9.2 Fundamental Theorem Of Calculus For Radon Measures
In this section the Besicovitch covering theorem will be used to prove a differentiation theorem. The
Lebesgue fundamental theorem of calculus says essentially that for x off a set of measure zero and
f ∈ L^{1}
(ℝp)
1 ∫
rli→m0 m--(B(x,r)) |f (x)− f (y)|dmp = 0
p B(x,r)
In particular, for almost every x,
∫
lim -----1----- f (y)dm = f (x)
r→0 mp (B(x,r)) B(x,r) p
If you consider the fundamental theorem of calculus from calculus, it ends up reducing to the above and it
holds for any f continuous. However, the above pertains to a function which might not be continuous
anywhere. What follows will include this as a very special case. The reason this can be done is the
Besicovitch covering theorem. Define
Z ≡ {x ∈ ℝp : μ(B (x,r)) = 0 for some r > 0},
Lemma 9.2.1Z is measurable and μ
(Z )
= 0.
Proof:For each x ∈Z, there exists a ball B
(x,r)
with μ
(B (x,r))
= 0. Let C be the
collection of these balls. Since ℝ^{p} has a countable basis, a countable subset,
^
C
, of C also covers Z.
Let
^C = {Bi}∞i=1.
Then letting μ denote the outer measure determined by μ,
-- ∑∞ -- ∑∞
μ(Z) ≤ μ (Bi) = μ (Bi) = 0
i=1 i=1
Therefore, Z is measurable and has measure zero as claimed. ■
Let Mf : ℝ^{p}→
[0,∞ ]
by
{ sup ---1---∫ |f|dμ if x ∕∈ Z
M f (x) ≡ r≤1μ(B(x,r)) B (x,r) .
0 if x ∈ Z
I will write
[M f > ε]
instead of the longer
{x ∈ ℝp : M f (x ) > ε}
with other situations being
similar.
Theorem 9.2.2Let μ be a Radon measureand let f ∈ L^{1}