Integration

4. Integration The (µ)-integral of a non-negative simple function is by definition � � (4.1) f dµ = ai µ(Y � Ei ) , Y ≥ M . Y i Here the convention is that if µ(Y � Ei ) = ⊂ but ai = 0 then ai · µ(Y � Ei ) = 0. Clearly this integral takes values in [0, ⊂]. More significantly, if c → 0 is a constant and f and g are two non-negative (µ-measurable) simple functions then � � cf dµ = c f dµ Y � � Y � (f + g)dµ = f dµ + gdµ (4.2) Y Y Y � � g dµ . f dµ ∼ 0∼f ∼g ⇔ Y Y (See [1] Proposition 2.13 on page 48.) To see this, observe that (4.1) holds for any presentation (3.5) of f with all ai → 0. Indeed, by restriction to Ei and division by ai (which can be assumed non-zero) it is enough to consider the special case � �E = b j �Fj . j The Fj can always be written as the union of a finite number, N ≥ , of disjoint measurable sets, Fj = ∀l→Sj Gl where j = 1, . . . , N and Sj ↑ {1, . . . , N ≥ }. Thus � � � bj bj µ(Fj ) = µ(Gl ) = µ(E) j j l→Sj � since {j;l→Sj } bj = 1 for each j. From this all the statements follow easily. Definition 4.1. For a non-negative µ-measurable extended function f : X −← [0, ⊂] the integral (with respect to µ) over any measurable set E ↑ X is � � f dµ = sup{ hdµ; 0 ∼ h ∼ f, h simple and measurable.} (4.3) E E � By taking suprema, E f dµ has the first and last properties in (4.2). It also has the middle property, but this is less obvious. To see this, we shall prove the basic ‘Monotone convergence theorem’ (of Lebesgue). Before doing so however, note what the vanishing of the integral means. Lemma 4.2. If f : X −← [0, ⊂] is measurable then measurable set E if and only if {x ≥ E; f (x) > 0} has measure zero. (4.4) � E f dµ = 0 for a Proof. If (4.4) holds, then any positive simple function bounded above by f must also vanish � outside a set of measure zero, so its integral must be zero and hence E f dµ = 0. Conversely, observe that the set in (4.4) can be written as  En = {x ≥ E; f (x) > 1/n}. n Since these sets increase with n, if (4.4) does not hold then one of these In that case the simple function n−1 �En must have positive measure. � � has positive integral so E f dµ > 0. Notice the fundamental difference in approach here between Rie­ mann and Lebesgue integrals. The Lebesgue integral, (4.3), uses ap­ proximation by functions constant on possibly quite nasty measurable sets, not just intervals as in the Riemann lower and upper integrals. Theorem 4.3 (Monotone Convergence). Let fn be an increasing se­ quence of non-negative measurable (extended) functions, then f (x) = limn�∗ fn (x) is measurable and � � (4.5) f dµ = lim fn dµ E n�∗ E for any measurable set E ↑ X. Proof. To see that f is measurable, observe that  fn−1 (a, ⊂]. (4.6) f −1 (a, ⊂] = n Since the sets (a, ⊂] generate the Borel χ-algebra this shows that f is measurable. So we proceed to prove the main part of the proposition, which is (4.5). Rudin has quite a nice proof of this, [5] page 21. Here I paraphrase it. We can easily see from (4.1) that � � � fn dµ ∼ f dµ. � = sup fn dµ = lim E n�∗ E E Given a simple measurable function g with 0 ∼ g ∼ f and 0 < c < 1 consider the sets En = {x ≥ E; fn (x)�→ cg(x)}. These are measurable and increase with n. Moreover E = n En . It follows that � � � � ai µ(En � Fi ) gdµ = (4.7) fn dµ → fn dµ → c E En En i � in terms of the natural presentation of g = i ai �Fi . Now, the fact that the En are measurable and increase to E shows that µ(En � Fi ) ← µ(E � Fi ) � as n ← ⊂. Thus the right side of (4.7) tends to c gdµ as n ← ⊂. E � Hence � → c E gdµ for all 0 < c < 1. Taking the supremum over c and then over all such g shows that � � � f dµ. fn dµ → sup gdµ = � = lim n�∗ E E E They must therefore be equal. � Now for instance the additivity in (4.1) for f → 0 and g → 0 any measurable functions follows from Proposition 3.3. Thus if f → 0 is measurable and fn is an �approximating sequence as in the Proposition � then E f dµ = limn�∗ E fn dµ. So if f and g are two non-negative measurable functions then fn (x) + gn (x) ∩ f + g(x) which shows not only that f + g is measurable by also that � � � gdµ. f dµ + (f + g)dµ = E E E As with the definition of u+ long ago, this allows us to extend the definition of the integral to any integrable function. Definition 4.4. A measurable extended function f : X −← [−⊂, ⊂] is said to be integrable on E if its positive and negative parts both have finite integrals over E, and then � � � f− dµ. f dµ = f+ dµ − E E E Notice if f is µ-integrable then so is |f |. One of the objects we wish to study is the space of integrable functions. The fact that the integral of |f | can vanish encourages us to look at what at first seems a much more complicated object. Namely we consider an equivalence relation between integrable functions (4.8) ⇐ f2 (x)}) = 0. f1 ∈ f2 ∞⇔ µ({x ≥ X; f1 (x) = That is we identify two such functions if they are equal ‘off a set of measure zero.’ Clearly if f1 ∈ f2 in this sense then � � � � |f1 |dµ = |f2 |dµ = 0, f1 dµ = f2 dµ. X X X X A necessary condition for a measurable function f → 0 to be inte­ grable is µ{x ≥ X; f (x) = ⊂} = 0. Let E be the (necessarily measureable) set where f = ⊂. Indeed, if this does not have measure zero, then the sequence of simple functions n�E ∼ f has integral tending to infinity. It follows that each equiva­ lence class under (4.8) has a representative which is an honest function, i.e. which is finite everywhere. Namely if f is one representative then � /E f (x) x ≥ f ≥ (x) = x≥E 0 is also a representative. We shall denote by L1 (X, µ) the space consisting of such equivalence classes of integrable functions. This is a normed linear space as I ask you to show in Problem 11. The monotone convergence theorem often occurrs in the slightly dis­ guised form of Fatou’s Lemma. Lemma 4.5 (Fatou). If fk is a sequence of non-negative integrable functions then � � lim inf fn dµ ∼ lim inf fn dµ . n�∗ n�∗ Proof. Set Fk (x) = inf n�k fn (x). Thus Fk is an increasing sequence of non-negative functions with limiting function lim inf n�∗ fn and Fk (x) ∼ fn (x) � n → k. By the monotone convergence theorem � � � Fk (x) dµ ∼ lim inf fn dµ. lim inf fn dµ = lim n�∗ n�∗ k�∗ � We further extend the integral to complex-valued functions, just say­ ing that f :X←C is integrable if its real and imaginary parts are both integrable. Then, by definition, � � � Im f dµ Re f dµ + i f dµ = E E E for any E ↑ X measurable. It follows that if f is integrable then so is |f |. Furthermore � � �� � � � f dµ� ∼ |f | dµ . � � E E � This is obvious if E f dµ = 0, and if not then � f dµ = Rei� R > 0 , α ↑ [0, 2�) . E Then � �� � � � � f dµ� = e−i� f dµ � � E E � e−i� f dµ = E � = Re(e−i� f ) dµ �E � � �Re(e−i� f )� dµ ∼ � �E � −i� � �e f � dµ = |f | dµ . ∼ E E The other important convergence result for integrals is Lebesgue’s Dominated convergence theorem. Theorem 4.6. If fn is a sequence of integrable functions, fk ← f a.e.5 and |fn | ∼ g for some integrable g then f is integrable and � � fn dµ . f dµ = lim n�∗ Proof. First we can make the sequence fn (x) converge by changing all the fn (x)’s to zero on a set of measure zero outside which they converge. This does not change the conclusions. Moreover, it suffices to suppose that the fn are real-valued. Then consider hk = g − f k → 0 . Now, lim inf k�∗ hk = g − f by the convergence of fn ; in particular f is integrable. By monotone convergence and Fatou’s lemma � � � (g − f )dµ = lim inf hk dµ ∼ lim inf (g − fk ) dµ k�∗ k�∗ � � = g dµ − lim sup fk dµ . k�∗ Similarly, if Hk = g + fk then � � � � (g + f )dµ = lim inf Hk dµ ∼ g dµ + lim inf fk dµ. k�∗ k�∗ It follows that lim sup k�∗ 5Means � fk dµ ∼ � f dµ ∼ lim inf k�∗ on the complement of a set of measure zero. � fk dµ. Thus in fact � fk dµ ← � f dµ . � Having proved Lebesgue’s theorem of dominated convergence, let me use it to show something important. As before, let µ be a positive measure on X. We have defined L1 (X, µ); let me consider the more general space Lp (X, µ). A measurable function f :X←C is said to be ‘L ’, for 1 ∼ p < ⊂, if |f |p is integrable6, i.e., � |f |p dµ < ⊂ . p X As before we consider equivalence classes of such functions under the equivalence relation (4.9) f ∪ g ≤ µ {x; (f − g)(x) = ⇐ 0} = 0 . We denote by Lp (X, µ) the space of such equivalence classes. It is a linear space and the function �� 1/p p (4.10) ≡f ≡p = |f | dµ X is a norm (we always assume 1 ∼ p < ⊂, sometimes p = 1 is excluded but later p = ⊂ is allowed). It is straightforward to check everything except the triangle inequality. For this we start with Lemma 4.7. If a → 0, b → 0 and 0 < π < 1 then a� b1−� ∼ πa + (1 − π)b (4.11) with equality only when a = b. Proof. If b = 0 this is easy. So assume b > 0 and divide by b. Taking t = a/b we must show t� ∼ πt + 1 − π , 0 ∼ t , 0 < π < 1 . (4.12) The function f (t) = t� − πt is differentiable for t > 0 with derivative πt�−1 − π, which is positive for t < 1 and negative for t > 1. Thus f (t) ∼ f (1) with equality only for t = 1. Since f (1) = 1 − π, this is (4.12), proving the lemma. � We use this to prove Hölder’s inequality 6Check p that |f | is automatically measurable. Lemma 4.8. If f and g are measurable then �� � � � � f gdµ� ∼ ≡f ≡p ≡g≡q (4.13) � � for any 1 < p < ⊂, with 1 p + 1 q = 1. Proof. If ≡f ≡p = 0 or ≡g≡q = 0 the result is trivial, as it is if either is infinite. Thus consider � � � � � g(x) �q � f (x) �p � � � � ,b=� a=� ≡f ≡p � ≡g≡q � and apply (4.11) with π = 1p . This gives |f (x)|p |g(x)|q |f (x)g(x)| ∼ + . ≡f ≡p ≡g≡q p≡f ≡pp q≡g≡qq Integrating over X we find 1 ≡f ≡p ≡g≡q � |f (x)g(x)| dµ X ∼ 1 1 + = 1. p q �� � � Since � X f g dµ� ∼ X |f g| dµ this implies (4.13). � The final inequality we need is Minkowski’s inequality. Proposition 4.9. If 1 < p < ⊂ and f, g ≥ Lp (X, µ) then (4.14) ≡f + g≡p ∼ ≡f ≡p + ≡g≡p . Proof. The case p = 1 you have already done. It is also obvious if f + g = 0 a.e.. If not we can write |f + g|p ∼ (|f | + |g|) |f + g|p−1 and apply Hölder’s inequality, to the right side, expanded out, 1/q �� � p q(p−1) |f + g| . |f + g| dµ ∼ (≡f ≡p + ≡g≡p) , dµ Since q(p − 1) = p and 1 − 1 q = 1/p this is just (4.14). � So, now we know that Lp (X, µ) is a normed space for 1 ∼ p < ⊂. In particular it is a metric space. One important additional property that a metric space may have is completeness, meaning that every Cauchy sequence is convergent. Definition 4.10. A normed space in which the underlying metric space is complete is called a Banach space. Theorem 4.11. For any measure space (X, M, µ) the spaces Lp (X, µ), 1 ∼ p < ⊂, are Banach spaces. Proof. We need to show that a given Cauchy sequence {fn } converges in Lp (X, µ). It suffices to show that it has a convergent subsequence. By the Cauchy property, for each k � n = n(k) s.t. ≡fn − f� ≡p ∼ 2−k � θ → n . (4.15) Consider the sequence g1 = f1 , gk = fn(k) − fn(k−1) , k > 1 . � By (4.15), ≡gk ≡p ∼ 2−k , for k > 1, so the series k ≡gk ≡p converges, say to B < ⊂. Now set n ∗ � � |gk (x)| , n → 1 , h(x) = gk (x). hn (x) = k=1 k=1 Then by the monotone convergence theorem � � p h dµ = lim |hn |p dµ ∼ B p , n�∗ X X where we have also used Minkowski’s inequality. Thus h ≥ Lp (X, µ), so the series ∗ � f (x) = gk (x) k=1 converges (absolutely) almost everywhere. Since �p � n �� � � � gk � ∼ h p |f (x)|p = lim � n�∗ � � k=1 p ≥ with h ≥ L (X, µ), the dominated convergence theorem applies and shows that f ≥ Lp (X, µ). Furthermore, � � k=1 �p � gk (x) = fn(�) (x) and �f (x) − fn(�) (x)� ∼ (2h(x))p so again by the dominated convergence theorem, � � � �f (x) − fn(�) (x)�p ← 0 . X Thus the subsequence fn(�) ← f in Lp (X, µ), proving its completeness. � Next I want to return to our starting point and discuss the Riesz representation theorem. There are two important results in measure theory that I have not covered — I will get you to do most of them in the problems — namely the Hahn decomposition theorem and the Radon-Nikodym theorem. For the moment we can do without the latter, but I will use the former. So, consider a locally compact metric space, X. By a Borel measure on X, or a signed Borel measure, we shall mean a function on Borel sets µ : B (X) ← R which is given as the difference of two finite positive Borel measures (4.16) µ(E) = µ1 (E) − µ2 (E) . Similarly we shall say that µ is Radon, or a signed Radon measure, if it can be written as such a difference, with both µ1 and µ2 finite Radon measures. See the problems below for a discussion of this point. Let Mfin (X) denote the set of finite Radon measures on X. This is a normed space with (4.17) ≡µ≡1 = inf(µ1 (X) + µ2 (X)) with the infimum over all Radon decompositions (4.16). Each signed Radon measure defines a continuous linear functional on C0 (X): � � f · dµ . (4.18) · dµ : C0 (X) � f ⇒−← X Theorem 4.12 (Riesz representation.). If X is a locally compact met­ ric space then every continuous linear functional on C0 (X) is given by a unique finite Radon measure on X through (4.18). Thus the dual space of C0 (X) is Mfin (X) – at least this is how such a result is usually interpreted (4.19) (C0 (X))≥ = Mfin (X), see the remarks following the proof. Proof. We have done half of this already. Let me remind you of the steps. We started with u ≥ (C0 (X))≥ and showed that u = u+ − u− where u± are positive continuous linear functionals; this is Lemma 1.5. Then we showed that u → 0 defines a finite positive Radon measure µ. Here µ is defined by (1.11) on open sets and µ(E) = µ� (E) is given by (1.12) on general Borel sets. It is finite because (4.20) µ(X) = sup {u(f ) ; 0 ∼ f ∼ 1 , supp f X , f ≥ C(X)} ∼ ≡u≡ . From Proposition 2.8 we conclude that µ is a Radon measure. Since this argument applies to u± we get two positive finite Radon measures µ± and hence a signed Radon measure (4.21) µ = µ+ − µ− ≥ Mfin (X). In the problems you are supposed to prove the Hahn decomposition theorem, in particular in Problem 14 I ask you to show that (4.21) is the Hahn decomposition of µ — this means that there is a Borel set E ↑ X such that µ− (E) = 0 , µ+ (X \ E) = 0. What we have defined is a linear map (4.22) (C0 (X))≥ ← M (X), u ⇒−← µ . We want to show that this is an isomorphism, i.e., it is 1 − 1 and onto. We first show that it is 1 − 1. That is, suppose µ = 0. Given the uniqueness of the Hahn decomposition this implies that µ+ = µ− = 0. So we can suppose that u → 0 and µ = µ+ = 0 and we have to show that u = 0; this is obvious since µ(X) = sup {u(f ); supp u X, 0 ∼ f ∼ 1 f ≥ C(X)} = 0 (4.23) ⇔ u(f ) = 0 for all such f . X then f ≥ = f /≡f ≡∗ is of this type If 0 ∼ f ≥ C(X) and supp f so u(f ) = 0 for every 0 ∼ f ≥ C(X) of compact support. From the decomposition of continuous functions into positive and negative parts it follows that u(f ) = 0 for every f of compact support. Now, if f ≥ Co (X), then given n ≥ N there exists K X such that |f | < 1/n on X \ K. As you showed in the problems, there exists � ≥ C(X) with supp(�) X and � = 1 on K. Thus if fn = �f then supp(fn ) X and ≡f − fn ≡ = sup(|f − fn | < 1/n. This shows that C0 (X) is the closure of the subspace of continuous functions of compact support so by the assumed continuity of u, u = 0. So it remains to show that every finite Radon measure on X arises from (4.22). We do this by starting from µ and constructing u. Again we use the Hahn decomposition of µ, as in (4.21)7. Thus we assume µ → 0 and construct u. It is obvious what we want, namely � (4.24) u(f ) = f dµ , f ≥ Cc (X) . X 7Actually we can just take any decomposition (4.21) into a difference of positive Radon measures. Here we need to recall from Proposition 3.2 that continuous functions on X, a locally compact metric space, are (Borel) measurable. Further­ more, we know that there is an increasing sequence of simple functions with limit f , so � �� � � � f dµ� ∼ µ(X) · ≡f ≡∗ . (4.25) � � X This shows that u in (4.24) is continuous and that its norm ≡u≡ ∼ µ(X). In fact (4.26) ≡u≡ = µ(X) . Indeed, the inner regularity of µ implies that there is a compact set K X with µ(K) → µ(X)− n1 ; then there is f ≥ Cc (X) with 0 ∼ f ∼ 1 and f = 1 on K. It follows that µ(f ) → µ(K) → µ(X) − n1 , for any n. This proves (4.26). We still have to show that if u is defined by (4.24), with µ a finite positive Radon measure, then the measure µ̃ defined from u via (4.24) is precisely µ itself. This is easy provided we keep things clear. Starting from µ → 0 a finite Radon measure, define u by (4.24) and, for U ↑ X open �� � f dµ, 0 ∼ f ∼ 1, f ≥ C(X), supp(f ) U . (4.27) µ(U ˜ ) = sup X U By the properties of the integral, µ̃(U ) ∼ µ(U ). Conversely if K there exists an element f ≥ Cc (X), 0 ∼ f ∼ 1, f = 1 on K and supp(f ) ↑ U. Then we know that � (4.28) µ(U ˜ )→ f dµ → µ(K). X By the inner regularity of µ, we can choose K U such that µ(K) → µ(U ) − σ, given σ > 0. Thus µ̃(U ) = µ(U ). This proves the Riesz representation theorem, modulo the decompo­ sition of the measure - which I will do in class if the demand is there! � In my view this is quite enough measure theory. Notice that we have in fact proved something stronger than the state­ ment of the theorem. Namely we have shown that under the correspon­ dence u ∃← µ, (4.29) ≡u≡ = |µ| (X) =: ≡µ≡1 . Thus the map is an isometry.