9.4 Differentiation Of Radon Measures
This section Involves differentiating a Radon measure with respect to another Radon measure. In this
section, B
will denote a ball with center
x and radius
r. Also, let
λ and
μ be Radon
measures and as above,
Z will denote a
μ measure zero set off of which
μ > 0 for all
r > 0
.
Definition 9.4.1 For x
Z, define the upper and lower symmetric derivatives as
--
D μλ(x) ≡ lim sup λ(B-(x,r)),D-μλ(x) ≡ lim inf λ-(B-(x,r)).
r→0 μ(B (x,r)) r→0 μ (B (x,r))
respectively. Also define
--
D μλ(x) ≡ Dμλ (x ) = Dμλ (x)
in the case when both the upper and lower derivatives are equal.
Lemma 9.4.2 Let λ and μ be Radon measures. If A is a bounded subset of
,
then
and if A is a bounded subset of
, then
Proof: Suppose first that A is a bounded subset of
{ -- }
x ∕∈ Z :Dμλ (x) ≥ a
, let
ε > 0
, and let
V be a bounded open set with
V ⊇ A and
λ − ε < λ ,μ − ε < μ . Then if
x ∈ A,
λ (B (x,r))
----------> a − ε,B (x,r) ⊆ V,
μ (B (x,r))
for infinitely many values of r which are arbitrarily small. Thus the collection of such balls constitutes a
Vitali cover for A. By Corollary 9.3.4 there is a disjoint sequence of these balls
such
that
μ(A ∖∪ ∞ B ) = 0. (9.13)
i=1 i
(9.13)
Therefore,
∑∞ ∞∑ --
(a− ε) μ (Bi) < λ(Bi) ≤ λ (V ) < ε +λ (A)
i=1 i=1
and so
∞∑ --
a μ(Bi) ≤ ε +εμ (V)+ λ (A )
i=1
≤ ε +ε (μ-(A)+ ε)+ λ(A ) (9.14)
Now
μ(A ∖∪ ∞i=1Bi)+ μ-(∪ ∞i=1Bi) ≥ μ(A)
and so by 9.13 and the fact the B _{i} are disjoint, it follows from 9.14 ,
-- -- ∞ ∑∞
aμ(A) ≤ aμ (∪i=1Bi) = a μ (Bi)
-- i=1-
≤ ε + ε(μ(A)+ ε)+ λ (A ). (9.15)
Hence
a μ ≤ λ since
ε > 0 was arbitrary
.
Now suppose A is a bounded subset of
and let
V be a bounded open set
containing
A with
μ − ε < μ . Then if
x ∈ A,
λ(B (x,r))
---------< a + ε,B(x,r) ⊆ V
μ(B (x,r))
for values of r which are arbitrarily small. Therefore, by Corollary 9.3.4 again, there exists a disjoint
sequence of these balls,
satisfying this time,
Then by arguments similar to the above,
-- ∞∑ --
λ(A) ≤ λ(Bi) < (a+ ε)μ (V) < (a +ε)(μ (A )+ ε).
i=1
Since ε was arbitrary, this proves the lemma. ■
Theorem 9.4.3 There exists a set of measure zero N containing Z such that for x
N, D _{μ} λ
exists and also X _{NC} D _{μ} λ is a μ measurable function. Furthermore, D _{μ} λ < ∞ μ a.e.
Proof: First I show D _{μ} λ
exists a.e. Let 0
≤ a < b < ∞ and let
A be any bounded subset
of
N (a,b) ≡ {x ∕∈ Z : D-λ (x) > b > a > D λ(x)}.
μ --μ
By Lemma 9.4.2 ,
-- -- --
aμ(A ) ≥ λ(A) ≥ bμ(A)
and so μ
= 0 and
A is
μ measurable. It follows
μ = 0 because
∑∞
μ (N (a,b)) ≤ μ(N (a,b)∩ B (0,m)) = 0.
m=1
Define
{ -- }
N0 ≡ x ∕∈ Z :D μλ(x) > D-μλ(x) .
Thus μ
= 0 because
N0 ⊆ ∪ {N (a,b) : 0 ≤ a < b, and a,b ∈ ℚ }
Therefore, N _{0} is also μ measurable and has μ measure zero. Letting N ≡ N _{0} ∪ Z, it follows D _{μ} λ
exists
on
N ^{C} . It remains to verify
X _{NC} D _{μ} λ is finite a.e. and is
μ measurable.
Let
Then by Lemma 9.4.2
--
λ (I ∩B (0,m )) ≥ aμ(I ∩ B (0,m))
for all a and since λ is finite on bounded sets, the above implies μ
= 0 for each
m which
implies that
I is
μ measurable and has
μ measure zero since
Letting η be an arbitrary Radon measure, let r > 0, and suppose η
= 0
. (Since
η is finite on
every ball, there are only countably many
r such that
η > 0
. ) and let
V be an open set
containing
B . Then whenever
y is close enough to
x , it follows that
B is also a subset of
V.
Thus
η(V) ≥ lim sup η(B (y,r))
y→x
Since V is an arbitrary open set containing B
, it follows
(------)
η(B (x,r)) = η B (x,r) ≥ lim syu→px η(B (y,r))
and so y → η
an upper semicontinuous real valued function of
x , one which satisfies
f (x) ≥ lim sup f (xn )
n→∞
whenever x _{n} → x . Now it is routine to verify that a function f is upper semicontinuous if and only if
f ^{−1}
is open for all
a ∈ ℝ . Therefore,
f ^{−1} is a Borel set for all
a ∈ ℝ and
so
f is Borel measurable by Lemma
5.1.2 . Now the measurability of
X _{NC} D _{μ} λ follows
from
λ(B (x,ri))
XNC (x)D μλ(x) = lrim→0 ----------XNC (x)
i μ(B (x,ri))
where r _{i} is such that ∂B
has
μ and
λ measure zero.
■
Typically I will write D _{μ} λ
rather than the more precise
X _{NC} D _{μ} λ since the
values on the set of measure zero
N are not important due to the completeness of the measure
μ .