Measure and Integral (Ukázka, strana 99)

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23. Indefinite Lebesgue Integral and Absolute Continuity

make this definition reasonable, we should add the assumption that f is continuous in order to guarantee that all “antiderivatives” of f differ up a constant. An analogous problem appears when assuming f = ϕ only almost everywhere. Now, the example of the Cantor function shows that continuity of f does not guarantee that the increment f (b)−f (a) does not depend on choice of the “generalized antiderivative”. However, it is possible to give an alternative definition of the Lebesgue integral of a function ϕ as the increment of an absolutely continuous function f on [a, b] with f = ϕ almost everywhere. (Notice that this definition does not lead to a true generalization — some antiderivative are not absolutely continuous, see Example 25.1). The definitions of various integrals based on the idea of (in some way) generalized antiderivatives are called the descriptive ones. Let us note that these generalizations may consist in omitting “small sets” or in a “generalized differentiation”. Furthermore, in Chapter 25 we mention Perron’s method which is also included among descriptive approaches.

23.8. Lebesgue Points. Let I ⊂ R be an interval and x ∈ I. We say that x is a Lebesgue point for a locally integrable function f if h 1 lim |f (x + t) − f (x)| dt = 0. h→0 2h −h If F is an indefinite Lebesgue integral of f on an interval I and x ∈ I is a point where F (x) = f (x), then 1 h→0 2h

h

lim

−h

(f (x + t) − f (x)) dt = 0

but x does not need to be a Lebesgue point for f . However, it is clear that F = f at each Lebesgue point for f . The following theorem is thus a strengthening of Theorem 23.4.

23.9. Lebesgue Differentiation Theorem. Let f be a locally integrable function on an interval I. Then almost every point of I is a Lebesgue point for f.

Proof. For a fixed r ∈ R, the function x → |f (x) − r| is locally integrable on I. By virtue of Theorem 23.4 there exists a set Er ⊂ I of Lebesgue measure zero such that h 1 lim |f (x + t) − r| dt = |f (x) − r| h→0 2h −h

for every x ∈ I \ Er . If E := Er , then λE = 0. Now, if x ∈ I \ E and ε > 0, r∈Q

then there exists r ∈ Q with |f (x) − r| < ε. Consequently, |f (x + t) − f (x)| ≤ |f (x + t) − r| + ε and lim sup h→0

1 2h

h

−h

|f (x + t) − f (x)| dt ≤ |f (x) − r| + ε ≤ 2ε.

23.10. Remark.

Every continuity point of f is a Lebesgue point for f . An interesting relationship between Lebesgue points and points of approximate continuity will be given in Exercise 29.11.

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D. Integration on R

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23.11. Banach-Zarecki Theorem. Let f be a real-valued function on an interval [a, b]. The following assertions are equivalent: (i) f is absolutely continuous on [a, b]; (iv) f is continuous on [a, b], f is of bounded variation and has Luzin’s (N)-property.

Hint. We already know that any absolutely continuous function f is continuous and of finite

variation. According to Theorem 23.2, λf (G) ≤ G |f | for any open set G ⊂ [a, b], and a moment’s reflection shows that f has Luzin’s (N)-property. Conversely, suppose that (iv) holds and let D be the set of all points where f has a finite derivative. By Exercise 20.8, D is measurable. Since λf ([a, b] \ D) = 0, Theorem 20.4 yields

t

|f (s) − f (t)| ≤

f (ξ) dξ

s

for every interval [s, t] ⊂ [a, b]. Now it is sufficient to notice that the function x → absolutely continuous.

x a

|f | is

23.12. A Characterizations of Lipschitz Functions. Show that for a real-valued function f on an interval [a, b] the following conditions are equivalent: (i) f is Lipschitz on [a, b]; (ii) for any ε > 0 there exists δ > 0 so that for every finite collection of intervals [aj , bj ] ⊂ [a, b] with (bj − aj ) < δ the inequality j |f (bj ) − f (aj )| < ε holds; j

(iii) f is absolutely continuous on [a, b] and f is bounded on [a, b] \ N , where λN = 0.

(Compare with a similar characterization of absolutely continuous functions.)

23.13. Integration by Parts for Lebesgue Integral. functions on an interval [a, b]. Then

b

f g = f (b)g(b) − f (a)g(a) −

a

Let f , g be absolutely continuous

b

f g.

a

23.14. Notes. Fundamental theorem (23.5) of “the calculus for Lebesgue integrals” was proved

by H. Lebesgue [*1904]. The implication (iv) =⇒ (i) of Theorem 23.11 is due to S. Banach [1925].

24. Radon Measures on R and Distribution Functions 24.1. Distribution Functions.

A distribution function of a Radon measure ν on R is a nondecreasing and right continuous function F on R such that F (b) − F (a) = ν(a, b] for every interval (a, b] ⊂ R. It is obvious that any other distribution function of ν can differ from F by only a constant.

If ν is a probability measure, then a distribution function of ν, given as Gν (x) := ν(−∞, x] , is normalized so that lim Gν (x) = 0 and lim Gν (x) = 1. x→−∞

x→+∞

24.2. Theorem. (a) Let F be a nondecreasing and right continuous function on R. Then there exists a unique Radon measure νF on B(R) such that F is the distribution function of νF . (b) Let ν be a Radon measure on R. Then there is a distribution function F of ν.

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24. Radon Measures on R and Distribution Functions

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Proof. (a) As for the uniqueness, any two such measures agree on all open (or compact) sets in R. The existence can be proved in various ways, depending on what kind of general theorem we wish to use. For instance, we can use Hopf’s extension theorem 5.5 or start from the covering collection {(a, b] : a < b, a, b ∈ R} and the set function νF (a, b] := F (b) − F (a), create the outer measure and to restrict it to measurable sets. In any case, we have to prove the following property: ∞ ∞

(an , bn ], then F (b) − F (a) ≤ (F (bn ) − F (an )). To see this, given If (a, b] ⊂ n=1

n=1

ε > 0, let δn > 0, δ > 0 be so that F (bn +δn ) < F (bn )+ε2−n, F (a+δ) < F (a)+ε.

Then consider the cover {(an , bn + δn )} of the interval [a + δ, b]. (b) Setting

⎧ for x > 0, ⎪ ⎨ ν(0, x] for x = 0, Fν (x) := 0 ⎪ ⎩ −ν(x, 0] for x < 0,

Fν is nondecreasing, right continuous, Fν (0) = 0 and ν(a, b] = Fν (b) − Fν (a) for every interval (a, b] ⊂ R. 24.3. Theorem. defined on B(R):

The following conditions are equivalent for a measure ν

(i) ν is a Radon measure; (ii) there exists a distribution function F such that ν(a, b] = F (b) − F (a) for every bounded interval (a, b]; (iii) there exists a nondecreasing function ϕ on R such that ν is the LebesgueStieltjes measure λϕ of Example 14.8.b.

Proof. Use previous results and the uniqueness part of the Riesz representation theorem 16.5.

24.4. Remarks.

1. According to Remark 15.2 any finite measure on (R) is a Radon measure. However, there exist σ-finite measures on (R) which are not Radon measures (for δ1/k of Dirac measures). Hence, these measures are not Lebesgue-Stieltjes example, the sum k

measures and have not distribution functions.

2. The function ϕ from the last theorem need not be right continuous. However, as a monotone function it has a right limit at each point. If ϕ(x) ˜ := limt→x+ ϕ(t), then ϕ ˜ is a distribution function of ν and determines the same Lebesgue-Stieltjes measure as ϕ. The functions ϕ ˜ and F of Theorem 24.3.ii differ by a constant (cf. Exercise 24.6).

24.5. Exercise. Let F be a distribution function of a Radon measure μ on R and let μF correspond to F as in Theorem 24.2. Show that μ = μF on (R).

24.6. Exercise. constant.

Show that any two distribution functions of the same measure differ by a

24.7. Exercise. Let F be a distribution function of a Radon measure μ. Show that: (a) F is locally absolutely continuous on R if and only if μ is absolutely continuous with ym derivative dμ/ dλ. respect to the Lebesgue measure λ. In this case, F is the Radon-Nikod´ (b) Measures μ and λ are mutually singular if and only if F = 0 almost everywhere. (c) μ({x}) = F (x) − lim F (t) for every x ∈ R. In particular, F is continuous at x if and t→x−

only if μ({x}) = 0.

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