I saw this material first in the book [46]. 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 in
complete generality rather than through a tedious concatenation of special cases and dubious
assumptions.
Let μ be a finite Radon measure. I will show here that a formula of the following form
holds.
∫ ∫ ∫
μ(F) = dμ = XF (x,y)dνx(y)dα (x)
F ℝn ℝm
where α
(E )
= μ
(E × ℝm )
. When this is done, the measures, νx, 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, νx. 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 14.6.1The space Cc
(ℝm )
with the norm
m
||f|| ≡ sup {|f (y)| : y ∈ ℝ }
is separable.
Proof:Let Dl consist of all functions which are of the form
∑ a yα (dist(y,B (0,l+ 1)C))n α
|α|≤N α
where aα∈ ℚ, α is a multi-index, and nα is a positive integer. Consider D≡∪lDl. Then D is
countable. If f ∈ Cc
n
(ℝ )
, then choose l large enough that spt
(f)
⊆ B
(0,l+ 1)
, a locally compact
space, f ∈ C0
(B (0,l +1))
. Then since Dl separates the points of B
(0,l+ 1)
is closed with
respect to conjugates, and annihilates no point, it is dense in C0
(B (0,l+ 1))
by the Stone
Weierstrass theorem. Alternatively, D is dense in C0
n
(ℝ )
by Stone Weierstrass and Cc
n
(ℝ )
is a subspace so it is also separable. So is Cc
n
(ℝ )
+, the nonnegative functions in Cc
n
(ℝ )
.
■
From the regularity of Radon measures, the following lemma follows.
Lemma 14.6.2If μ and ν are two Radon measures defined on σ algebras, Sμand Sν, of subsetsof ℝnand if μ
(V )
= ν
(V )
for all V open, then μ = ν and Sμ = Sν.
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 Gδ and Fσ set. (Recall Gδ sets are
countable intersections of open sets and Fσ sets are countable unions of closed sets.) Now suppose E ∈Sν
is a bounded set. Then by regularity of ν there exists G a Gδ set and F, an Fσ 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 ∈Sμ
and
μ(E ) = μ(F ) = ν(F ) = ν(E ).
If E ∈Sν not necessarily bounded, let Em = E ∩ B
(0,m )
and then Em∈Sμ and μ
(E )
m
= ν
(E )
m
.
Letting m →∞,E ∈Sμ and μ
(E )
= ν
(E)
. Similarly, Sμ⊆Sν and the two measures are equal on Sμ.
■
The main result in the section is the following theorem.
Theorem 14.6.3Let μ be a finite Radon measure on ℝn+mdefined on a σ algebra, ℱ. Thenthere exists a unique finite Radon measure α, defined on a σ algebra S, of sets of ℝnwhichsatisfies
α(E) = μ(E × ℝm ) (14.23)
(14.23)
for all E Borel. There also exists a Borel set of α measure zero N, such that for each x
∈∕
N, there exists aRadon probability measure νxsuch that if f is a nonnegative μ measurable function or a μ measurablefunction in L1
(μ)
,
y → f (x,y) is νx measurable α a.e.
∫
x → f (x,y)dν (y) is α measurable (14.24)
ℝm x
(14.24)
and
∫ ∫ (∫ )
n+m f (x,y )dμ = n m f (x,y)dνx(y) dα(x). (14.25)
ℝ ℝ ℝ
(14.25)
If
^ν
xis any other collection of Radon measures satisfying 14.24and 14.25, then
^ν
x = νxfor α a.e.x.
Proof:
Existence and uniqueness of α
First consider the uniqueness of α. Suppose α1 is another Radon measure satisfying 14.23. Then in
particular, α1 and α agree on open sets and so the two measures are the same by Lemma
14.6.2.
To establish the existence of α, define α0 on Borel sets by
α0(E ) = μ(E × ℝm ).
Thus α0 is a finite Borel measure and so it is finite on compact sets. Lemma 14.5.11 on Page 1163 implies
the existence of the Radon measure α extending α0.
Uniqueness of νx
Next consider the uniqueness of νx. Suppose νxand
^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 9.2.1, that for x
∕∈
N ∪
^
N
, α
(B (x,r))
> 0 whenever r > 0. Now
let
∏m
A = (ai,bi]
i=1
where ai and bi are rational. Thus there are countably many such sets. Then from the conclusion of the
theorem, if x0
∈∕
N ∪
^
N
,
----1-----∫ ∫
α(B (x0,r)) B (x0,r) ℝm XA (y )dνx (y)dα
and by the Lebesgue Besicovitch Differentiation theorem, there exists a set of α measure zero, EA, such
that if x0
∕∈
EA∪ 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 EA for A as described above, it follows that E is a
set of measure zero and if x0
∕∈
E ∪ N ∪
N^
then νx0
(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 x0,
νx0
(V )
=
^ν
x0
(V )
for all V open. By Lemma 14.6.2 this shows the two measures are equal
and proves the uniqueness assertion for νx. It remains to show the existence of the measures
νx.
Existence of νx
For f ≥ 0, f,g ∈ Cc
(ℝm )
and Cc
(ℝn)
respectively, define
∫
g → g (x )f (y)dμ
ℝn+m
Since f ≥ 0, this is a positive linear functional on Cc
(ℝn )
. Therefore, there exists a unique Radon measure
νf such that for all g ∈ Cc
(ℝn)
,
∫ ∫
n+m g(x)f (y)dμ = n g (x )dνf.
ℝ ℝ
I claim that νf≪ α, the two being considered as measures on ℬ
∫
≤ XV ×ℝm (x,y) f (y)dμ ≤ ||f|| μ(V × ℝm ) = ∥f∥ α (V )
ℝm+m ∞ ∞
Then for any ε > 0, one can choose V such that the right side is less than ε. Therefore, νf
(K )
= 0 also. By
regularity considerations, νf≪ α as claimed.
It follows from the Radon Nikodym theorem, either Theorem 9.5.2 or the more abstract Radon
Nikodym theorem which is presented later, there exists a function hf∈ L1
(α)
such that for all
g ∈ Cc
(ℝn)
,
∫ ∫ ∫
g(x)f (y)dμ = g (x)dν = g(x)h (x)dα. (14.26)
ℝn+m ℝn f ℝn f
(14.26)
It is obvious from the formula that the map from f ∈ Cc
m
(ℝ )
to L1
(α)
given by f → hf is linear.
However, this is not sufficiently specific because functions in L1
(α)
are only determined a.e. However, for
hf∈ L1
(α)
, you can specify a particular representative α a.e. By the fundamental theorem of
calculus,
exists off some set of measure zero Zf. Note that since this involves the integral over a ball, it does not
matter which representative of hf is placed in the formula. Therefore,
^hf
(x)
is well defined pointwise for all
x not in some set of measure zero Zf. Since
^
hf
= hf a.e. it follows that
^
hf
is well defined and will work in
the formula 14.26. Let
Z = ∪{Zf : f ∈ D}
where D is a countable dense subset of Cc
(ℝm )
+. Of course it is desired to have the limit 14.27
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 14.27 holds for all f ∈ Cc
(ℝm)
+ whenever z
∕∈
Z, the
above set of measure zero.
Now for f an arbitrary real valued function of Cc
(ℝn )
, simply apply the above result to positive and
negative parts to obtain hf≡ hf+−hf− and
^hf
≡
^hf+
−
^hf−
. Then it follows that for all f ∈ Cc
(ℝm )
and
g ∈ Cc
(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 νx such that for all
x