I saw this material first in the book [10]. It can be presented as an application of the theory
of differentiation of Radon measures and the Riesz representation theorem for positive linear
functionals. It is an amazing theorem and can be used to understand conditional
probability.
Let μ be a finite Radon measure. I will show here that a formula of the following form
holds.
∫ ∫ ∫
μ(F) = dμ = X (x,y)dν (y)dα(x)
F ℝn ℝm F x
where α
(E )
= μ
(E × ℝm )
. 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. This can be tied to the
concept of independence or not of random variables. First here are two technical
lemmas.
Lemma 21.7.1The space C_{c}
m
(ℝ )
with the norm
||f|| ≡ sup {|f (y )| : y ∈ ℝm }
is separable.
Proof:Let D_{l} consist of all functions which are of the form
∑ α ( ( C))nα
aαy dist y,B (0,l+ 1)
|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}
(ℝn)
, 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}
(ℝn )
by Stone Weierstrass and
C_{c}
(ℝn)
is a subspace so it is also separable. So is C_{c}
(ℝn )
^{+}, the nonnegative functions in
C_{c}
(ℝn)
. ■
From the regularity of Radon measures, the following lemma follows.
Lemma 21.7.2If 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 21.7.3Let 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 of ℝ^{n}which satisfies
α (E) = μ(E × ℝm) (21.13)
(21.13)
for all E Borel. There also exists a Borel set of α measure zero N, such that for each x
∈∕
N,there exists a Radon probability measure ν_{x}such that if f is a nonnegative μ measurablefunction 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 (21.14)
(21.14)
and
( )
∫ ∫ ∫
ℝn+m f (x,y)dμ = ℝn ℝm f (x,y)dνx(y) dα (x). (21.15)
(21.15)
If
^ν
_{x }is any other collection of Radon measures satisfying 21.14and 21.15, then
^ν
_{x} = ν_{x}forα a.e.x.
Proof:
Existence and uniqueness of α
First consider the uniqueness of α. Suppose α_{1} is another Radon measure satisfying 21.13.
Then in particular, α_{1} and α agree on open sets and so the two measures are the same by
Lemma 21.7.2.
To establish the existence of α, define α_{0} on Borel sets by
m
α0(E) = μ(E × ℝ ).
Thus α_{0} is a finite Borel measure and so it is finite on compact sets. Lemma 22.3.9 on Page
1560 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 18.2.1, 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}
∫ ∫
= ----1------ XA (y)d^νx(y)dα,
α(B (x0,r)) B(x0,r) ℝm
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
νx0 (A) = ^νx0 (A).
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 21.7.2 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 → g (x)f (y)dμ
ℝn+m
Since f ≥ 0, this is a positive linear functional on C_{c}
(ℝn )
. Therefore, there exists a unique
Radon measure ν_{f} such that for all g ∈ C_{c}
(ℝn)
,
∫ ∫
ℝn+m g (x )f (y)dμ = ℝn g(x)dνf.
I claim that ν_{f}≪ α, the two being considered as measures on ℬ
Then for any ε > 0, one can choose V such that the right side is less than ε. Therefore,
ν_{f}
(K )
= 0 also. By regularity considerations, ν_{f}≪ α as claimed.
It follows from the Radon Nikodym theorem, either Theorem 18.5.2 or the more abstract
Radon Nikodym theorem which is presented later, there exists a function h_{f}∈ L^{1}
(α )
such
that for all g ∈ C_{c}
(ℝn)
,
∫ ∫ ∫
g(x)f (y)dμ = g (x )dνf = g (x )hf (x)dα. (21.16)
ℝn+m ℝn ℝn
(21.16)
It is obvious from the formula that the map from f ∈ C_{c}
(ℝm)
to L^{1}
(α )
given by f → h_{f} is
linear. However, this is not sufficiently specific because functions in L^{1}
(α )
are only
determined a.e. However, for h_{f}∈ L^{1}
(α )
, you can specify a particular representative α a.e.
By the fundamental theorem of calculus,
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 21.16.
Let
Z = ∪{Zf : f ∈ D }
where D is a countable dense subset of C_{c}
(ℝm )
^{+}. Of course it is desired to have the limit
21.17 hold for all f, not just f ∈D. We will show that this limit holds for all x
and since f^{′} is arbitrary, it follows that the limit of 21.17 holds for all f ∈ C_{c}
(ℝm )
^{+}
whenever z
∕∈
Z, the above set of measure zero.
Now for f an arbitrary real valued function of C_{c}
(ℝn )
, 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}
(ℝm )
and g ∈ C_{c}
(ℝm )
∫ ∫
g(x)f (y)dμ = g(x)^hf (x)dα.
ℝn+m ℝn
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