MATH41112/61112 Lecture 9

9. Integration

§9.1 The Lebesgue Let (X, B, µ) be a measure . We are interested in how to integrate functions defined on X with respect to the measure µ. In the special case when X = [0, 1], B is the Borel σ- and µ is , this will extend the definition of the to functions that are not Riemann integrable.

Definition. A f : X → R is measurable if f −1(D) ∈ B for every Borel D of R, or, equivalently, if f −1(c, ∞) ∈ B for all c ∈ R. A function f : X → C is measurable if both the real and imaginary parts, Ref and Imf, are measurable.

We define integration via simple functions.

Definition. A function f : X → R is simple if it can be written as a of characteristic functions of sets in B, i.e.:

r

f = aiχAi , Xi=1 for some ai ∈ R, Ai ∈ B, where the Ai are pairwise disjoint. For a f : X → R we define

r f dµ = aiµ(Ai) Z Xi=1 (which can be shown to be independent of the representation of f as a simple function). Thus for simple functions, the integral can be thought of as being defined to be the underneath the graph. If f : X → R, f ≥ 0, is measurable then one can show that there exists 1 an increasing of simple functions fn such that fn ↑ f pointwise as n → ∞ and we define

f dµ = lim fn dµ. Z n→∞ Z

This can be shown to be independent of the choice of sequence fn.

1 fn ↑ f pointwise means: for every x, fn(x) is an increasing sequence and fn(x) → f(x) as n → ∞.

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For an arbitrary f : X → R, we write f = f + − f −, where f + = max{f, 0} ≥ 0 and f − = max{−f, 0} ≥ 0 and define

f dµ = f + dµ − f − dµ. Z Z Z Finally, for a measurable function f : X → C, we define

f dµ = Ref dµ + i Imf dµ. Z Z Z We say that f is integrable if

|f| dµ < +∞. Z

Denote the space of C-valued integrable functions by L1(X, B, µ).

§9.2 Examples

Lebesgue measure. Let X = [0, 1] and let µ denote Lebesgue measure on the Borel σ-algebra. If f : [0, 1] → R is Riemann integrable then it is also Lebesgue integrable and the two definitions agree.

The Steiltjes integral. Let ρ : [0, 1] → R+ and suppose that ρ is differen- tiable. Then 0 f dµρ = f(x)ρ (x) dx. Z Z

Integration with respect to Dirac measures. Let x ∈ X and recall that we defined the by

1 if x ∈ A δ (A) = x  0 if x 6∈ A.

If χA denotes the characteristic function of A then

1 if x ∈ A χA dδx = Z  0 if x 6∈ A.

Hence if f = aiχAi is a simple function then f dδx = ai where x ∈ Ai. Now let f : XP→ R. By choosing an increasing sequenceR of simple functions, we see that

f dδx = f(x). Z

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§9.3 The Lp Spaces Let us say that two measurable functions f, g : X → C are equivalent if f = g µ-a.e. We shall write L1(X, B, µ) (or L1(µ)) for the of equivalence classes of integrable functions on (X, B, µ). We define

kfk1 = |f| dµ. Z

1 Then d(f, g) = kf − gk1 is a on L (X, B, µ). More generally, for any p ≥ 1, we can define the space Lp(X, B, µ) con- sisting of (equivalence classes of) measurable functions f : X → C such that |f|p is integrable. We can again define a metric on Lp(X, B, µ) by defining d(f, g) = kf − gkp where

1/p p kfkp = |f| dµ . Z  It is worth remarking that convergence in Lp neither implies nor is im- plied by convergence . If (X, B, µ) is a finite and if 1 ≤ p < q then

Lq(X, B, µ) ⊂ Lp(X, B, µ).

Apart from L1, the most interesting Lp space is L2(X, B, µ). This is a with the inner product

hf, gi = fg¯ dµ. Z Remark. We have and shall continue to abuse notation by saying that, for example, a function f ∈ L1(X, B, µ) when, strictly speaking, we mean that the equivalence class of f lies in L1(X, B, µ).

Exercise 9.1 1 Give an example of a sequence of functions fn ∈ L ([0, 1], B, µ) (µ = Lebesgue) 1 such that fn → 0 µ-a.e. but fn 6→ 0 in L .

Exercise 9.2 Give an example to show that

L2(R, B, µ) 6⊂ L1(R, B, µ) where µ is Lebesgue measure.

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§9.4 Convergence theorems We state the following two convergence theorems for integration.

Theorem 9.1 (Monotone Convergence Theorem) Suppose that fn : X → R is an increasing sequence of integrable functions on (X, B, µ). If fn dµ is a bounded sequence of real numbers then limn→∞ fn exists µ-a.e. Rand is integrable and

lim fn dµ = lim fn dµ. Z n→∞ n→∞ Z

Theorem 9.2 (Dominated Convergence Theorem) Suppose that g : X → R is integrable and that fn : X → R is a sequence of measurable functions functions with |fn| ≤ g µ-a.e. and limn→∞ fn = f µ-a.e. Then f is integrable and

lim fn dµ = f dµ. n→∞ Z Z

Remark. Both the Monotone Convergence Theorem and the Dominated Convergence Theorem fail for Riemann integration.

§9.5 The Riesz Representation Theorem Let X be a compact and let

C(X, R) = {f : X → R | f is continuous} denote the space of all continuous functions on X. Equip C(X, R) with the metric d(f, g) = kf − gk∞ = sup |f(x) − g(x)|. x∈X Let B denote the Borel σ-algebra on X and let µ be a measure on (X, B). Then we can think of µ as a functional that acts on C(X, R), namely C(X, R) → R : f 7→ f dµ. Z We will often write µ(f) for f dµ. Notice that this map enjoRys several natural properties:

(i) the functional defined by µ is continuous: i.e. if fn ∈ C(X, R) and fn → f then µ(fn) → µ(f). (i’) the functional defined by µ is bounded: i.e. if f ∈ C(X, R) then |µ(f)| ≤ kfk∞.

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(ii) the functional defined by µ is linear:

µ(λ1f1 + λ2f2) = λ1µ(f1) + λ2µ(f2)

where λ1, λ2 ∈ R and f1, f2 ∈ C(X, R).

(iii) if f ≥ 0 then µ(f) ≥ 0 (i.e. the map µ is positive);

(iv) consider the function 1 defined by 1(x) ≡ 1 for all x; then µ(1) = 1 (i.e. the map µ is normalised).

Exercise 9.3 Prove the above assertions.

Remark. It can be shown that a linear functional if continuous if and only if it is bounded. Thus in the presence of (ii), we have that (i) is equivalent to (i’).

The Riesz Representation Theorem says that the above properties char- acterise all Borel probability measures on X. That is, if we have a map w : C(X, R) → R that satisfies the above four properties, then w must be given by integrating with respect to a Borel . This will be a very useful method of constructing measures: we need only construct continuous positive normalised linear functionals!

Theorem 9.3 (Riesz Representation Theorem) Let w : C(X, R) → R be a functional such that:

(i) w is bounded: i.e. for all f ∈ C(X, R) we have |w(f)| ≤ kfk∞;

(ii) w is linear: i.e. w(λ1f1 + λ2f2) = λ1w(f1) + λ2w(f2);

(iii) w is positive: i.e. if f ≥ 0 then w(f) ≥ 0;

(iv) w is normalised: i.e. w(1) = 1.

Then there exists a Borel probability measure µ ∈ M(X) such that

w(f) = f dµ. Z Moreover, µ is unique.

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