350 CHAPTER 11. REGULAR MEASURES

Next suppose f (y)− f (x) =∫ y

x f ′ (t)dt where f ′ ∈ L1. If f is not absolutely continu-ous, there exists ε > 0 and sets Vn each of which is the union of non-overlapping intervalssuch that m(Vn)< 2−n but

∫Vn| f ′ (t)|dt ≥ ε . However, by the Borel Cantelli lemma, there

exists a set of measure zero N such that for x /∈N, it follows that x is in only finitely many ofthe Vn. Thus XVn (x)→ 0. Then a use of the dominated convergence theorem implies thatlimn→∞

∫Vn| f ′ (t)|dt = 0 which is a contradiction. Thus f must be absolutely continuous.

■The following picture illustrates the main items shown so far about functions of one

variable.

Lipschitzabsolutely continuous

∫ yx f ′(t)dt = f (y)− f (x)

11.16 Exercises1. Show that if f is absolutely continuous on [a,b] and if V (x) is the total variation of

f on [0,x] , then V is also absolutely continuous.

2. In Problem 5 on Page 310, you showed that if f ∈ L1 (Rp) , there exists h whichis continuous and equal to 0 off some compact set such that

∫| f −h|dm < ε. De-

fine fy (x) ≡ f (x−y) . Explain why fy is Lebesgue measurable and∫| fy|dmp =∫

| f |dmp. Now justify the following formula.∫| fy− f |dmp ≤

∫| fy−hy|dmp +∫

|hy−h|dmp +∫|h− f |dmp ≤ 2ε +

∫|hy−h|dmp. Now explain why the last term

is less than ε if ∥y∥ is small enough. Explain continuity of translation in L1 (Rp)which says that limy→0

∫Rp | fy− f |dmp = 0

3. This problem will help to understand that a certain kind of function exists. Let f (x)=e−1/x2

if x ̸= 0 and let f (x) = 0 if x = 0. Show that f is infinitely differentiable. Notethat you only need to be concerned with what happens at 0. There is no questionelsewhere. This is a little fussy but is not too hard.

4. ↑Let f (x) be as given above. Now let f̂ (x) = f (x) if x ≤ 0 and let f̂ (x) = 0 ifx > 0. Show that f̂ (x) is also infinitely differentiable. Now let r > 0 and defineg(x) ≡ f̂ (−(x− r)) f̂ (x+ r). Show that g is infinitely differentiable and vanishesfor |x| ≥ r. Let ψ (x) = ∏

pk=1 g(xk). For U = B(0,2r) with the norm given by ∥x∥=

max{|xk| ,k ≤ p} , show that ψ ∈C∞c (U).

5. ↑Using the above problem, show there exists ψ ≥ 0 such that ψ ∈C∞c (B(0,1)) and∫

ψdmp = 1. Now define ψn (x) ≡ npψ (nx). Show that ψn equals zero off a com-pact subset of B

(0, 1

n

)and

∫ψndmp = 1. We say that spt(ψn)⊆ B

(0, 1

n

). spt( f ) is

defined as the closure of the set on which f is not equal to 0. Such a sequence of func-tions as just defined {ψn} where

∫ψndmp = 1 and ψn ≥ 0 and spt(ψn) ⊆ B

(0, 1

n

)is called a mollifier.

6. ↑It is important to be able to approximate functions with those which are infinitelydifferentiable. Suppose f ∈ L1 (Rp) and let {ψn} be a mollifier as above. We de-fine the convolution as follows. f ∗ψn (x) ≡

∫f (x−y)ψn (y)dmp (y) Here the

notation means that the variable of integration is y. Show that f ∗ψn (x) exists