Chapter 3 Lebesgue Integration
3.3 Lebesgue’s Differentiation Theorem
3.3.2. The Absolutely Continuous Case
at −∞.
In this subsection I will treat F’s that are (cf. Exercise 2.2.38) absolutely continuous. However, before I can do so, I need to show that every absolutely continuous F is the indefinite integral of a non-negative function f in the sense that
F(x) = Z
(−∞,x]
f dλR for allx∈R,
and I begin with the case in which F is uniformly Lipschitz continuous.
3 Lebesgue Integration
Lemma 3.3.4. For any F there is a most one f ∈ L1(λR;R)of which it is the indefinite integral, and, if it exists, then µF(Γ) =R
Γf dλRfor allΓ∈ BR and thereforef ≥0 (a.e., λR). Moreover, if F(y)−F(x)≤L(y−x)for some L ∈[0,∞)and all x < y, then there is an f ∈L1(λR;R)for which F is the indefinite integral, and f can be chosen to take values between 0andL.
Proof: Suppose that F is the indefinite integral off ∈L1(λR;R). ThenF is continuous and so
µF (a, b)
=F(b)−F(a) = Z
(a,b)
f dλR
for all open intervals (a, b). Furthermore, it is easy to check that the set of Γ∈ BRfor whichµF(Γ) =R
Γf dλRis a Λ-system. Thus, since the set of open intervals is a Π-system that generates BR, it follows that this equality holds for all Γ∈ BR. As a consequence, Exercise 3.1.14 guarantees that, up to a set ofλR-measure 0,f is unique and non-negative.
Turning to the second part, for eachn≥0, definefn:R−→[0,∞) so that fn(x) = 2n F((k+ 1)2−n)−F(k2−n)
whenk2−n≤x <(k+ 1)2−n. Obviously, 0 ≤ fn ≤ L and kfnkL1(λR;R) = F(∞). Hence, fnfm ≥ 0 and R fnfmdλR≤LF(∞) for allm, n∈N. Moreover, if m < n, then
Z
[k2−m,(k+1)2−m)
fnfmdλR=fm(k2−m)
(k+1)2n−m−1
X
j=k2n−m
F((j+ 1)2−n)−F(j2−n)
= 2−mfm(k2−n)2= Z
[k2−m,(k+1)2−m)
fm2 dλR, and thereforeR
fnfmdλR=R
fm2 dλRfor allm≤n. In particular, this means that
(∗)
Z
(fn−fm)2dλR= Z
fn2λR− Z
fm2 dλR, and so the sequence of integrals R
fn2dλR is non-decreasing, bounded above by LF(∞), and therefore convergent as n→ ∞. Hence, by (∗) and (3.1.7), we now know that, for any >0,
sup
n≥m
λR |fn−fm| ≥
≤−2 sup
n≥m
Z
(fn−fm)2dλR−→0 as m→ ∞, and therefore, by Lemma 3.2.13, that there exists aBRmeasurablef to which {fn : n≥ 1} converges in λR-measure. Furthermore, because 0 ≤fn ≤L, we may assume that 0 ≤ f ≤ L. Also, Lebesgue’s Dominated Convergence
Theorem for convergence in measure (cf. Theorem 3.2.12) implies that, for all m∈Z+ andk < `,
F(`2−m)−F(k2−m) = Z
(k2−m,`2−m]
fn∨mdλR−→
Z
(k2−m,`2−m]
f dλR as n→ ∞, and therefore that
F(y)−F(x) = Z
(x,y]
f dλR,
first forx < ythat are dyadic numbers (i.e., of the formk2−m) and then, by continuity, for all x < y. Thus, after letting x→ −∞, we see that F is the indefinite integral off.
Lemma 3.3.5. GivenL∈[0,∞), defineFL:R−→Rby FL(x) =µF (−∞, x]∩ {LF ≤L}
forx∈R.
Then FL is a bounded, non-decreasing function that tends to 0 at −∞and satisfiesFL(y)−FL(x)≤L(y−x)for allx < y.
Proof: Clearly, it suffices to prove the final inequality.
For anyx < y, FL(y)−FL(x) =µF (x, y]∩ {LF ≤L}
. Thus, if (x, y]∩ {LF ≤L}=∅, thenFL(y)−FL(x) = 0; and ifc∈(x, y]∩ {LF ≤L}, then
0≤FL(y)−FL(x)≤F(y)−F(x) = F(y)−F(c)
+ F(c)−F(x)
≤L(y−c) +L(c−x) =L(y−x).
Theorem 3.3.6. The following properties are equivalent.
(a)F is absolutely continuous.
(b)µF(LF=∞) = 0.
(c) There exists a sequence {Fn : n ≥ 1} of bounded, non-decreasing, uni- formly Lipschitz continuous functions, each of which tends to0 at−∞, such that Fn+1−Fn is non-decreasing for eachn∈Z+ andFn(∞)%F(∞).
(d) There is a non-negativef ∈L1(λR;R)for whichF is the indefinite inte- gral.
Moreover, the f in (d) is the only element in L1(λR;R) of which F is its indefinite integral. (The implication (a)=⇒(d) is a special case of Theorem 8.1.3.)
Proof: IfFis absolutely continuous, then (cf. Exercise 2.2.38)µF λR, and therefore, becauseλR(LF =∞) = 0,µF(LF =∞) = 0. Hence, (a) =⇒(b).
Next, assume thatµF(LF =∞) = 0, and set Fn(x) =µF (−∞, x]∩ {LF ≤n}
forx∈R.
3 Lebesgue Integration
Then it is clear that, for each n ∈ Z+, Fn is a bounded, non-decreasing function that tends to 0 at−∞. In addition, by Lemma 3.3.5,
0≤Fn(y)−Fn(x)≤n(y−x) for allx < y, and
Fn+1(x)−Fn(x) =µF (−∞, x]∩ {n <LF ≤(n+ 1)}
is non-negative and non-decreasing. Finally,
Fn(∞) =µF(LF ≤n)%µF(LF <∞) =µF(R) =F(∞).
Thus (b) =⇒(c).
To prove that (c) =⇒ (d), for each n ∈ Z+ use Lemma 3.3.4 to find a non-negative fn ∈ L1(λR;R) for which Fn is the indefinite integral. Then, becauseFn+1(x)−Fn(x) =R
(−∞,x](fn+1−fn)dλR, we know thatfn+1−fn is the unique element ofL1(λR;R) for whichFn+1−Fnis the indefinite integral and, as such, is non-negative (a.e., λR). Hence,fn+1≥fn (a.e., λR). In other words, we can now assume that the fn’s are non-negative and fn+1 ≥ fn everywhere. Finally, set f = limn→∞fn. Because Fn −→ F pointwise (in fact, uniformly), it follows from the Monotone Convergence Theorem that
F(x) = lim
n→∞Fn(x) = lim
n→∞
Z
(−∞,x]
fndλR= Z
(−∞,x]
f dλR for allx∈R. That (d) =⇒(a) is (cf. Exercise 3.1.13) trivial, and the concluding unique- ness statement is covered by Lemma 3.3.4.
Now that we know that all absolutely continuousF’s are indefinite integrals, the λR-almost everywhere existence of F0 for suchF’s becomes a matter of extending the Fundamental Theorem of Calculus to measurable functions.
To this end, for f ∈ L1(λR;R), define the Hardy–Littlewood Maximal FunctionM f off by
M f(x) = sup
I3x
− Z
I
|f|dλR, x∈R,
where the supremum is over closed intervalsI3xwith ˚I6=∅ and
− Z
I
f dλR≡ 1 vol(I)
Z
I
f dλR
is the average value off onI. Obviously, ifF is the indefinite integral of|f|, thenM f =LF, and so Corollary 3.3.3 says that
(3.3.7) λR(M f > R)≤ 2 R
Z
{M f >R}
|f|dλR≤2kfkL1(λR;R)
R for allR >0.
(See Exercise 6.2.13 for an important consequence.)
The inequality (3.3.7) is the famousHardy–Littlewood inequalitythat plays a crucial role in the analysis of almost everywhere convergence results.
For us, its importance is demonstrated in the following statement of the Fun- damental Theorem of Calculus.
Theorem 3.3.8. For each f ∈L1(λR;R),
(3.3.9) lim
I&{x}− Z
I
|f−f(x)|dλR= 0 forλR-almost everyx∈R,
where the limit is taken over intervals I 3 x as vol(I) & 0. In particular, ifF is the indefinite integral of f, thenF0(x)exists and is equal tof(x) for λR-almost everyx∈R.
Proof: There is nothing to do whenf ∈C(R;R)∩L1(λR;R). Furthermore, by Corollary 3.2.15, we know thatC(R;R)∩L1(λR;R) is dense inL1(λR;R).
Hence, we will be done once we show that the set of f’s for which (3.3.9) holds is closed under convergence in L1(λR;R). To prove this, suppose that {fn : n≥1} ⊆L1(λR;R) is a sequence of functions for which (3.3.9) holds and that fn−→f inL1(λR;R). Then, for any >0 andn∈Z+,
λR
x: lim
I&{x}− Z
I
|f −f(x)|dλR≥3
≤λR
x: lim
I&{x}− Z
I
|f −fn|dλR≥
+λR
x: lim
I&{x}− Z
I
|fn−fn(x)|dλR≥
+λR |fn−f| ≥ .
By (3.3.7) and (3.1.7) applied tof−fn, the first and third terms on the right are dominated by, respectively, 2−1kf−fnkL1(λR;R)and−1kf−fnkL1(λR;R), and, by assumption, the second term on the right vanishes. Hence,
λR
x: lim
I&{x}− Z
I
|f −f(x)|dλR≥3
≤ 3kf−fnkL1(λR;R)
for everyn∈Z+, and so the desired conclusion follows after we letn→ ∞.
Before closing this subsection, there are several comments that should be made. First, one should recognize that the conclusions drawn in Theorem 3.3.8 remain true for any Lebesgue measurable f that is integrable on each compact subset ofR. Indeed, all the assertions there are completely local and therefore follow by replacing f with f1(−R,R), restricting ones attention to x∈(−R, R), and then letting R% ∞.
Second, one should notice that (3.3.7) would be a trivial consequence of Markov’s inequality if we had the estimatekM fkL1(λR;R)≤CkfkL1(λR;R) for some C <∞. Thus, one is tempted to ask whether such an estimate is true.
3 Lebesgue Integration
That the answer is a resoundingnocan be most easily seen from the observa- tion that, ifkfkL1(λR;R)6= 0, thenα≡R
(−r,r)
f(t)|λR(dt)>0 for somer >0 and therefore M f(x) ≥ |x|+rα for allx∈ R. In particular, if f ∈ L1(λR;R) does not vanish almost everywhere, thenM f is not integrable. (To see that the situation is even worse and that, in general,M f need not be integrable over bounded sets, see Exercise 3.3.16 below.) Thus, in a very real sense, (3.3.7) is about as well as one can do. Because this sort of situation arises quite often, inequalities of the form in (3.3.7) have been given a special name: they are calledweak-type inequalitiesto distinguish them from inequalities of the form kM fkL1(λR;R) ≤ CkfkL1(λR;R), which is called a strong-type inequality. See Exercise 6.2.11 below for related information.
Finally, it should be clear that, except for the derivation of (3.3.7), the arguments given in the proof of Theorem 3.3.8 would work equally well in RN. Thus, we would know that, for each Lebesgue integrablef onRN, (3.3.10) lim
B&{x}− Z
B
f(y)−f(x)
λRN(dy) = 0 forλRN-almost everyx∈RN if we knew that, for some C <∞,
(3.3.11) λRN(M f > )≤C
kfkL1(λ
RN;R), where M f is the Hardy–Littlewood maximal function
M f(x)≡sup
B3x
− Z
B
f(y)
λRN(dy)
and B denotes a generic open ball in RN. It turns out that (3.3.11), and therefore (3.3.10), are both true. However, because there is no multidimen- sional analogue of the Sunrise Lemma, the proof2 of (3.3.11) for N ≥ 2 is somewhat more involved than the one that I have given of (3.3.7).
§3.3.3. The General Case: In this subsection I will complete the program