1086 CHAPTER 31. DIFFERENTIATION, RADON MEASURES

where ai and bi are rational. Thus there are countably many such sets. Then from theconclusion of the theorem, if x0 /∈ N∪ N̂,

1α (B(x0,r))

∫B(x0,r)

∫Rm

XA (y)dνx (y)dα

=1

α (B(x0,r))

∫B(x0,r)

∫Rm

XA (y)dν̂x (y)dα,

and by the Lebesgue Besicovitch Differentiation theorem, there exists a set of α measurezero, EA, such that if x0 /∈ EA∪N∪ N̂, then the limit in the above exists as r→ 0 and yields

νx0 (A) = ν̂x0 (A).

Letting E denote the union of all the sets EA for A as described above, it follows that E isa set of measure zero and if x0 /∈ E ∪N ∪ N̂ then νx0 (A) = ν̂x0 (A) for all such sets A. Butevery open set can be written as a disjoint union of sets of this form and so for all suchx0, νx0 (V ) = ν̂x0 (V ) for all V open. By Lemma 31.2.2 this shows the two measures areequal and proves the uniqueness assertion for νx. It remains to show the existence of themeasures νx.

Existence of νx

For f ≥ 0, f ,g ∈Cc (Rm) and Cc (Rn) respectively, define

g→∫Rn+m

g(x) f (y)dµ

Since f ≥ 0, this is a positive linear functional on Cc (Rn). Therefore, there exists a uniqueRadon measure ν f such that for all g ∈Cc (Rn) ,∫

Rn+mg(x) f (y)dµ =

∫Rn

g(x)dν f .

I claim that ν f ≪ α, the two being considered as measures on B (Rn) . Suppose then thatK is a compact set and α (K) = 0. Then let K ≺ g≺V where V is open.

ν f (K) =∫Rn

XK (x)dν f (x)≤∫Rn

g(x)dν f (x) =∫Rn+m

g(x) f (y)dµ

≤∫Rm+m

XV×Rm (x,y) f (y)dµ ≤ || f ||∞

µ (V ×Rm) = ∥ f∥∞

α (V )

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 the existence of a function h f ∈ L1 (α)such that for all g ∈Cc (Rn) ,∫

Rn+mg(x) f (y)dµ =

∫Rn

g(x)dν f =∫Rn

g(x)h f (x)dα. (31.2.8)