where each μ_{i} = m_{1} and ℱ_{i}≡ℱ, the σ algebra of
Lebesgue measurable sets on ℝ, and let ℱ^{p} denote the product measurable sets ℱ×
⋅⋅⋅
×ℱ. Thus
using the typical notation from calculus rather than m_{1}, if f is a nonnegative ℱ^{p} measurable
function,
∫ ∫ ∫
p fdmp = ⋅⋅⋅ f (x1,⋅⋅⋅,xp)dxi1 ⋅⋅⋅dxip
ℝ ℝ ℝ
also m_{p}
∏p
( i=1Ei)
= ∏_{i=1}^{p}m_{1}
(Ei)
whenever E_{i} is Lebesgue measurable. Then one can obtain various
regularity properties of this measure space.
Proposition 8.1.1ℱ^{p}contains the Borel sets and for every Borel set E, there exist F an F_{σ}set and Ga G_{δ}set such that
F ⊆ E ⊆ G
and m_{p}
(G)
= m_{p}
(E )
= m_{p}
(F)
. In addition to this, the measure is translation invariant. Thatis
mp (E) = mp(x + E)
for all E a Borel set.
Proof: Each set of the form ∏_{k=1}^{p}U_{k} where U_{k} is an open set is in K. Also every open set in ℝ^{p} is a
countable union of open sets of this form. In fact, one can assume each U_{k} is an open interval. This follows
from the observation that ℝ^{p} with the norm
∥x∥∞ ≡ sup {|xi|,i = 1,⋅⋅⋅,p}
is a separable normed linear space. Hence, it is completely separable by Theorem 2.1.12 on Page 113.
Therefore, every open set is in ℱ^{p}. Thus σ
(K)
⊇ℬ
(ℝp)
, the Borel sets of ℝ^{p} because ℬ
(ℝp)
is by
definition the smallest σ algebra which contains the open sets.
By Corollary 5.4.7 this measure is automatically inner and outer regular on every Borel set. This is
because it is finite on closed balls. This corollary implies that if E is Borel, then there is F an F_{σ} set
contained in E and a G_{δ} set G containing E such that
mp(F ) = mp (E) = mp (G)
The translation invariance is obvious for any set of the form ∏_{i=1}^{p}
(ai,bi)
. Open rectangles like this
obviously form a pi system and by earlier results all open sets are countable unions of these so the σ
algebra generated by these open rectangles is the Borel sets. Now letting K denote the set of open
rectangles, let G be those Borel sets E such that for each k ∈ ℕ,
mp(x + E ∩Rk ) = mp (E ∩ Rk ) (8.1)
(8.1)
and there are no measurability issues. Here
∏p
Rk ≡ (− k,k)
i=1
Then G⊇K as just noted. If E ∈G, what about E^{C}?
( C )
mp (Rk ) = mp (x( + Rk) = mp) x + E ∩ Rk + mp (x+ E ∩ Rk)
= mp x+ EC ∩ Rk +mp (E ∩ Rk)
As to countable disjoint unions of sets of G if one considers the translations of these sets by x the resulting
translated sets are still disjoint. Hence it is clear that countable disjoint unions of G remain in G. Thus by
Dynkin’s lemma, G = ℬ the Borel sets. Now let k →∞ in 8.1 and use the properties of measures to obtain
the conclusion. ■
The Borel sets with the Lebesgue measure just described are mostly sufficient for this book but for the
sake of completeness (pun intended), we give the following definition.
Definition 8.1.2Let ℱ_{p}denote the completion of the Borel sets with respect to Lebesgue measurejustdescribed. Thus, for ℬ
(ℝp)
the Borel sets of ℝ^{p},
mˆp (F) ≡ inf{mp (E) : E ⊇ F and E ∈ ℬ (ℝp)}
Then ℱ_{p}consists of all sets of ℝ^{p}, F such that for all S ⊆ ℝ^{p},
is a complete measure space which contains the Borel sets. To save onnotation, I will use m_{p}for either of these measures.
Proposition 8.1.3For E ∈ℱ_{p}, there exist F an F_{σ}set and G a G_{δ}set such that
F ⊆ E ⊆ G
and m_{p}
(F)
= m_{p}
(E)
= m_{p}
(G )
. If f ∈ L^{1}
(ℝp,ℱp,mp)
, then there exists g ∈ L^{1}
(ℝp,ℬ (ℝp),mp)
such that
|g(x)|
≤
|f (x)|
and g
(x)
= f
(x)
for a.e. x.
Proof:Let Ê∈ℬ
(ℝp )
such that Ê⊇ E and m_{p}
(E)
= m_{p}
( )
Eˆ
. Then by the regularity part of
Proposition 8.1.3, there exists G a G_{δ} set such that G ⊇Ê and m_{p}
(G)
= m_{p}
( )
ˆE
. Then G ⊇ E and
m_{p}
(E)
= m_{p}
(ˆE)
= m_{p}
(G )
.
Next consider E^{C} and first suppose m_{p}
(E )
< ∞. E^{C} is also a Borel set and by what was just
shown, there is G a G_{δ} set such that G ⊇ E^{C} and m_{p}
(G)
= m_{p}
( )
EC
. Then G^{C}⊆ E and
G^{C} is a countable union of closed sets. Then m_{p}
( )
E ∖GC
= m_{p}
( )
G∖ EC
= 0 and so in this
case,
( C) ( C) ( C)
mp(E ) = mp G + mp E ∖ G = mp G
Now G^{C} is a countable union of closed sets. Each of these closed sets is a countable union of compact sets
so G^{C} is a countable union of compact sets.
The remaining case is m_{p}
(E )
= ∞. Let E_{k} =
[− k,k]
^{p}∩E so each E_{k} has finite measure. By what was
just shown, there is a F_{σ} set F_{k}⊆ E_{k} such that m_{p}
(Fk)
= m_{p}
(Ek)
so m_{p}
(Ek ∖Fk )
= 0. Then F ≡∪_{k}F_{k}
is an F_{σ} set and it is contained in E.
This proves the first part. For the remaining part, it suffices to consider only f
(x)
≥ 0 because you can
reduce to positive and negative parts of real and imaginary parts of f. By Theorem 5.1.9, there is an
increasing sequence of simple functions s_{k} such that for all x,
sk (x ) ↑ f (x )
Now for s_{k}
(x )
= ∑_{i=1}^{mk}a_{
i}^{k}X_{
Eik},(a_{i}^{k}> 0) replace each E_{
i}^{k} with F_{
i}^{k} an F_{
σ} set having the same
measure as E_{i}^{k} but contained in it. Each m_{p}
( )
Eki
< ∞ because f is given to be in L^{1}. Let
Nˆ
≡∪_{k=1}^{∞}∪_{i=1}^{mk}
( )
Eki ∖ Fki
a set of measure zero. Thus there exists a Borel set N ⊇
ˆN
which also has
measure zero. Let ŝ_{k}
(x)
= s_{k}
(x)
off N and let ŝ
(x)
= 0 on N. Thus ŝ_{k}
(x)
is an increasing function which
converges to f
(x)
off the set of measure zero N and converges to 0 on N. Each ŝ_{k} is Borel measurable and
so letting g be the pointwise limit, it follows from Corollary 5.1.3 that g is Borel measurable and 0 ≤ g ≤ f.
■
Another useful and yet simple geometric observation is in the following lemma which says you can write
any open set as the union of half open disjoint rectangles.
Lemma 8.1.4Every open set in ℝ^{p}is the countable disjoint union of half open boxes of theform
p∏
(ai,ai + 2−k]
i=1
where a_{i} = l2^{−k}for some integers, l,k where k ≥ m. If ℬ_{m}denotes this collection of half open boxes, thenevery box of ℬ_{m+1}is contained in a box of ℬ_{m}or equals a box of ℬ_{m}.
Proof: Let m ∈ ℕ be given and let k ≥ m.
p∏
Ck = {All half open boxes (ai,ai + 2−k] where
i=1
ai = l2−k for some integer l.}
Thus C_{k} consists of a countable disjoint collection of boxes whose union is ℝ^{p}. This is sometimes called a
tiling of ℝ^{p}. Think of tiles on the floor of a bathroom and you will get the idea. Note that each box has
diameter no larger than 2^{−k}
√p
. This is because if we have two points,
∏p − k
x,y ∈ (ai,ai +2 ],
i=1
then
|xi − yi|
≤ 2^{−k}. Therefore,
(∑p )1∕2
|x − y| ≤ (2−k)2 = 2− k√p.
i=1
Also, a box of C_{k+1} is either contained in a box of C_{k} or it has empty intersection with this box of
C_{k}.
Let U be open and let ℬ_{1}≡ all sets of C_{1} which are contained in U. If ℬ_{1},
⋅⋅⋅
,ℬ_{k} have been chosen,
ℬ_{k+1}≡ all sets of C_{k+1} contained in
( k )
U ∖ ∪ ∪ i=1ℬi .
Let ℬ_{∞} = ∪_{i=1}^{∞}ℬ_{i}. We claim ∪ℬ_{∞} = U. Clearly ∪ℬ_{∞}⊆ U because every box of every ℬ_{i} is contained in
U. If p ∈ U, let k be the smallest integer such that p is contained in a box from C_{k} which is also a subset of
U. Thus
p ∈ ∪ ℬk ⊆ ∪ ℬ∞.
Hence ℬ_{∞} is the desired countable disjoint collection of half open boxes whose union is U. The last claim
follows from the construction. ■