0 and for any x ∈ R, λ∗ ({x}) = 0.

2. If E1 ⊂ E2 , then λ∗ (E1 )         λ∗ (E2 ).
∞                 ∞
3. If (Ek )k    1   is a countable collection of sets, then λ∗ (           Ek )             λ∗ (Ek ).
k=1            k=1

4. For any E ⊂ R and x ∈ R we have λ∗ (E + x) = λ∗ (E), where E + x = {y + x, y ∈ E} is the
translation of E.

Proof.       1. Indeed, take collection of empty intervals In = (0, 0), then                            In ⊃ ∅, so 0
∗
λ (∅)       0.
If x ∈ R is an arbitrary point, the interval Jε = (x − ε/2, x + ε/2) covers {x} and has
length ε. Hence if I1 = Jε and Ik = (0, 0) = ∅, we get 0 λ∗ ({x}) ε for all ε > 0. Thus
λ∗ ({x}) = 0.
∞                          ∞
2. If    j=1 Ij     ⊃ E2 , then     j=1 Ij   ⊃ E1 . Hence the inﬁmum for E1 is less than or equal to
the inﬁmum for E2 .

3. The statement is obviously satisﬁed if there exists k such that λ∗ (Ek ) = +∞. We
therefore assume λ∗ (Ek ) < +∞ for all k. Let ε > 0 be arbitrary. For each k, choose a
(k)                         (k)                      (k)
sequence of open intervals (Ij )∞ such that Ek ⊂ ∞ Ij and λ∗ (Ek )
j=1                 j=1
∞
j=1 |Ij | −
(k)
ε/2k . Then if we take all intervals {Ij ; j 1 and k 1}, we get a countable collection
(k)
of open intervals that cover ∞ Ek and ∞ λ∗ (Ek )
k=1           k=1
∞
k=1
∞
j=1 |Ij | − ε. The
RHS of the latter inequality is at least λ∗ (              ∞
k=1   Ek ) − ε. Since ε > 0 is arbitrary, we
get ∞ λ∗ (Ek ) λ∗ ( ∞ Ek .
k=1                k=1

4. Any cover by open intervals (Ij ) of E corresponds to the cover by open intervals (x + Ij )
of x + E.

Corollary 2.2. If E ⊂ R is ﬁnite or countable, then λ∗ (E) = 0. In particular, λ∗ (Q) = 0.
§2. Outer Lebesgue measure                                                     7

Proof. Let E = {q1 , q2 , . . . }. We know 0                               λ∗ (E); by Proposition 2.1 we have λ∗ (E)
∞
n=1     λ∗ ({qn }) = 0 as λ∗ ({qn }) = 0 for every n by Proposition 2.1. Thus λ∗ (E) = 0.

Lemma 2.3 (Outer Lebesgue measure of an interval). If I is an interval then λ∗ (I) coincides
with its length.

Proof. Assume ﬁrst I = [a, b] is a closed ﬁnite interval.
Since I ⊂ (a − ε/2, b + ε/2) for any ε > 0, we conclude λ∗ (I)                              b − a + ε for every ε > 0 and
so λ∗ (I)     b − a = |I|.
Assume now (Ij )j            1   is any collection of open intervals with union sontaining I. We are going
to prove that the sum of their lengths ∞ |Ij | is at least |I|. This would imply that every
j=1
element of the set in the right hand side of the deﬁnition of the outer Lebesgue measure of
I is at least |I|, and hence λ∗ (I)                   |I|.
By Heine-Borel lemma, we can choose ﬁnitely many of intervals among (Ij )j                                         1   with union
containing I; thus ∃N such that N Ij ⊃ I = [a, b]. Denote
j=1

F = {Ij , 1     j       N}.

Let J1 = (a1 , b1 ) ∈ F be an interval containing point a (then a1 < a). If b1                                            b then
b1 ∈ [a, b] \ J1 . Therefore
b1 ∈                  Ij .
Ij ∈F,Ij =J1

Let J2 ∈ F be an interval, such that b1 ∈ J2 . Assuming Jm = (am , bm ) ∈ F is deﬁned and
bm b we conclude bm ∈ [a, b] \ m Ji ; choose Jm+1 ∈ F, such that bm ∈ Jm+1 . Since we
i=1
have only ﬁnitely many intervals, this process must stop. We will therefore ﬁnd M such
that bM > b. Note that J1 , . . . , JM ∈ F are all different intervals (1                                i1 < i2       M implies
bi1 < bi2 ), therefore,
∞                N             M               M                                 M−1
|Ij |             |Ij |         |Ji | =         (bi − ai ) = (bM − a1 ) +         (bi − ai+1 ).
j=1               j=1           i=1             i=1                               i=1

Since ai+1 < bi for each 1     i    M − 1, we conclude that the latter sum is greater than
bM − a1 , which is bigger than b − a as bM > b and a1 < a. This proves ∞ |Ij | b − a = |I|.
j=1

Assume now I is any bounded interval a, b (could be [a, b], (a, b), [a, b) or (a, b]) with
a = b. Since [a + ε/2, b − ε/2] ⊂ I ⊂ [a − ε/2, b + ε/2] for any ε > 0, Proposition 2.1 implies
|I| − ε     λ∗ (I)      |I| + ε for all ε > 0; thus λ∗ (I) = |I|.
Finally, consider a case when I is an unbounded interval. In such a case, for any ∆ > 0
there exists an interval J of length ∆ such that J ⊂ I. Using again Proposition 2.1 we get
λ∗ (I) ∆ for any ∆ > 0, thus λ∗ (I) = +∞.

Remark. Property 3 from Proposition 2.1 implies that λ∗ (A ∪ B)          λ∗ (A) + λ∗ (B). Our
intuitive feeling is that in case A and B are disjoint, equality should hold in this relation.
8                                                Measure and Integration

Unfortunately, this is not the case that equality holds for an arbitrary pair of disjoint sub-
sets of R. However, we show that this desired equality relation is true in the case the sets
A and B are at positive distance from each other.

Lemma 2.4. If A, B ⊂ R are disjoint such that dist(A, B) > 0, then λ∗ (A∪B) = λ∗ (A)+λ∗ (B).

Proof. We know λ∗ (A ∪ B)                 λ∗ (A) + λ∗ (B). We need to show λ∗ (A ∪ B)                        λ∗ (A) + λ∗ (B).
Assume λ∗ (A ∪ B) < ∞ and d = dist(A, B) > 0. Let ε > 0. Consider any covering ∞ Ij ⊃
j=1
A ∪ B by open intervals, such that ∞ |Ij | < λ∗ (A ∪ B) + ε. We may cover each interval
j=1
Ij by ﬁnitely many open intervals of length less than d with sum of their lengths less
than |Ij | + ε/2j . Thus we may assume that we have a cover of A ∪ B by open intervals
Uk with        ∞
k=1   |Uk | < λ∗ (A ∪ B) + 2ε. Note that for each k, exactly one of the following
three is satisﬁed: Uk ∩ A = ∅, Uk ∩ B = ∅, Uk ∩ (A ∪ B) = ∅. The sum of lengths of
those Uk that intersect A is at least λ∗ (A), those that intersect B is at least λ∗ (B). Thus
λ∗ (A ∪ B) + 2ε >        ∞
k=1      |Uk |   λ∗ (A) + λ∗ (B). Since ε > 0 is arbitrary, we are done.

k              k
Corollary 2.5. If I1 , . . . , Ik are disjoint intervals, then λ∗ (                     j=1 Ij )   =   j=1   |Ij |.

Proof. Follows from Lemma 2.3 and Lemma 2.4.

Example (Non-additivity of outer Lebesgue measure)

For x, y ∈ R, deﬁne x ∼ y if x − y is rational. This is an equivalence relationship on
[0, 1]. For each equivalence class, pick an element out of that class (by the axiom of choice).
Call the collection of such points A. Note λ∗ (A + q) = λ∗ (A) for any q by Proposition 2.1.
Moreover, the sets A + q are disjoint for different rationals q as a1 , a2 ∈ A, q1 , q2 ∈ Q and
a1 + q1 = a2 + q2 implies a1 − a2 = q2 − q1 ∈ Q, which is impossible unless a1 = a2 . In this
case q1 = q2 .
Let [−1, 1] ∩ Q = Q[−1,1] = {q1 , q2 , . . . }. We see                      q∈Q[−1,1] (A   + q) ⊂ [−1, 2], so if λ∗ were
additive, we would have for every n 1,
∞                             n                  n
3     λ∗ (       (A + qi ))         λ∗ (       (A + qi )) =         λ∗ (A + qi ) = nλ∗ (A),
i=1                           i=1                    i=1

which implies λ∗ (A)              3/n for every n               1 and so λ∗ (A) = 0. But [0, 1] ⊂              q∈Q[−1,1] (A   + q),
∗                  ∗
so using λ (A + q) = λ (A) = 0 for every q ∈ Q, we get by Proposition 2.1
∞                     ∞
1     λ∗ (       (A + qi ))             λ∗ (A + qi ) = 0,
i=1                      i=1

§2. Outer Lebesgue measure                                       9

Example (Calculating outer Lebesgue measure)

1. λ∗ ([0, 1] \ Q) = λ∗ ({irrational numbers between 0 and 1}) = 1.
Proof. Let E = [0, 1] \ Q and Q1 = Q ∩ [0, 1]. Then
1 = λ∗ ([0, 1]) = λ∗ (E ∪ Q1 ) λ∗ (E) + λ∗ (Q1 ) = λ∗ (E) λ∗ ([0, 1]) = 1.
Note that the set of irrational numbers between 0 and 1 contains no intervals, still its
outer Lebesgue measure is strictly positive.

2. There exist uncountable sets of outer Lebesgue measure zero.
Proof. Consider the Cantor set C = x ∈ [0, 1] | x = ∞ ak , ak ∈ {0, 2} for all k
k=1 3
k
1 ,
∞   ak
let An = x ∈ [0, 1] | x =      k=1 3k , ak   ∈ {0, 1, 2} for all k   1 and ∃k   n : ak = 1 .
Clearly C ∩ An = ∅ for all n, therefore, λ∗ (C)        λ∗ ([0, 1] \ An ) for all n. Note that
[0, 1]\ An is a union of 2n disjoint closed intervals of equal length. This length is equal
to (1/3)n , and so by Corollary 2.5, λ∗ ([0, 1] \ An ) = 2n /3n . Thus 0 λ∗ (C) (2/3)n
for any n, so λ∗ (C) = 0. It remains to say that the Cantor set is uncountable.

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