. When this is done, the measures, ν_{x}, are called slicing
measures and this shows that an integral with respect to μ can be written as an iterated
integral in terms of the measure α and the slicing measures, ν_{x}. This is like going
backwards in the construction of product measure. One starts with a measure μ, defined
on the Cartesian product and produces α and an infinite family of slicing measures from
it whereas in the construction of product measure, one starts with two measures and
obtains a new measure on a σ algebra of subsets of the Cartesian product of two
spaces. These slicing measures are dependent on x. Later, this will be tied to the
concept of independence or not of random variables. First here are two technical
lemmas.
Lemma 29.3.1The space C_{c}
(ℝm )
with the norm
||f|| ≡ sup{|f (y)| : y ∈ ℝm }
is separable.
Proof:Let D_{l} consist of all functions which are of the form
∑ ( ( ) )
aαyα dist y,B (0,l+ 1)C nα
|α|≤N
where a_{α}∈ ℚ, α is a multi-index, and n_{α} is a positive integer. Consider D≡∪_{l}D_{l}. Then
D is countable. If f ∈ C_{c}
n
(ℝ )
, then choose l large enough that spt
(f )
⊆ B
(0,l+ 1)
, a
locally compact space, f ∈ C_{0}
(B (0,l+ 1))
. Then since D_{l} separates the points of
B
(0,l +1)
is closed with respect to conjugates, and annihilates no point, it is dense in
C_{0}
(B (0,l+ 1))
by the Stone Weierstrass theorem. Alternatively, D is dense in C_{0}
n
(ℝ )
by Stone Weierstrass and C_{c}
n
(ℝ )
is a subspace so it is also separable. So is C_{c}
n
(ℝ )
^{+},
the nonnegative functions in C_{c}
n
(ℝ )
. ■
From the regularity of Radon measures, the following lemma follows.
Lemma 29.3.2If μ and ν are two Radon measures defined on σ algebras, S_{μ}andS_{ν}, of subsets of ℝ^{n}and if μ
(V )
= ν
(V )
for all V open, then μ = ν and S_{μ} = S_{ν}.
Proof: Every compact set is a countable intersection of open sets so the two measures
agree on every compact set. Hence it is routine that the two measures agree on every G_{δ}
and F_{σ} set. (Recall G_{δ} sets are countable intersections of open sets and F_{σ} sets are
countable unions of closed sets.) Now suppose E ∈S_{ν} is a bounded set. Then by
regularity of ν there exists G a G_{δ} set and F, an F_{σ} set such that F ⊆ E ⊆ G and
ν
(G ∖ F)
= 0. Then it is also true that μ
(G ∖ F)
= 0. Hence E = F ∪
(E ∖F)
and E ∖F
is a subset of G ∖ F, a set of μ measure zero. By completeness of μ, it follows E ∈S_{μ}
and
μ (E ) = μ (F ) = ν (F ) = ν(E ).
If E ∈S_{ν} not necessarily bounded, let E_{m} = E ∩ B
(0,m )
and then E_{m}∈S_{μ} and
μ
(Em )
= ν
(Em )
. Letting m →∞,E ∈S_{μ} and μ
(E )
= ν
(E )
. Similarly, S_{μ}⊆S_{ν} and
the two measures are equal on S_{μ}.
The main result in the section is the following theorem.
Theorem 29.3.3Let μ be a finite Radon measure on ℝ^{n+m}defined on a σ algebra, ℱ.Then there exists a unique finite Radon measure α, defined on a σ algebra S, of sets ofℝ^{n}which satisfies
α(E ) = μ(E × ℝm ) (29.3.12)
(29.3.12)
for all E Borel. There also exists a Borel set of α measure zero N, such that for eachx
∕∈
N, there exists a Radon probability measure ν_{x}such that if f is a nonnegative μmeasurable function or a μ measurable function in L^{1}
(μ)
,
y → f (x,y) is νx measurable α a.e.
∫
x → ℝm f (x,y)dνx (y) is α measurable (29.3.13)
(29.3.13)
and
∫ ∫ ( ∫ )
f (x,y)dμ = f (x,y )dν (y) dα (x) . (29.3.14)
ℝn+m ℝn ℝm x
(29.3.14)
If
^ν
_{x }is any other collection of Radon measures satisfying 29.3.13and 29.3.14, then
^ν
_{x} = ν_{x}for α a.e.x.
Proof:
Existence and uniqueness of α
First consider the uniqueness of α. Suppose α_{1} is another Radon measure satisfying
29.3.12. Then in particular, α_{1} and α agree on open sets and so the two measures are the
same by Lemma 29.3.2.
To establish the existence of α, define α_{0} on Borel sets by
α0(E ) = μ(E × ℝm ).
Thus α_{0} is a finite Borel measure and so it is finite on compact sets. Lemma
12.2.3 on Page 1087 implies the existence of the Radon measure α extending
α_{0}.
Uniqueness of ν_{x}
Next consider the uniqueness of ν_{x}. Suppose ν_{x }and
^ν
_{x} satisfy all conclusions of the
theorem with exceptional sets denoted by N and
N^
respectively. Then, enlarging N and
N^
, one may also assume, using Lemma 29.2.1, that for x
∕∈
N ∪
^N
, α
(B (x,r))
> 0
whenever r > 0. Now let
∏m
A = (ai,bi]
i=1
where a_{i} and b_{i} are rational. Thus there are countably many such sets. Then from the
conclusion of the theorem, if x_{0}
and by the Lebesgue Besicovitch Differentiation theorem, there exists a set of α measure
zero, E_{A}, such that if x_{0}
∈∕
E_{A}∪ N ∪
N^
, then the limit in the above exists as r → 0 and
yields
ν (A ) = ^ν (A).
x0 x0
Letting E denote the union of all the sets E_{A} for A as described above, it follows that E
is a set of measure zero and if x_{0}
∕∈
E ∪N ∪
^
N
then ν_{x0}
(A )
=
^ν
_{x0}
(A)
for all such sets A.
But every open set can be written as a disjoint union of sets of this form and so for all
such x_{0}, ν_{x0}
(V)
=
^ν
_{x0}
(V )
for all V open. By Lemma 29.3.2 this shows the two measures
are equal and proves the uniqueness assertion for ν_{x}. It remains to show the existence of
the measures ν_{x}.
Existence of ν_{x}
For f ≥ 0, f,g ∈ C_{c}
(ℝm )
and C_{c}
(ℝn)
respectively, define
∫
g → ℝn+m g(x)f (y)dμ
Since f ≥ 0, this is a positive linear functional on C_{c}
n
(ℝ )
. Therefore, there exists a
unique Radon measure ν_{f} such that for all g ∈ C_{c}
n
(ℝ )
,
∫ ∫
g(x)f (y)dμ = g (x)dνf.
ℝn+m ℝn
I claim that ν_{f}≪ α, the two being considered as measures on ℬ
exists off some set of measure zero Z_{f}. Note that since this involves the integral over a
ball, it does not matter which representative of h_{f} is placed in the formula. Therefore,
^
hf
(x )
is well defined pointwise for all x not in some set of measure zero Z_{f}. Since
^
hf
= h_{f} a.e. it follows that
^
hf
is well defined and will work in the formula 29.3.15.
Let
Z = ∪{Zf : f ∈ D}
where D is a countable dense subset of C_{c}
m
(ℝ )
^{+}. Of course it is desired to have the
limit 29.3.16 hold for all f, not just f ∈D. We will show that this limit holds for all
x
∕∈
Z. Thus, we will have x →
^hf
(x)
defined by the above limit off Z and so, since
^hf
(x )
= h_{f}
(x)
a.e., it follows that
∫ ∫ ∫
g (x)f (y)dμ = g (x)dνf = g (x) ^hf (x)dα
ℝn+m ℝn ℝn
One could then take
^hf
(x)
to be defined as 0 for x
∈∕
Z.
For f an arbitrary function in C_{c}
(ℝm )
^{+} and f^{′}∈D, a dense countable subset of
C_{c}
and since f^{′} is arbitrary, it follows that the limit of 29.3.16 holds for all f ∈ C_{c}
m
(ℝ )
^{+}
whenever z
∕∈
Z, the above set of measure zero.
Now for f an arbitrary real valued function of C_{c}
n
(ℝ )
, simply apply the above result
to positive and negative parts to obtain h_{f}≡ h_{f+}− h_{f−} and
^
hf
≡
^
hf+
−
^
hf−
. Then it
follows that for all f ∈ C_{c}
m
(ℝ )
and g ∈ C_{c}
m
(ℝ )
∫ ∫
n+m g (x )f (y)dμ = n g(x)^hf (x)dα.
ℝ ℝ
It is obvious from the description given above that for each x
∈∕
Z,the set of
measure zero given above, that f →
^
hf
(x)
is a positive linear functional. It is
clear that it acts like a linear map for nonnegative f and so the usual trick just
described above is well defined and delivers a positive linear functional. Hence by
the Riesz representation theorem, there exists a unique ν_{x} such that for all
x