Calculus of Real and Complex Variables

14.6 Representing a Radon Measure by Slicing

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

= μ. 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.

Proof: Let D_l consist of all functions which are of the form

∑ a yα (dist(y,B (0,l+ 1)C))n α |α|≤N α

where a_α ∈ ℚ, α is a multi-index, and n_α is a positive integer. Consider D≡∪_lD_l. Then D is countable. If f ∈ C_c

, then choose l large enough that spt ⊆ B, a locally compact space, f ∈ C₀. Then since D_l separates the points of B is closed with respect to conjugates, and annihilates no point, it is dense in C₀ by the Stone Weierstrass theorem. Alternatively, D is dense in C₀ by Stone Weierstrass and C_c is a subspace so it is also separable. So is C_c⁺, the nonnegative functions in C_c. ■

From the regularity of Radon measures, the following lemma follows.

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

= 0. Then it is also true that μ = 0. Hence 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

and then E_m ∈S_μ and μ = ν. Letting m →∞,E ∈S_μ and μ = ν. Similarly, S_μ ⊆S_ν and the two measures are equal on S_μ. ■

The main result in the section is the following theorem.

Proof: 

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.

Next consider the uniqueness of ν_x. Suppose ν_x and

_x satisfy all conclusions of the theorem with exceptional sets denoted by N and respectively. Then, enlarging N and , one may also assume, using Lemma 9.2.1, that for x N ∪, α > 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 x₀

N ∪,

----1-----∫ ∫ α(B (x0,r)) B (x0,r) ℝm XA (y )dνx (y)dα

∫ ∫ -----1----- = α (B (x0,r)) B(x0,r) ℝm XA (y)d^νx(y)dα,

and by the Lebesgue Besicovitch Differentiation theorem, there exists a set of α measure zero, E_A, such that if x₀

E_A ∪ 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 x₀

E ∪ N ∪ then ν_x0 = _x0 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, ν_x0 = _x0 for all V open. By Lemma 14.6.2 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.

For f ≥ 0, f,g ∈ C_c

and C_c respectively, define

∫ g → g (x )f (y)dμ ℝn+m

Since f ≥ 0, this is a positive linear functional on C_c

. Therefore, there exists a unique Radon measure ν_f such that for all g ∈ C_c,

∫ ∫ n+m g(x)f (y)dμ = n g (x )dνf. ℝ ℝ

I claim that ν_f ≪ α, the two being considered as measures on ℬ

. Suppose then that K is a compact set and α = 0. Then let K ≺ g ≺ V where V is open.

∫ ∫ ∫ νf (K ) = ℝn XK (x)dνf (x ) ≤ ℝn g (x)dνf (x ) = ℝn+m g (x )f (y)dμ

∫ ≤ 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, ν_f

= 0 also. By regularity considerations, ν_f ≪ 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 h_f ∈ L¹

such that for all g ∈ C_c,

∫ ∫ ∫ 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 ∈ C_c

to L¹ given by f → h_f is linear. However, this is not sufficiently specific because functions in L¹ are only determined a.e. However, for h_f ∈ L¹, you can specify a particular representative α a.e. By the fundamental theorem of calculus,

1 ∫ ^hf (x ) ≡ lrim→0 α(B-(x,r))-B (x,r)hf (z)dα (z) (14.27)

(14.27)

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,

is well defined pointwise for all x not in some set of measure zero Z_f. Since = h_f a.e. it follows that is well defined and will work in the formula 14.26. Let

Z = ∪{Zf : f ∈ D}

where D is a countable dense subset of C_c

⁺. 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 xZ. Thus, we will have x → defined by the above limit off Z and so, since = h_f a.e., it follows that

∫ ∫ ∫ ^ ℝn+m g(x)f (y )dμ = ℝn g(x)dνf = ℝn g(x)hf (x)dα

One could then take

to be defined as 0 for xZ.

For f an arbitrary function in C_c

⁺ and f^′∈D, a dense countable subset of C_c⁺, it follows from 14.26,

|∫ | ∫ || g(x)(hf (x)− hf′ (x))dα||≤ ||f − f′|| |g (x )|dμ | ℝn | ∞ ℝn+m

Let g_k

↑XB

(z,r)

where zZ. Then by the dominated convergence theorem, the above implies

| | ||∫ || ′ ∫ ′ ||B (z,r)(hf (x)− hf′ (x))dα|| ≤ ||f − f ||∞ B(z,r)×ℝm dμ = ||f − f ||∞ α(B (z,r)).

Dividing by α

, it follows that if α > 0 for all r > 0, then for all r > 0,

|| 1 ∫ || ||---------- (hf (x )− hf′ (x ))dα ||≤ ||f − f′||∞ |α (B (z,r)) B(z,r) |

It follows that for f ∈ C_c

⁺ arbitrary and zZ,

and since f^′ is arbitrary, it follows that the limit of 14.27 holds for all f ∈ C_c⁺ whenever zZ, the above set of measure zero.

Now for f an arbitrary real valued function of C_c

, simply apply the above result to positive and negative parts to obtain h_f ≡ h_f⁺ −h_f⁻ and ≡ −. Then it follows that for all f ∈ C_c and g ∈ C_c

∫ ∫ 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 → 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

∫ ^h (x) = f (y )dν (y). f ℝm x

It follows that

∫ ∫ ∫ g(x)f (y)dμ = g(x)f (y )dνx(y)dα(x) (14.28) ℝn+m ℝn ℝm

(14.28)

and x →∫ _ℝ^mf

dν_x is α measurable and ν_x is a Radon measure.

Now let f_k ↑X_ℝ^m and g ≥ 0. Then by monotone convergence theorem,

∫ ∫ ∫ g(x)dμ = g (x ) dν dα ℝn+m ℝn ℝm x

If g_k ↑X_ℝⁿ, the monotone convergence theorem shows that x →∫ _ℝ^mdν_x is L¹

.

Next let g_k ↑X_B

and use monotone convergence theorem to write

∫ ∫ ∫ α(B (x,r)) ≡ m dμ = m dνxdα B(x,r)×ℝ B(x,r) ℝ