9.1.2 n Dimensional Lebesgue Measure And Integrals
With Dynkin’s Lemma about π systems given above and the monotone convergence theorem,
it is possible to give a very elegant and fairly easy definition of the Lebesgue integral of a
function of n real variables. This is done in the following proposition.
Notation 9.1.2A set in ℝ^{n}is called F_{σ}if it is a countable union of compact sets. Aset is called G_{δ}if it is a countable intersection of open sets.
Proposition 9.1.3There exists a σ algebra ℱ^{n}of sets of ℝ^{n}whichcontains the opensets and a measure m_{n}defined on this σ algebra such that if f : ℝ^{n}→ [0,∞) ismeasurable with respect to ℱ^{n}then for any permutation
∫ ∫ ∫ ∫
⋅⋅⋅ X Cdx ⋅⋅⋅dx = ⋅⋅⋅ (X − X )dx ⋅⋅⋅dx .
Rp∩F i1 in Rp Rp∩F i1 in
Since R_{p}∈G, the iterated integrals on the right and hence on the left make sense. Then
continuing with the expression on the right and using that F ∈G,
_{i=1}^{∞} is a sequence of disjoint sets in G. Let F = ∪_{i=1}^{∞}F_{i}. I need to
show F ∈G. Since the sets are disjoint,
∫ ∫ ∫ ∫ ∞
⋅⋅⋅ X dx ⋅⋅⋅dx = ⋅⋅⋅ ∑ X dx ⋅⋅⋅dx
Rp∩F i1 in k=1 Rp∩Fk i1 in
∫ ∫ ∑N
= ⋅⋅⋅ lim XRp ∩Fkdxi1 ⋅⋅⋅dxin
N →∞ k=1
Do the iterated integrals make sense? Note that the iterated integral makes sense for
∑_{k=1}^{N}X_{Rp∩Fk} as the integrand because it is just a finite sum of functions for which the
iterated integral makes sense. Therefore,
∑∞
xi1 → XRp ∩Fk (x)
k=1
is measurable and by the monotone convergence theorem,
is measurable and so the limit of these functions,
∫ ∞
∑ X (x )dx
k=1 Rp∩Fk i1
is also measurable. Therefore, one can do another integral to this function. Continuing this
way using the monotone convergence theorem, it follows the iterated integral makes
sense. The same reasoning shows the iterated integral makes sense for any other
permutation.
Now applying the monotone convergence theorem as needed,
. Each set of the form ∏_{k=1}^{n}U_{k} where U_{k} is an open set is in K. Also
every open set in ℝ^{n} 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^{n} with the
norm
∥x∥∞ ≡ sup{|xi|,i = 1,⋅⋅⋅,n}
is a separable normed linear space. Hence, it is completely separable by Theorem 2.4.2 on
Page 62. Therefore, every open set is in ℱ^{n}. Thus σ
(K)
⊇ℬ
(ℝn)
, the Borel sets of
ℝ^{n}.
For F ∈ℱ^{n} define
∫ ∫
mn (F) ≡ lpim→∞ ⋅⋅⋅ XRp∩Fdxj1 ⋅⋅⋅dxjn
where
(j1,⋅⋅⋅,jn)
is a permutation of
{1,⋅⋅⋅,n}
. It doesn’t matter which one. It was shown
above they all give the same result. I need to verify m_{n} is a measure. Let
Applying the monotone convergence theorem repeatedly on the right, this yields that the
iterated integral makes sense and
∫ ∫ ∫
X dm = ⋅⋅⋅ X dx ⋅⋅⋅dx
ℝn F n F j1 jn
It follows 9.2 holds for every nonnegative simple function in place of f because these are
just linear combinations of functions, X_{F}. Now taking an increasing sequence of
nonnegative simple functions,
{sk}
which converges to a measurable nonnegative function
f
By Corollary 7.5.8 this measure is automatically inner and outer regular on every Borel
set. This is because it is finite on closed balls. ■
Next is the notion of the completion of a measure space. The idea is that the
one defined above is not complete. However, one can easily complete it. This is a
general idea so the following theorem is presented in a general setting. Recall that
(Ω, ℱ,λ)
is σ finite if there is a countable set
{Ωn }
_{n=1}^{∞} such that ∪_{n}Ω_{n} = Ω and
λ
(Ωn )
< ∞.
Theorem 9.1.4Let
(Ω,ℱ,μ)
be a measure space. Then there exists a measurespace,
(Ω,G,λ)
satisfying
(Ω,G,λ)
is a complete measure space.
λ = μ on ℱ
G⊇ℱ
For every E ∈G there exists G ∈ℱ such that G ⊇ E and μ
(G)
= λ
(E )
.
In addition to this, if
(Ω,ℱ, μ)
is σ finite, then the following approximation resultholds.
For every E ∈G there exists F ∈ℱ and G ∈ℱ such that F ⊆ E ⊆ G and
Denote by G the σ algebra of λ measurable sets. Then I claim that λ = μ on ℱ. It is clear
that λ ≤ μ on ℱ directly from the definition. Now if A ∈ℱ and λ
(A)
= ∞, there is
nothing to prove. Therefore, assume λ
(A )
< ∞ and suppose A ⊆ E such that E ∈ℱ
and
λ(A )+ ε > μ (E ) ≥ μ (A )
Then since ε is arbitrary, λ ≥ μ also.
From the definition, there exists E ⊇ S such that if λ
(S)
< ∞,
λ (S )+ ε > μ (E ) ≥ λ (S )
Let E_{n}⊇ S, be a decreasing sequence of sets of ℱ such that
λ(S) ≤ μ (E ) ≤ λ (S)+ 2−n
n
Now let G = ∩_{n=1}^{∞}E_{n}. It follows that G ⊇ S and λ
(S)
= μ
(G)
.
Why is ℱ⊆G? Letting λ
(S )
< ∞, (There is nothing to prove if λ
(S)
= ∞.) let G ∈ℱ be
such that G ⊇ S and λ
(S )
= μ
(G )
. Then if A ∈ℱ,
( C) ( C)
λ (S ) ≤ λ(S ∩A )+ λ(S ∩A )≤ λ(G ∩A )+ λ G ∩ A
= μ(G ∩A )+ μ G ∩ AC = μ (G ) = λ(S) .
Thus ℱ⊆G..
Finally suppose μ is σ finite. Let Ω = ∪_{n=1}^{∞}Ω_{n} where the Ω_{n} are disjoint sets of ℱ and
μ
(Ωn )
< ∞. Letting A ∈G, consider A_{n}≡ A ∩ Ω_{n}. From what was just shown, there exists
G_{n}⊇ A^{C}∩ Ω_{n}, G_{n}⊆ Ω_{n} such that μ
(Gn )
= λ
(AC ∩ Ωn )
.
PICT
Since μ
(Ωn )
< ∞, this implies
( ( )) ( )
λ Gn ∖ AC ∩ Ωn = λ (Gn )− λ AC ∩Ωn = 0.
Now G_{n}^{C}⊆ A ∪ Ω_{n} and so F_{n}≡ G_{n}^{C}∩ Ω_{n}⊆ A_{n} and λ
(An ∖ Fn)
=
λ (A∩ Ωn ∖(GC ∩ Ωn)) = λ (A ∩Ωn ∩ Gn) = λ(A ∩Gn ) = λ (Gn ∖ AC)
n ( ( C ))
≤ λ Gn ∖ A ∩ Ωn = 0.