Kenneth Kuttler

1088 CHAPTER 31. DIFFERENTIATION, RADON MEASURES

It follows that for f ∈Cc (Rm)+ arbitrary and z /∈ Z,

limsupr→0

1α (B(z,r))

∫B(z,r)

h f (x)dα− lim infr→0

1α (B(z,r))

∫B(z,r)

h f (x)dα

= limsupr→0

1α (B(z,r))

∫B(z,r)

(h f (x)−h f ′ (x)

)dα (x)

− lim infr→0

1α (B(z,r))

∫B(z,r)

(h f (x)−h f ′ (x)

)dα (x)

≤∣∣∣∣limsup

r→0

1α (B(z,r))

∫B(z,r)

(h f (x)−h f ′ (x)

)dα (x)

∣∣∣∣+

∣∣∣∣lim infr→0

1α (B(z,r))

∫B(z,r)

(h f (x)−h f ′ (x)

)dα (x)

∣∣∣∣≤ 2

∣∣∣∣ f − f ′∣∣∣∣

∞

and since f ′ is arbitrary, it follows that the limit of 31.2.9 holds for all f ∈Cc (Rm)+ when-ever z /∈ Z, the above set of measure zero.

Now for f an arbitrary real valued function of Cc (Rn) , simply apply the above result topositive and negative parts to obtain h f ≡ h f+ −h f− and ĥ f ≡ ĥ f+ − ĥ f− . Then it followsthat for all f ∈Cc (Rm) and g ∈Cc (Rm)∫

Rn+mg(x) f (y)dµ =

∫Rn

g(x) ĥ f (x)dα.

It is obvious from the description given above that for each x /∈ Z, the set of measure zerogiven above, that f → ĥ f (x) is a positive linear functional. It is clear that it acts like alinear map for nonnegative f and so the usual trick just described above is well definedand delivers a positive linear functional. Hence by the Riesz representation theorem, thereexists a unique νx such that for all x

ĥ f (x) =∫Rm

f (y)dνx (y) .

It follows that∫Rn+m

g(x) f (y)dµ =∫Rn

∫Rm

g(x) f (y)dνx (y)dα (x) (31.2.10)

and x→∫Rm f (y)dνx is α measurable and νx is a Radon measure.

Now let fk ↑XRm and g≥ 0. Then by monotone convergence theorem,∫Rn+m

g(x)dµ =∫Rn

g(x)∫Rm

dνxdα

If gk ↑XRn , the monotone convergence theorem shows that x→∫Rm dνx is L1 (α).

Next let gk ↑XB(x,r) and use monotone convergence theorem to write

α (B(x,r))≡∫

B(x,r)×Rmdµ =

∫B(x,r)

∫Rm

dνxdα

1088 CHAPTER 31. DIFFERENTIATION, RADON MEASURESIt follows that for f € C, (R”)* arbitrary and z ¢ Z,1—________ h x) da—lim inf ———_— |SUP OBEN) [. trod ro (Baar) I(x”hy (x)da= lim sup (B(z.r)) Love (hy (x) — hp (x)) da (x)1— lim inf Ber) Inter (hp (x) — Ip (x)) da (x)1< lim sup (Bin) in (hy (x) — hp (x)) da (x)1+ |lim inf Ban) Loan (hy (x) — hy (x)) da (x)< 2\|f-F'll.and since f’ is arbitrary, it follows that the limit of 31.2.9 holds for all f € C, (IR’”)* when-ever z ¢ Z, the above set of measure zero.Now for f an arbitrary real valued function of C, (IR") , simply apply the above result topositive and negative parts to obtain hy = h y+ —hy- and hy = h y+ —hy-. Then it followsthat for all f © C, (R”) and g EC, (R”)| g(x) f(y)du= | g(x)hp(x)da.Rem RaIt is obvious from the description given above that for each x ¢ Z, the set of measure zerogiven above, that f — hy (x) is a positive linear functional. It is clear that it acts like alinear map for nonnegative f and so the usual trick just described above is well definedand delivers a positive linear functional. Hence by the Riesz representation theorem, thereexists a unique Vx such that for all x“~h(x) = [ £(v)dvs(y).It follows thatI am XS (Y) dH = | ; | 8X) Fy) dVx (y) da (x) (31.2.10)and x — Jam f (y)dvVx is & measurable and vx is a Radon measure.Now let f; ¢ Zigm and g > 0. Then by monotone convergence theorem,Fons aH = [800 [avsaeIf gx + 2ign, the monotone convergence theorem shows that x —> fin dVx is L' (a).Next let gx ¢ 2% 5(x,) and use monotone convergence theorem to writea (B(x,r)) =/ du - | dVxdaB(x,r) xR” B(x,r) JR”