For a given nonnegative integrable function, ℝn can be decomposed into a set where the
function is small and a set which is the union of disjoint cubes on which the average of
the function is under some control. The measure in this section will always be
Lebesgue measure on ℝn. This theorem depends on the Lebesgue theory of
differentiation.
Theorem 31.2.1Let f ≥ 0,∫fdx < ∞, and let α be a positive constant. Then thereexist sets F and Ω such that
n
ℝ = F ∪ Ω,F ∩ Ω = ∅ (31.2.3)
(31.2.3)
f (x) ≤ α a.e. on F (31.2.4)
(31.2.4)
Ω = ∪k=1∞Qkwhere the interiors of the cubes are disjoint and for each cube,Qk,
∫
--1--- n
α < m (Qk) Qk f (x)dx ≤ 2 α. (31.2.5)
(31.2.5)
Proof:Let S0 be a tiling of ℝn into cubes having sides of length M where M is
chosen large enough that if Q is one of these cubes, then
∫
--1-- fdm ≤ α. (31.2.6)
m (Q) Q
(31.2.6)
Suppose S0,
⋅⋅⋅
,Sm have been chosen. To get Sm+1, replace each cube of Sm by the 2n
cubes obtained by bisecting the sides. Then Sm+1 consists of exactly those cubes of Sm
for which 31.2.6 holds and let Tm+1 consist of the bisected cubes from Sm for which
31.2.6 does not hold. Now define
F ≡ {x : x is contained in some cube from Sm for all m} ,
Ω ≡ ℝn ∖F = ∪∞ ∪ {Q : Q ∈ T }
m=1 m
Note that the cubes from Tm have pair wise disjoint interiors and also the interiors of
cubes from Tm have empty intersections with the interiors of cubes of Tk if
k≠m.
Let x be a point of Ω and let x be in a cube of Tm such that m is the first index for
which this happens. Let Q be the cube in Sm−1 containing x and let Q∗ be the cube in
the bisection of Q which contains x. Therefore 31.2.6 does not hold for Q∗.
Thus
≤α
∫ ◜----◞◟∫----◝
α < --1--- f dx ≤-m-(Q) --1-- fdx ≤ 2nα
m (Q ∗) Q∗ m (Q ∗) m (Q ) Q
which shows Ω is the union of cubes having disjoint interiors for which 31.2.5
holds.
Now a.e. point of F is a Lebesgue point of f. Let x be such a point of F and suppose
x∈ Qk for Qk∈ Sk. Let dk≡ diameter of Qk. Thus dk→ 0.
∫ ∫
---1-- |f (y)− f (x)|dy ≤--1--- |f (y)− f (x)|dy
m (Qk) Qk m (Qk) B (x,dk)
∫
= m-(B(x,dk))----1------ |f (x) − f (y)|dy
m (Qk) m (B (x,dk)) B(x,dk)
1 ∫
≤ Kn m-(B-(x,d-))- |f (x) − f (y)|dy
k B(x,dk)
where Kn is a constant which depends on n and measures the ratio of the volume of a
ball with diamiter 2d and a cube with diameter d. The last expression converges to 0
because x is a Lebesgue point. Hence