Introductory Analysis 2–Spring 2010 Concordance with Chapter 4 of Royden

1. Definition of simple function. Page 77. 2. Canonical representation of a simple function. Page 77. Z Z Z 3. Definition of φ(x) dx = φ dm = φ if φ is a simple function such that the sets on which φ assumes a non-zero value have finite measure. Or, as Royden puts it, φ vanishes outside of a set of finite measure. 4. Lemma 1 on page 78. 5. Proposition 2 on page 78. 6. We now need a result that is part of Royden’s proof his Proposition 3 (sort of), but I want to have it as a separate result.

Lemma 1 Let E be a measurable subset of R and let f : E → [0, ∞] be measurable. There exists a sequence {ϕn} of non-negative simple functions such that:

(a) ϕ1(x) ≤ ϕ2(x) ≤ · · · for all x ∈ E.

(b) limn→∞ ϕn(x) = f(x) for all x ∈ E. (The sequence converges pointwise to f) If f is bounded, then the sequence converges uniformly to f on E.

Proof. For n = 1, 2,... deﬁne ϕn as follows. Let k k + 1 E = {x ∈ E : ≤ f(x) < } nk 2n 2n for k = 0, 1, 2 .... Actually, we only consider these sets for 0 ≤ k ≤ n2n − 1. k k + 1 Let x ∈ E. If f(x) ≥ n, set ϕ (x) = n. If 0 ≤ f(x) < n there exists a unique k such that ≤ f(x) < , n 2n 2n n n 0 ≤ k ≤ n2 − 1. Set ϕn(x) = k/2 . An equivalent deﬁnition is

n nX2 −1 k ϕ = χ + nχ n 2n Enk Fn k=0 Ã ! 2n[n−1 where Fn = E\ Enk . The sets Enk,Fn are measurable because f is measurable, thus ϕn is indeed a simple k=0 function for each n; incidentally it vanishes outside of E. Obviously ϕn(x) ≥ 0 for all n. It is also quite easy to see that ϕn(x) ≤ ϕn+1(x) ≤ f(x) for all x ∈ E. In fact, In the first place the definition makes it quite clear that ϕn(x) ≤ f(x) for all n ∈ N, x ∈ E. To see that ϕn ≤ ϕn+1 assume first that f(x) > n+1. Then ϕn(x) = n < n+1 = ϕn+1(x) < f(x). n+1 n+1 n+1 Assume n < f(x) ≤ n + 1. In this case ϕn(x) = n, ϕn+1(x) = k/2 , where k/2 ≤ f(x) < (k + 1)/2 . Then n+1 n+1 n+1 n+1 (k + 1)/2 > n, so k + 1 > n2 , hence k ≥ n2 and ϕn+1(x) = k/2 ≥ n = ϕn(x). Finally, assume f(x) < n. n n n n n+1 Let k be such that k/2 ≤ f(x) < (k + 1)/2 , so ϕn(x) = k/2 . Then either k/2 ≤ f(x) < (2k + 1)/2 or (2k + 1)/2n+1 ≤ f(x) < (k + 1)/2n. (Notice that (2k + 1)/2n+1 is the midpoint of the interval [k/2n, (k + 1)/2n).) n+1 n+1 n+1 n The former case can be written as (2k)/2 ≤ f(x) < (2k + 1)/2 , hence ϕn+1(x) = 2k/2 = k/2 = ϕn(x); the n+1 n+1 n+1 n latter case can be written as (2k + 1)/2 ≤ f(x) < (2k + 2)/2 , hence ϕn+1(x) = (2k + 1)/2 > k/2 = ϕn(x). This completes the proof that the sequence {ϕn} is increasing in the sense that {ϕn(x)} is an increasing sequence of real numbers for each x ∈ E.

Let x ∈ E. If f(x) = ∞ then ϕn(x) = n for all n ∈ N and limn→∞ ϕn(x) = ∞ = f(x). Assume f(x) ≤ M, some M ∈ R. Given ² > 0 let N ∈ N satisfy N > M and 2−N < ². If n ≥ N then f(x) < n and there exists k ∈ {0, 1, . . . n2n − 1} n n n such that k/2 ≤ f(x) < (k + 1)/2 . Then ϕn(x) = k/2 and k + 1 k 1 0 ≤ f(x) − ϕ (x) < − = < ². n 2n 2n 2n We see that limn→∞ ϕn(x) = f(x) also in this case. This completes the proof that the sequence {ϕn} converges pointwise to f. If f is bounded, then there is M ∈ R such that f(x) ≤ M for all x ∈ E and the argument given above proves the sequence converges uniformly to f

7. Now come some major differences for a while. We extend the definition of the integral of a simple non-negative function Pm to include the case in which the function assumes a positive value on a set of infinite measure. If ϕ = j=1 cjχEj where cj ≥ 0 for j = 1, . . . , m and E1,...,Em are measurable, then Z X∞ ϕ dm = cjm(Ej) j=1

with the convention that 0×∞ = 0. This is obviously the same definition as before if all the sets Ej have finite measure (or only a set Ej for which cj = 0 has infinite measure) and extends the integral by setting it equal to ∞ in case there ϕ assumes a positive value on a set of infinite measure. 8. We now define, and this differs for a while from Royden: If f : R → [0, ∞] is measurable, then Z Z f dm = sup{ ϕ dm : ϕ simple,0 ≤ ϕ ≤ f}

Note: If f, g : X → [−∞, ∞], X a set, then we deﬁne f ≤ g to mean f(x) ≤ g(x) for all x ∈ X. We also deﬁne, if E is a measurable subset of R, f : E → [0, ∞] measurable, Z Z f dm = f˜ dm E where f˜(x) = f(x) if x ∈ E, f˜(x) = 0 if x ∈ R\E.

Exercise 1 f˜ : R → [0, ∞] is measurable.

Exercise 2 An equivalent deﬁnition is Z Z

f dm = sup{ χEϕ dm : ϕ simple,0 ≤ ϕ(x) ≤ f(x), x ∈ E} E

It should be mentioned that these definitions make sense even if f is not measurable, but not much can be proved if we cannot invoke Lemma 1, which requires measurability. While the theorems are the same as in the text, our proofs could be different. So take notes! R Task number one is to realize that if ϕ is simple, non-negative, we have two different definitions of E ϕ. The fact that they are equal, while very easy, is perhaps less obvious than one might think. My hope was that one of the exercises in Exam 2 would make you aware of the factR thatR certain “obvious” things do need proofs. Here it is very easy because we doR have that if 0 ≤ ψ ≤ ϕ simple, then E ψ ≤ E ϕ (old definition ofR integral of a simple function), thus the supremum of E ψ as ψ ranges over all simple functions such that 0 ≤ ψ ≤ ϕ is E ϕ. We need to supplement Royden’s Proposition 2 with the following simple result

Lemma 2 Let ϕ be a non-negative simple function. The map Z

mϕ : E 7→ ϕ : M → [0, ∞] E

is a measure on M; that is, mϕ(∅) = 0 and if {En} is a family of pairwise disjoint measurable sets, then [∞ X∞ mϕ( En) = mϕ(En). n=1 n=1

Moreover, mϕ(E) = 0 for all Lebesgue null sets E.

2 Pm Proof. Write ϕ = j=1 cjχAj where cj ≥ 0, Aj measurable for j = 1, . . . , m. Then X∞ mϕ(E) = cjm(Aj ∩ E) j=1

for all E ∈ M and everything is fairly obvious. We just have to be a bit careful with the possible appearances of ∞. S∞ If m(E) = 0 it should be obvious that mϕ(E) = 0. If E = n=1 En is a disjoint union, then we have Ã ! [∞ X∞ m(Aj ∩ E) = m (Aj ∩ En) = m(Aj ∩ En) n=1 n=1

from which the σ-additivity of mϕ follows at once.

Corollary 3 Let ϕ be a non-negative simple function. Then for every increasing sequence {En} of measurable sets we have Ã ! [∞ mϕ En = lim mϕ(En). n→∞ n=1

9. Parts i and iii of Proposition 8 of page 85, but part iii without the a.e. That is Z Z Proposition 4 i. cf = c f for all c ∈ R, c ≥ 0, f : E → [0, ∞] a non-negative measurable function. Z Z E E iii. f ≤ g for all measurable f, g : E → [0, ∞] such that f ≤ g. E E

Please notice that our definition of integral of a non-negative measurable function differs from that of Royden (that is, eventually we should be able to prove it is the same, but for now it is different) and we do NOT have his Proposition 5. Our proof is a consequence of Royden’s Proposition 2, which we do have, and of our Lemma 1. Other than that, the proof is identical. Proof. i. If c = 0, and we stick to the convention c × ∞ = 0 (a convention that has to be used with care, only when ∗ justified ), theZ result isZ obvious. Assume, thus, c > 0. If ϕ is simple, 0 ≤ ϕ ≤ fZ, then 0Z≤ cϕ ≤ cf. By the definition of R E cf we have cϕ ≤ cf; by Proposition 2 of Royden, this translates to c ϕ ≤ cf and since c > 0 we proved E E E E Z Z 1 ϕ ≤ cf E c E Z for all simple ϕ, 0 ≤ ϕ ≤ f. By the definition of f, it follows that E Z Z Z Z 1 f ≤ cf; i.e., c f ≤ cf E c E E E To get the inverse inequality, let g = cf. Then g ≥ 0 is measurable, 1/c > 0 and by what we proved (replacing f by g and c by 1/c) Z Z Z Z 1 1 1 g ≤ g; i.e., cf ≤ f. c E E c c E E R R iii. Assume now 0 ≤ f ≤ g; f, g measurable on E. If ϕ is simple, 0 ≤ ϕ ≤ f, then 0 ≤ ϕ ≤ g, thus E ϕ ≤ E g. Since ϕ was arbitrary subject to 0 ≤ ϕ ≤ f, we are done.

10. Lebesgue’s Monotone Convergence Theorem and its corollaries. (Points 10, 11, 12 in Royden, pages 87-88.)

∗See note about inﬁnity at the end of these notes.

3 Theorem 5 (Lebesgue’s Monotone Convergence Theorem.) Assume E is a measurable subset of R and for n ∈ N let fn : E → [0, ∞] be measurable and satisfy fn ≤ fn+1 for n = 1, 2,... (That is, fn(x) ≤ fn+1(x) for all x ∈ E, n ∈ N). Then Z Z

( lim fn) = lim fn. E n→∞ n→∞ E

Proof. Let us notice ﬁrst that both limits exist. limn→∞ fn exists because at every x ∈ E, the sequence of (extended) real numbers {fn(x)} is increasing, thus (limn→∞ fn)(x) = limn→∞ fn(x) exists (possibly equal to ∞). Let us call it f; that is, let f = limn→∞ fn. By results from Chapter 3, f is measurable.R By our limited version of PropositionR 8 in Royden (our Proposition 4), the sequence of extended real numbers { E fn} is increasing, thusR J = (limR n→∞ E fn exists (possibly equal to ∞). Moreover, because we must have f ≤ f (I hope this is clear), thus f ≤ f for all n Z n E n E and hence also in the limit; that is J ≤ f. All we need to prove is E Z (1) f ≤ J. E

We may assume that J < ∞, otherwise we are done. Here is where things get nice and ingenious,R sort of. To prove (1) we need to see (if we do it in the obvious way, and it is always good to begin with the obvious) E ϕ ≤ J Z (2) ϕ ≤ J. E

Rfor all simpleR ϕ, 0 ≤ ϕ ≤ f. So pick such a ϕ. It would be nice if one could say that ϕ ≤ fn for some n, because then E ϕ ≤ E fn ≤ J. But a moment’s reﬂection shows this is not necessarily true. But here is the trick. Let a ∈ (0, 1). For n = 1, 2,... deﬁne En = {x ∈ E : aϕ(x) ≤ fn(x)}. S∞ Notice that E1 ⊂ E2 ⊂ E3 ··· and E = n=1 En. In fact, if x ∈ En, then aϕ(x) ≤ fn(x) ≤ fn+1(x), hence x ∈ En+1. If x ∈ E and f(x) = 0, then fn(x) = 0 for all n and ϕ(x) = 0, thus 0 = aϕ(x) ≤ fn(x) and x ∈ En for all n. If f(x) > 0, then (since 0 < a < 1), aϕ(x) < f(x) = limn→∞ fn(x) and there is n such that fn(x) > aϕ(x), hence x ∈ En. Now Z Z Z

J fn ≥ fn ≥ aϕ = amϕ(En). E En En By the corollary to our Lemma 2, (2 follows taking limits for n → ∞.

11. Now we can complete the proof of Proposition 8 of Royden:

Proposition 6 Let E be a measurable subset of R. Z Z Z ii. (f + g) = f + g for all non-negative measurable functions f, g deﬁned on E (at least). E E E

Proof. By Lemma 1 we can ﬁnd increasing sequences {ϕn}, {ψn} of non-negative simple functions converging at every point to f, g, respectively.Z ThenZ{ϕn + ψZn} is an increasing sequence of non-negative simple functions converging to

f + g and since (ϕn + ψn) = ϕn + ψn for all n, the result follows from Theorem 5 E E E

Corollary 7 Lebesgue’s monotone convergence theorem remains valid if fn ≤ fn+1 and the existence of limn→∞ fn only hold a.e. That is:

Let E be a measurable subset of R and for n ∈ N let fn : E → [0, ∞] be measurable and satisfy fn ≤ fn+1 for n = 1, 2,... a.e.; that is, fn(x) ≤ fn+1(x) for all x ∈ E\Nn, Nn a null set, n ∈ N), and assume that f = limn→∞ fn a.e.; i.e., there is a null set N such that f(x) = limn→∞ fn(x) for all x ∈ E\N. Then Z Z

f = lim fn. E n→∞ E

4 S∞ Proof. Let C = N ∪ n=1 Nn. Then C is a null set; by Theorem 5 we have Z Z

f = lim fn, E\C n→∞ E\C

and all integrals over C are zero†. We also have the following corollary. Since it is one of the most common ways one applies Lebesgue’s Monotone Theorem we call it a theorem.

Theorem 8 Let E be a measurable subset of R and assume fn : E → [0, ∞] is measurable for n = 1, 2,.... Then Z X∞ X∞ fn = fn. E n=1 n=1

P∞ P∞ Proof. Notice that both limits exist. n=1 fn is the function which at every x ∈ E is deﬁned by ( n=1 fn)(x) = P∞ n=1(fn(x)) and it exists (possibly equal to ∞) because we have a series of non-negative terms; for a similar reason the series on the right hand side above makes sense. Pn P∞ Let gn = k=1 fk. Then g1 ≤ g2 ≤ · · · and limn→∞ gn = n=1 fn. By Theorem 5, Z Z X∞ lim gn = fn. n→∞ E E n=1 By Proposition 6, Z X∞ Z lim gn = fn. n→∞ E n=1 E The result follows. This could be the point to throw in a very simple but actually quite frequently used result. Z Lemma 9 Let f : E → [0, ∞] be measurable. If f < ∞, then f is ﬁnite a.e.; i.e., the set {x ∈ E : f(x) = ∞} is a E Lebesgue null set.

Proof. Let F = {x ∈ E : f(x) = ∞}. Then F is measurable and cχF ≤ f for all c ∈ (0, ∞). Thus Z Z

cµ(F ) = cχF ≤ f E and since the right hand side of the last equality is ﬁnite, the only way this can hold for arbitrary c > 0 is if m(F ) = 0.

12. Fatou’s Lemma, in a slightly more general version:

Lemma 10 Let {fn} be a sequence of non-negative measurable functions on a set E. Then Z Z

(lim inf fn) dm ≤ lim inf fn dm. E n→∞ n→∞ E

Proof. We use again the characterization of the lim inf of a sequence {an} of extended real numbers as a sup of an inf: µ ¶ µ ¶

lim inf an = sup inf ak = lim inf ak . n→∞ n∈N k≥n n→∞ k≥n

†To be precise, we did not prove yet that this holds for a general non-negative measurable function, but it is an immediate consequence of the fact that it holds for non-negative simple functions.

5 For n ∈ N set gn = (infk≥n fk) (that is, gn is the function on the measurable set E deﬁned by gn(x) = (infk≥n fk(x)) for each x ∈ E); then {gn} is a sequence of non-negative measurable functions on E satisfying g1 ≤ g2 ≤ · · · and limn→∞ gn = lim infn→∞ fn. Theorem 5 (Lebesgue’s Monotone Convergence Theorem) implies Z Z

lim inf fn = lim gn,. E n→∞ n→∞ E Z Z

Obviously (I hope) gn ≤ fn for all n, thus gn ≤ fn and we must have E E Z Z

lim gn ≤ lim inf fn . n→∞ E n→∞ E We are done.

13. Deﬁnition. A measurable function f : E → [0, ∞] is said to be integrable (or Lebesgue integrable) iﬀ Z f dm < ∞. E (Same as in Royden) 14. Royden Propositions 13, 14, page 88. And more.

Proposition 11 (Royden, Chapter 4, Proposition 13) Let f, g : E → [0, ∞] be measurable and assume 0 ≤ f ≤ g. If g is integrable, so is f.

Proof. By Proposition 4 of these notes, part (iii), we have Z Z f ≤ g < ∞.

The same result holds if we replace f ≤ g by f ≤ g a.e.

We can extend now Lemma 2 to the case of an arbitrary non-negative measurable function. But ﬁrst a little exercise.

Exercise 3 Let X be a set, let S be a σ-algebra in X. If Y ∈ S, deﬁne

SY = {A : A ∈ S,A ⊂ Y }

Prove that SY is a σ-algebra in Y and that one has

SY = {E ∩ Y : E ∈ S}.

Proposition 12 Let E be a measurable subset of R and let f : E → [0, ∞] be measurable. Let ME be the family of all Lebesgue subsets of E. Deﬁne mf : ME → [0, ∞] by Z Z Z

mf (A) = f dm = f dm = χAf dm A∩E A E

if A ∈ ME. Then mf is a measure on the σ-algebra ME.

6 R Proof. We already remarked that if A is a null set, then A f dm = 0 for all measurable non-negative functions deﬁned on A. In particular, mf (∅) = 0. To see that mf is σ-additive, let {An} be a sequence of pairwise disjoint measurable [∞ subsets of E, and let A = An. Because of the pairwise disjointness, one sees that n=1 X∞

χA = χAn . n=1 I hope this is evident and the proof seen as being trivial (”hope springs eternal!”. Then X∞

χAf = χAn f. n=1 and by the Theorem 8 (the corollary to LMCT), Z Z X∞ Z X∞

mf (A) = f = χAf = χAn f = mf (An) A E n=1 E n=1

For the next lemma we need the following observation. We don’t state it as a lemma or proposition because it is an obvious consequence of linearity; except that we don’t have linearity yet. Let f, g be measurable and nonnegative on E. If 0 ≤ f ≤ g, then g − f ≥ 0 measurable; if f is integrable, then Z Z Z (g − f) dm = g dm − f dm

Of course g − f ≥ 0 measurable. Now g = f + (g − f) so that Z Z Z (g − f) dm + f dm = g dm. R If f is integrable we can subtract f dm from both sides.

Proposition 13 Let f : E → [0, ∞Z ] be integrable. Then for every ² > 0 there is δ > 0 such that if A is a measurable subset of E of measure < δ, then f dm < ². A

Proof. One can get this as a consequence of Proposition 12. But I’ll do it more or less like RoydenR (though it may not seem so) and derive it from Lemma 1 in these notes. Let ² > 0 be given. By the deﬁnition of E f dm as a sup, there exists ϕ simple, 0 ≤ ϕ ≤ f such that Z Z ² f dm < ϕ dm + . E 2 Then for every measurable subset A of E, since ϕ ≤ f and f, hence also ϕ, is integrable Z Z Z Z ² (f − ϕ) dm ≤ (f − ϕ) dm = f dm − ϕ dm < , A E E E 2 thus Z Z ² f < ϕ dm + A A 2 Pn for all A ⊂ E. Write ϕ = i=1 ciχEi , where we will assume, as we may, ci > 0, Ei measurable, for all i = 1, . . . , n. Let Pn δ = ²/[2 i=1 ci]. If m(A) < δ, then Z Xn Xn Xn ² ϕ dm = c m(A ∩ E ) ≤ c m(A) < c δ = . i i i i 2 A i=1 i=1 i=1 The result follows.

7 15. We begin with the topics of Section 4. In general, given a function u : S → [−∞, ∞], S a set, one deﬁnes u+, u− by ½ ½ u(x) if x ∈ S, u(x) ≥ 0, −u(x) if x ∈ S, u(x) < 0, u+(x) = u−(x) = 0 if x ∈ S, u(x) < 0; 0 if x ∈ S, u(x) ≥ 0.

Brieﬂy: u+ = max(u, 0), u− = − min(u, 0).

Exercise 4 Prove the following properties. Assume u : S → R ∪ {−∞, ∞}, S some set. Just in case, assume S not empty (to avoid trivial nonsense). (a) u+ ≥ 0, u− ≥ 0 (as usual, a function from a set to the extended reals is non-negative iﬀ all its values are non- negative numbers or inﬁnity). (b) u = u+ − u−, |u| = u+ + u−. (c) The previous property characterizes u+, u−. That is, if v, w : S → [−∞, ∞], if u = v − w and |u| = v + w, then v = u+, w = u−.

Our ﬁrst result is quite immediate; the proof is left as an exercise.

Lemma 14 Let f : E → [−∞, ∞], where E is a measurable subset of R. Then f is measurable if and only if both f +and f − are measurable.

Exercise 5 Prove Lemma 14.

Suppose now f : E → [−∞, ∞] is measurable. Then f +, f − are measurable nonnegative functions, and their integral is defined. If and only if at least one of f +, f − is integrable, we define Z Z Z f dm = f + dm − f − dm. E E E The requirement that at least one of f ± be integrable avoids the possibility of having ∞ − ∞. Obviously, if f ≥ 0, this is the same integral as before. But we are going to be more interested in the finite valued cases and so we have a definition.

+ − Deﬁnition 1 Let f : ER→ [−∞, ∞] be measurable. We say f is (Lebesgue) integrable over E iﬀ both f , f are integrable. In this case, E f dm ∈ R.

Lemma 15 Let f : E → [−∞, ∞] be measurable. Then f is integrable if and only if |f| is integrable. Moreover ¯Z ¯ Z ¯ ¯ ¯ ¯ ¯ f dm¯ ≤ |f| dm. E E R Exercise 6 Prove Lemma 15. Should be very easy! The displayed inequality is actually true as long as E f makes sense. The proof does not require results yet to come; it requires knowing that if a real number (or an extended real number) has been written in the form a = c − d where c, d ≥ 0, then |a| ≤ c + d, and it does require knowing that for nonnegative functions the integral of the sum is the sum of the integrals.

One very common way of putting this is: Let f : E → [−∞, ∞] be measurable. Then f is integrable if and only if Z |f| dm < ∞. E

One of the really nice things about the Lebesgue integral is that, as long as the function is measurable, it is always deﬁned for nonnegative functions.

8 1 Deﬁnition 2 LetZ E be a measurable subset of R. If f : E → [−∞, ∞] we will say that f ∈ L (E) if and only if f is measurable and |f| dm < ∞. In other words, L1(E) is the set of all integrable functions of domain E. E

We have reached Proposition 15 in Royden, which I will divide up into three propositions; Propositions 16, 17, 18. The proofs will be identical, so I omit them from these notes.

Proposition 16 Let E be a measurable subset of R. Then L1(E) is a real vector space and the map Z f 7→ f dm : L1(E) → R E is a linear functional on L1(E).

(With a diﬀerent phrasing, this is parts i and ii of Proposition 15 in Royden) Z Z Proposition 17 Let E be a measurable subset of R and let f, g ∈ L1(E). If f ≤ g a.e., then f dm ≤ g dm. E E

Proposition 18 Let E be a measurable subset of R. Let A, B be measurable subsets of E such that A ∩ B = ∅. Then Z Z Z f dm = f dm + f dm A∪B A B for all f ∈ L1(E).

Exercise 7 Let f ∈ L1(E). [∞ (a) Assume {An} is a sequence of pairwise disjoint measurable subsets of E. Let A = An. Prove: The series n=1 X∞ Z f dm converges absolutely and, in fact, n=1 An X∞ Z Z f dm = f dm. n=1 An A

[∞ (b) Assume {An} is an increasing sequence of measurable subsets of E. Let A = An. Prove: n=1 Z Z lim f dm = f dm. n→∞ An A \∞ (c) Assume {An} is a decreasing sequence of measurable subsets of E. Let A = An. Prove: n=1 Z Z lim f dm = f dm. n→∞ An A

16. We have reached what is probably the most important theorem of this Chapter, namely Lebesgue’s Dominated Con- vergence Theorem (LDCT). Our proof diﬀers slightly from Royden’s proof. It doesn’t need two steps.

9 Theorem 19 Assume E is a measurable subset of R and let {fn} be a sequence of measurable function on E. If limn→∞ fn = f a.e., and there exists an integrable function g such that |fn| ≤ g a.e. for all n ∈ N, then Z Z

lim fn dm = f dm. n→∞ E E

Proof. Clearly |f| ≤ g a.e. so that |f − fn| ≤ 2g a.e. and 2g − |f − fn| is a nonnegative measurable function. Moreover

lim inf(2g − |f − fn|) = lim (2g − |f − fn|) = 2g. n→∞ n→∞ By Fatou’s Lemma, Z Z ³ ´ Z 2g dm = lim (2g − |f − fn|) dm ≤ lim inf (2g − |f − fn|) dm n→∞ n→∞ E E µZ Z ¶ ZE Z

= lim inf 2g dm − |f − fn| dm = 2g dm − lim sup |f − fn| dm, n→∞ E E E n→∞ E R where we used that lim infn(a − bn) = a − lim supn bn. All quantities involved here are ﬁnite, especially E 2g dm. So we can subtract it to get Z

lim sup |f − fn| dm ≤ 0. n→∞ E But if the limsup of a sequence of nonnegative quantities is ≤ 0, then the sequence converges to 0.

17. Section 2 of Chapter 4; with some big modifications of the proofs and without duplicate definitions. First, a summary of how things were defined in these notes. And in class. (a) The integral of a simple function that takes of all of its non-zero values on sets of finite measure was defined. Such simple functions form a real vector space and it was proved that the integral was a linear functional on such a space, and that it preserved order. That is, if φ ≤ ψ, meaning that φ(x) ≤ ψ(x) for all x, then the integral of φ Pn is less than or equal that of ψ. One conclusion obtained from all this is that if φ = i=1 ciχAi , where each Ai is measurable and m(Ai) < ∞ if ci 6= 0, then Z Xn (3) φ = cim(Ai). R i=1 (In all of this 0 · ∞ = 0; to be used with care.) Pn Because there are many ways in which a simple function can be written in the form i=1 ciχAi , one could not use (3) as a definition, but had to prove it as a lemma or proposition. So far, all this is as in Royden. (b) If E is a measurable subset of R, we define for a simple function φ Z Z

(4) φ = χEφ. E R A simple function times the characteristic function of a simple function is still simple. It is a trivial exercise to Pn Pn verify that if φ = i=1 ciχAi , where each Ai is measurable and m(Ai) < ∞ if ci 6= 0, then χEφ = i=1 ciχE∩Ai Z Xn (5) φ = cim(E ∩ Ai). R i=1 It is also trivial to see that χ φ is the simple function obtained from φ by setting φ = 0 outside of E. R E (c) The definition of E φ was extended a bit to include the possibility of φ taking a non-zeroR value on a set of infinite measure, in case that value was positive. That is, we notice that the original definition of φ for a simple function makes sense if we just set, as one should c × ∞ = ∞ if c > 0, as long as there are no negative values to cause the cataclysmic mass+antimass effect one gets when subtracting infinity from infinity. One needs to verify that (3) is still valid, but that’s a simple exercise.

10 (d) The integral is defined next for all measurable f : E → [0, ∞] by Z Z (6) f = sup{ φ : φ simple, 0 ≤ φ(x) ≤ f(x) ∀x ∈ E}. E E R The integral can, of course, be equal to ∞. At this point one has two definitions of E φ, if φ is simple. But because one knows that the integral is order preserving among simple functions, it is between very easy and trivial to prove that both definitions yield the same value for the integral. (e) In our development, the approximation lemma Lemma 1 is a key element. Every non-negative measurable function is the pointwise limit of an increasing sequence of non-negative simple functions; the limit is uniform if the function is bounded. (f) It is a trivial exercise to prove that if f : R → [0, ∞] is measurable, if E is a measurable subset of R, then Z Z

f = χEf. E R R (g) We now proveR for non-negativeR valued measurable functions that E cf = c E f if c ∈ R, c ≥ 0, and that 0 ≤ f ≤ g on E implies E f ≤ E g. Proving that the integral of a sum is the sum of the integrals is postponed. (h) A result we will need to prove the next important result, Lebesgue’s Monotone Convergence Theorem, is Lemma 2; more precisely its Corollary 3. (i) Lebesgue’s Monotone Convergence Theorem (LMCT) can be proved now. (j) Because every non-negative measurable function is the pointwise limit of an increasing sequence of non-negative simple functions, and the integral is linear on simple functions, thanks to LMCT it is now easy to prove that the integral of a sum of non-negative measurable functions is the sum of the integrals. (k) The next result is the series version of LMCT, Theorem 8. (l) One result that needs to be slipped into all this is that if E is a null set, then ∈E f =R 0 for all non-negative measurable f. The result is trivial if f is simple, thus is immediate from the definition of E f. Because of this, Theorems 5, 8 remain valid if all inequalities are replaced by inequalities a.e., and all pointwise limits are replaced by limits a.e. (m) An easy and useful result for a non-negative, measurable function: R If f < ∞ then f is finite a.e. on E. Combined with LMCT, the series version, one gets that if one has a series P E P R P n fn of non-negative measurable functions on the measurable set E and if n E fn < ∞, then the series n fn converges a.e. (n) We then proved Fatou’s Lemma, Lemma 10. R (o) A non-negative measurable function is said to be integrable over E iff E f < ∞. (p) We now move to general measurable functions. For a measurable f : E → [−∞, ∞] we define f +, f −, and verify the obvious facts that f is measurable if and only if both f +, f − are measurable; f = f + − f −, |f| = f + + f −. We define L1(E), the set of integrable functions over E as the set of all measurable f such that both f +, f − are integrable, and define for f ∈ L1(E), Z Z Z (7) f = f + − f −. E E E One should verify that if φ is simple, then φ ∈ L1(E) if and only if any set on which φ assumes a non-zero value has finite measure when intersected with E, and that the integral as defined by (3) is the same as by (7). R 1 Another important but trivial thing to verify is that f ∈ L (E) if and only if f is measurable and E |f| < ∞. From the definition of the integral, it is also obvious that ¯Z ¯ Z ¯ ¯ ¯ ¯ ¯ f¯ ≤ |f| E E for all f ∈ L1(E).

11 R 1 1 (q) One proves now that L (E) is a real vector space and the map f 7→R E f Ris an order preserving functional on L (E). 1 Also: If f ∈ L (E) and we define mf : ME → R by mf (A) = f = χAf if A ∈ ME = {A ∈ M : A ⊂ E}, S∞ A E P if {An} is a sequence of pairwise disjoint sets and A = n=1 An, then the series n mf (An) converges absolutely and its limit is mf (A). This is actually fairly immediate if one divides it into positive and negative parts. (r) The big result: Lebesgue’s Dominated Convergence Theorem (LDCT), Theorem 19. This is proved as a consequence of Fatou’s Lemma. Our proof uses essentially the same idea as Royden. And now we can get to Section 2. We do NOT redefine any integral. What we notice first (or should have done so) 1 is the very obvious fact that if f : E → R is bounded and m(E) < ∞, then f ∈ L (ER). In fact, if |f(x)| ≤ M for a.e. x ∈ E, then |f| ≤ MχE a.e.; integrability of f is a consequence of Proposition 11 and MχE = Mm(E) < ∞. Here is a fact mentioned in class:

Proposition 20 Let E be measurable, f : E → R measurable and bounded. Then there exist sequences {φn}, {ψn} such that φ1 ≤ φ2 ≤ · ≤ f ≤ · ≤ ψ2 ≤ ψ1 a.e. on E, and both sequences converge uniformly to f on E.

Proof. Assume |f(x)| ≤ M for x ∈ E. Then f +M (i.e., the function deﬁned by f(x)+M for all x ∈ E) is measurable and nonnegative; by Lemma 1 there exists a sequence {vn} of nonnegative simple functions increasing to f and converging uniformly to f on E. Let φn = vn − M. Since f bounded, measurable, implies −f is bounded measurable, by what we proved there is an increasing sequence {wn} of simple functions converging uniformly to f on E; set ψn = −wn. It is now quite trivial to PROVE

Proposition 21 If f : E → R is measurable and bounded; m(E) < ∞, then Z Z

(8) intEf = sup{ ϕ : ϕsimple, ϕ ≤ f} = inf{ ϕ : ϕsimple, ϕ ≥ f}. E E

Proof. I’ll do the first equality in (??). The second one can be proved similarly. Or one gets it as a consequence of the first one, by replacing f by −f. In the first place, by Proposition 17 it is clear that Z

intEf ≥ sup{ ϕ : ϕsimple, ϕ ≤ f}. E

To reverse the inequality, by Proposition 21 there is an increasing sequence of simple functions {φn} converging uniformly to f. We can now use either LDCT or LMCT to see that Z Z

(9) φn → f. E E To use LDCT, we use the fact that a uniformly convergent sequence whose limit is bounded and be uniformly bounded; there is thus M such that |φn| ≤ MχE, and because m(E) < ∞, MχE is integrable and (9) follows. If we prefer using LMCT, we can let M be a bound for f; |f(x)| ≤ M for all (or almost all) x ∈ E. Then f + M is non-negative. Letting A = {x ∈ E : φ (x) < −M}, we can replace φ by χ φ and thus can assume that φ + M ≥ 0 for all n; LMCT n n R R n E\An n n now applies to prove that E(φn + M) → E(f + M); subtracting Mm(E) we get (9). Having proved this we notice that if Z s < f, E then there is n such that Z Z

s < ϕn ≤ sup{ ϕ : ϕsimple, ϕ ≤ f}; E E completing the proof of the ﬁrst equality in (??).

A result that could have been done earlier is:

12 R Proposition 22 Let f : E → [0, ∞] be measurable. If E f = 0, then f = 0 a.e. on E.

S∞ 1 Proof. Let A = {x ∈ E : f(x) > 0}. Then A = n=1 An, where An = {x ∈ E : f(x) > 1/n}. Clearly 0 ≤ n χAn ≤ f, thus Z Z 1 1 0 ≤ m(An) = χAn ≤ f = 0. n E n E

Thus m(An) = 0 for all n, hence also m(A) = 0. It is now quite easy to prove:

Theorem 23 Let f : E → R be bounded, where E is a measurable set, m(E) < ∞. Then f is measurable if and only if Z Z (10) sup{ ϕ : ϕsimple, ϕ ≤ f} = inf{ ϕ : ϕsimple, ϕ ≥ f}. E E

Proof. Half of the theorem, that measurability implies (10) is the content of PropositionR 21. For theR converse, assume (10). For each n ∈ N there will exist simple functions φn, ψn such that φn ≤ f ≤ ψn and E ψn − E φn < 1/n. (Given sets A, B of real numbers, or of extended real numbers, such that a ≤ b for every a ∈ A, b ∈ B, then supA = infB if and only if for every n ∈ N there exist an ∈ A, bn ∈ B such that bn − an < 1/n.) Let

∗ ∗ φ = sup φn, ψ = inf ψn. n n

∗ ∗ ∗ ∗ Then φ , ψ are measurable, φn ≤ φ ≤ f ≤ ψ ≤ ψm for all n, m. All this should be evident (or not.). Now Z Z Z Z Z ∗ ∗ ∗ ∗ (ψ − φ ) = ψ − φ ≤ ψn − φn < 1/n E E E E E R ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ and it follows that E(ψ − φ ) = 0. Since ψ − φ ≥ 0, Proposition 22 implies ψ = φ a.e., forcing φ = f = psi a.e. Thus f is measurable being equal a.e. to a measurable function. This takes care of most of Section 2. Lebesgue’s bounded convergence theorem is an immediate consequence of LDCT. Or one can use Royden’s proof, that also has some instructive value. The theorem about Riemann integrability at the end of the section will be left as an exercise; same as in Royden.

Note on inﬁnity

We should distinguish perhaps between a static infinity or 0 and a potential infinity or 0. Static zero times a static infinity is 0. Example: If A is a set of infinite measure, then 0 · (A) = 0. Or, if lim an = ∞, then 0 limn→∞ an = 0. But if we have limn→∞ an = ∞, limn→∞ bn = 0, then limn→∞(anbn) is not necessarily 0 because we are in the potential case here. That is, if we define in this course 0 × ∞ = 0, we have to be very careful to realize that limn→∞(anbn) = (limn→∞ an)(limn→∞ bn) is not always true.