31.2. SLICING MEASURES 1087

It is obvious from the formula that the map from f ∈Cc (Rm) to L1 (α) given by f → h fis linear. However, this is not sufficiently specific because functions in L1 (α) are onlydetermined a.e. However, for h f ∈ L1 (α) , you can specify a particular representative α

a.e. By the fundamental theorem of calculus,

ĥ f (x)≡ limr→0

1α (B(x,r))

∫B(x,r)

h f (z)dα (z) (31.2.9)

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, ĥ f (x) iswell defined pointwise for all x not in some set of measure zero Z f . Since ĥ f = h f a.e. itfollows that ĥ f is well defined and will work in the formula 31.2.8. Let

Z = ∪{

Z f : f ∈D}

where D is a countable dense subset of Cc (Rm)+. Of course it is desired to have the limit31.2.9 hold for all f , not just f ∈D . We will show that this limit holds for all x /∈ Z. Thus,we will have x→ ĥ f (x) defined by the above limit off Z and so, since ĥ f (x) = h f (x) a.e.,it follows that

∫Rn+m

g(x) f (y)dµ =∫Rn

g(x)dν f =∫Rn

g(x) ĥ f (x)dα

One could then take ĥ f (x) to be defined as 0 for x /∈ Z.

For f an arbitrary function in Cc (Rm)+ and f ′ ∈ D , a dense countable subset ofCc (Rn)+ , it follows from 31.2.8,

∣∣∣∣∫Rng(x)

(h f (x)−h f ′ (x)

)dα

∣∣∣∣≤ ∣∣∣∣ f − f ′∣∣∣∣

∞

∫Rn+m|g(x)|dµ

Let gk (x) ↑ XB(z,r) (x) where z /∈ Z. Then by the dominated convergence theorem, theabove implies

∣∣∣∣∫B(z,r)

(h f (x)−h f ′ (x)

)dα

∣∣∣∣≤ ∣∣∣∣ f − f ′∣∣∣∣

∞

∫B(z,r)×Rm

dµ =∣∣∣∣ f − f ′

∣∣∣∣∞

α (B(z,r)) .

Dividing by α (B(z,r)) , it follows that if α (B(z,r))> 0 for all r > 0, then for all r > 0,

∣∣∣∣ 1α (B(z,r))

∫B(z,r)

(h f (x)−h f ′ (x)

)dα

∣∣∣∣≤ ∣∣∣∣ f − f ′∣∣∣∣

∞