Calculus of Real and Complex Variables

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 ∈ L¹

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},

Proof: For each x ∈ Z, there exists a ball B

with μ = 0. Let C be the collection of these balls. Since ℝ^p has a countable basis, a countable subset, , 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 →

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

instead of the longer with other situations being similar.

Proof: First consider the following claim which is called a weak type estimate.

Claim 1: The following inequality holds for N_p the constant of the Besicovitch covering theorem. 

-- −1 μ ([M f > ε]) ≤ Npε ||f||1

Proof: First note

∩Z = ∅ and without loss of generality, you can assume μ > 0. Next, for each x ∈  there exists a ball B_x = B with r_x ≤ 1 and

∫ μ (B )−1 |f|dμ > ε. x B (x,rx)

Let ℱ be this collection of balls so that

is the set of centers of balls of ℱ. By the Besicovitch covering theorem,

[M f > ε] ⊆ ∪Ni=p1 {B : B ∈ Gi}

where G_i is a collection of disjoint balls of ℱ. Now for some i,

μ([M f > ε])∕Np ≤ μ(∪{B : B ∈ Gi})

because if this is not so, then

a contradiction. Therefore for this i, This shows Claim 1.

Claim 2: If g is any continuous function defined on ℝ^p, then for x

Z,

∫ lim ----1----- |g(y)− g (x)|dμ(y) = 0 r→0 μ(B (x,r)) B (x,r)

and

1 ∫ rli→m0 μ(B-(x,r)) g (y )dμ(y) = g(x). (9.8) B (x,r)

(9.8)

Proof: Since g is continuous at x, whenever r is small enough,

∫ ∫ ----1----- |g (y )− g(x)|dμ (y) ≤----1---- εdμ (y) = ε. μ (B (x,r)) B(x,r) μ (B (x,r)) B(x,r)

9.8 follows from the above and the triangle inequality. This proves the claim.

Now let g ∈ C_c

and x Z. Then from the above observations about continuous functions,

([ ∫ ]) -- ----1----- μ x ∕∈ Z :lim sur→p0 μ (B (x,r)) B(x,r)|f (y)− f (x)|dμ(y) > ε (9.9)

(9.9)

--([ ε]) --([ ε]) ≤ μ M (f − g) > 2 +μ |f − g| > 2 (9.10)

(9.10)

Now

∫ ε ([ ε ]) |f − g|dμ ≥- μ- |f − g| >- [|f−g|>ε2] 2 2

and so from Claim 1 9.10 and hence 9.9 is dominated by

( 2 N ) - + -p- ||f − g||L1(ℝp,μ). ε ε

But by regularity of Radon measures, C_c

is dense in L¹, and so since g in the above is arbitrary, this shows 9.9 equals 0. Now

By completeness of μ this implies

[ ∫ ] x ∕∈ Z :lim sup----1----- |f (y)− f (x)|dμ(y) > 0 r→0μ (B(x,r)) B(x,r)

is a set of μ measure zero. ■

The following corollary is the main result referred to as the Lebesgue Besicovitch Differentiation theorem.

Proof: If f is replaced by fX_B

then the conclusion 9.11 holds for all xF_k where F_k is a set of μ measure 0. Letting k = 1,2,, and F ≡∪_k=1^∞F_k, it follows that F is a set of measure zero and for any xF, and k ∈, 9.11 holds if f is replaced by fXB

(0,k)

. Picking any such x, and letting k > + 1, this shows

∫ -----1---- lrim→0μ (B(x,r)) B(x,r)|f (y)− f (x)|dμ (y)

∫ = lim -----1---- ||fX (y)− fX (x )||dμ (y) = 0.■ r→0μ (B(x,r)) B(x,r) B(0,k) B(0,k)