Download 1.3 Equivalent Formulations of Lebesgue Measurability

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1 Lebesgue Measure
1.3 Equivalent Formulations of Lebesgue Measurability
The collection LRd of Lebesgue measurable subsets of Rd is closed under both
countable unions and complements. Since LRd contains all of the open and
closed subsets of Rd , it also contains all of the following types of sets.
Definition 1.34. (a) A set H ⊆ Rd is a Gδ set if there exist countably many
open sets Uk such that H = ∩Uk .
(b) A set H ⊆ Rd is an Fσ set if there exist countably many closed sets Fk
S
such that H = Fk . ♦
The symbol σ in this definition reminds us of the word “sums” and hence
unions, while δ suggests the word “difference” and hence intersections.1 Since
Definition 1.34 only uses the concepts of countable unions and intersections of
open and closed sets, it has a natural generalization to any topological space,
i.e., we can define Fσ and Gδ sets in any space X that has a topology. In
particular, every metric space has an associated topology. However, for the
most part we will only be concerned with X = Rd .
The half-open interval [a, b) is neither an open nor a closed subset of R,
but it is a Gδ set because we can write
[a, b) =
∞
T
k=1
a − k1 , b .
It is also an Fσ set because
[a, b) =
∞ S
k=1
a, b −
1
k
.
Here are some additional examples of Gδ and Fσ sets.
Example 1.35. (a) Since the set of rationals is countable, let Q = {rn }n∈N
be an enumeration of all the rationals. Then Un = R\{rn } is an open set
that does not contain the rational rn , and the intersection of these sets is
T
Un = R\Q, the set of irrationals. Hence the set of irrationals is a Gδ set.
(b) Could the set of rationals Q also be a Gδ set? If it was, then we could
T
write Q = Vn where each Vn is open. Each set Vn contains Q, and therefore
is dense in R. The set Un defined above is also dense in in R, and
∞
∞
T
T
T
T
Vn
Un = Q (R\Q) = ∅.
n=1
n=1
1
More precisely, Fσ is derived from the French words fermé (closed) and somme
(union), while Gδ is derived from the German Gebiet (area, neighborhood, open set)
and Durchschnitt (average, intersection).
c 2011 by Christopher Heil
1.3 Equivalent Formulations of Lebesgue Measurability
21
However, since the real line is a complete metric space, the Baire Category
Theorem [Hei11b, Thm. 3.96] implies that a countable intersection of open,
dense subsets of R must be nonempty. This is a contradiction, so we conclude
that Q cannot be a Gδ set. ⊓
⊔
We can keep going and define an Fσδ set to be a countable intersection
of Fσ sets, a Gδσ set to be a countable union of Gδ sets, an Fσδσ set to be a
countable union of Fσδ sets, and so forth. All of these are Lebesgue measurable,
though the collection of all of these sets does not exhaust the family LRd . Not
only are all the sets in the various Gδ and Fσ families measurable, but they all
belong to a special class of sets known as the Borel σ-algebra. We will explore
Borel sets in more detail in Chapter 2 (see Definition 2.4).
Our next lemma shows that every set E, measurable or not, can be surrounded by a Gδ set that has exactly the same measure as E.
Lemma 1.36. Given a set E ⊆ Rd , there exists a Gδ set H ⊇ E such that
|E|e = |H|.
Proof. For each k ∈ N there exists an open set Uk ⊇ E such that
|Uk | < |E|e +
1
.
k
Let H = ∩Uk . Then H is a Gδ set, and E ⊆ H ⊆ Uk for every k. By
monotonicity we therefore have |E|e ≤ |H| ≤ |Uk | ≤ |E|e + k1 . As this is true
for every k, we conclude that |E|e = |H|. ⊓
⊔
We cannot conclude from Lemma 1.36 that |H\E|e = 0! Indeed, the next
result shows that this is an equivalent way of distinguishing Lebesgue measurable sets from nonmeasurable sets.
Theorem 1.37. Given E ⊆ Rd , the following statements are equivalent.
(a) E is Lebesgue measurable.
(b) For every ε > 0, there exists a closed set F ⊆ E such that |E\F |e ≤ ε.
(c) E = H\Z where H is a Gδ set and |Z| = 0.
(d) E = H ∪ Z where H is an Fσ set and |Z| = 0.
Proof. (a) ⇒ (c). Suppose that E is measurable. Then for each k ∈ N we can
find an open set Uk ⊇ E such that |Uk \E| < k1 (and not just |Uk | < |E|e + k1
as in the proof of Lemma 1.36). Let H = ∩Uk . Then H is a Gδ set, H ⊇ E,
and Z = H\E ⊆ Uk \E for every k. Hence |Z|e ≤ |Uk \E| < k1 for every k, so
|Z| = 0.
Exercise: Complete the proof of the remaining implications. ⊓
⊔
Thus, every Lebesgue measurable set can be obtained by taking a Gδ or Fσ
set and adding or subtracting a set with measure zero. Some other equivalent
formulations of measurability are given in the Additional Problems for this
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1 Lebesgue Measure
section, and in Section 1.5 we will derive another equivalent formulation known
as Carathéodory’s Criterion.
The next exercise applies Theorem 1.37 to prove the seemingly “obvious”
fact that the measure of a Cartesian product |E × F | is the product of the
measures of E and F. One inequality relating |E × F | to |E| |F | is easy, for if
{Qk }k is a covering of E by boxes and {Rℓ }ℓ is a covering of F by boxes then
{Qk × Rℓ }k,ℓ is a covering of E × F by boxes, so
X
X
X
|E × F | ≤
vol(Qk × Rℓ ) =
vol(Qk )
vol(Rℓ ) .
k,ℓ
k
ℓ
Taking the infimum over all such coverings of E and F yields the inequality
|E × F | ≤ |E| |F |. However, it is not so easy to prove that equality holds. The
next exercise proceeds through cases to show that this is the case (recall that
we interpret 0 · ∞ and ∞ · 0 as 0).
Exercise 1.38. (a) Show that if U ⊆ Rm and V ⊆ Rn are open sets, then
U × V is an open subset of Rm+n and |U × V | = |U | |V |.
(b) Show that if G ⊆ Rm and H ⊆ Rn are bounded Gδ sets, then G × H is a
Gδ set and |G × H| = |G| |H|.
(c) Show that E is a bounded measurable subset of Rm and Z is a subset of
Rn with measure zero, then E ×Z is measurable and |E ×Z| = 0 = |E| |Z|.
(d) Show that E ⊆ Rm and F ⊆ Rn are bounded measurable sets, then E × F
is measurable and |E × F | = |E| |F |.
(e) Show that E ⊆ Rm and F ⊆ Rn are any measurable sets, then E × F is
measurable and |E × F | = |E| |F |. ♦
Additional Problems
1.13. Show that the complement of a Gδ set is an Fσ set, and the complement
of an Fσ set is a Gδ set.
1.14. (a) Show that every countable set is an Fσ set.
(b) Is any countable set a Gδ set? Is every countable set a Gδ set? Is
{ n1 }n∈N a Gδ set?
1.15. Exhibit a subset of Rd that belongs to one of the classes Gδσ , Fσδ , Gδσδ
Fσδσ , etc., but is not a Gδ or an Fσ set.
1.16. Given E ⊆ Rd , define dE (x) = dist(x, E) for x ∈ Rd . Prove the following
statements.
(a) dE is continuous on Rd .
(b) The set Er = {x ∈ Rd : dist(x, E) < r} is open for each r > 0.
1.3 Equivalent Formulations of Lebesgue Measurability
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(c) If E ⊆ Rd is closed, then dE (x) = 0 if and only if x ∈ E.
(d) Every closed set in Rd is a Gδ set.
(e) Every open set in Rd is an Fδ set.
1.17. Given a function f : Rd → R, the oscillation of f at x is defined to be
oscf (x) = inf sup |f (y) − f (z)| : y, z ∈ Bδ (x) ,
δ>0
where Bδ (x) is the open ball with radius δ centered at x. Prove the following
statements.
(a) f is continuous at x if and only if oscf (x) = 0.
(b) The set {x ∈ Rd : oscf (x) ≥ ε} is closed for each ε > 0.
(c) D = {x ∈ Rd : f is discontinuous at x} is an Fσ set, and therefore the
set of continuities of f is a Gδ set.
1.18. (a) Does there exist a function f : R → R that is continuous at each
rational point and discontinuous at each irrational?
(b) Does there exist a function f : R → R that is continuous at each
irrational point and discontinuous at each rational?
1.19. Define the inner Lebesgue measure of a set E ⊆ Rd to be
|E|i = sup |F | : F closed, F ⊆ E .
(a) Show that if |E|e < ∞, then E is Lebesgue measurable if and only if
|E|e = |E|i .
(b) Assuming that nonmeasurable subsets of Rd exist (see Problem 1.29),
show that part (a) can fail if |E|e = ∞.
1.20. Given a set E ⊆ Rd with |E|e < ∞, show that the following statements
are equivalent.
(a) E is Lebesgue measurable.
(b) For each ε > 0 we can write E = (S ∪ N1 ) \ N2 where S is a union
of finitely many nonoverlapping boxes and |N1 |e , |N2 |e < ε.
(c) For each ε > 0 there exists a set S that is a finite union of boxes and
satisfies |E△S|e < ε.