MA 2210 - Real Function Theory

Francesco Di Plinio

BROWN UNIVERSITY MATHEMATICS DEPARTMENT,BOX 1917, PROVIDENCE, RI 02912, USA E-mail address: [email protected] (F. Di Plinio)

CHAPTER 1

Outer measures, measures and the Lebesgue measure

1.1 Outer measure. Let X be a set. An outer measure µ is a monotone, countably subadditive function on the power set of X with values in the nonnegative extended real numbers and such that µ( ) = 0. In symbols µ : (X ) [0, ] satisfies ; P → ∞ (1) (normalization) µ( ) = 0; ; (2) (monotonicity) F, F 0 P(X ), F F 0 = µ(F) µ(F 0); (3) (countable subadditivity)∈ if Fn⊂is a countable⇒ ≤ collection of subsets of X then { } [∞ X∞ µ Fn µ(Fn). n=1 ≤ n=1 1.2 Outer measure from a premeasure. Let X be a set, E be a collection of subsets of X and σ : E [0, ] be any function. Define for F (X ) → ∞ ∈ P ¨ « X∞ [∞ µ(F) = inf σ(En) : over E0 = En : n N E with F En n=0 { ∈ } ⊂ ⊂ n=0 Then µ is an outer measure on X , generated by the premeasure σ. To see this, we need to verify the three properties of Definition 1.1. First of all, the empty set is covered by the empty subcollection of E. Thus µ( ) = 0. Secondly, if F F 0, then any countable E0 E ; ⊂ ⊂ covering F 0 covers F as well. It follows that µ(F) µ(F 0) since for F we are taking an infimum over a larger set. Finally, let Fn be a countable≤ collection of sets and F be its union. n Let " > 0 be given. For each Fn, we may find a subcollection En = Ej : j N of E which { ∈ } covers Fn and X n j σ(Ej ) µ(Fn) + "2− . j N ≤ ∈ n Then E0 = Ej : j, n N covers F, so that { ∈ } X X n X µ(F) σ(Ej ) µ(Fn) + " ≤ n N j N ≤ n N ∈ ∈ ∈ which yields the claim (3) due to " being arbitrary. We were able to exchange the summation order because the terms in the summation are nonnegative.

d 1.3 Lebesgue outer measure via dyadic cubes. A cube I R is the cartesian product ⊂ of bounded intervals (open, closed or half-open unless otherwise speciﬁed) I j, j = 1, . . . , d. d We now consider the problem of computing the volume of an arbitrary subset of R . Our d measuring tools will be dyadic cubes. Let Q0 = [0, 1) be the unit cube. The collection of 3 4 1. OUTER MEASURES, MEASURES AND THE LEBESGUE MEASURE

dilates and translates of Q0

[∞ n n n n = n, n = [k12− , (k1 + 1)2− ) [kd 2− , (kd + 1)2− ), k1,..., kd Z D n= D D × · · · ∈ −∞ d n is termed the standard dyadic grid on R . We denote by `(Q) = 2− the sidelength of Q if (k) Q n. We denote by Q the unique dyadic cube (k-th ancestor of Q) with sidelength k 2 `∈(Q D) that contains Q. We record the following property for further use d (1.1) x R , n Z, !Q n with x Q. ∀ ∈ ∀ ∈ ∃ ∈ D ∈ A consequence of this is that forms a grid in the sense that whenever Q,Q0 , either D ∈ D Q Q0 have trivial intersection or one of the inclusions Q Q0,Q0 Q holds. In symbols ∩ ⊂ ⊂ (1.2) Q,Q0 = Q Q0 Q,Q0, . ∈ D d⇒ ∩ ∈ { ;} We consider the outer measure µ : (R ) [0, ] generated by the volume premeasure P d →nd ∞ σ(Q) = `(Q) = 2− , Q n, n Z. ∈ D ∈ We call this outer measure the Lebesgue outer measure. The remainder of this section will be dedicated to the proof of the fact that the outer measure coincides with the Euclidean volume on dyadic cubes, and, more generally, on cartesian products of open, closed, half- open intervals. A preliminary but fundamental lemma is the following.

d 1.3.1 LEMMA. Let I = I1 Id be an axis-parallel cube in R with inf I j = aj, sup I j = bj. × · · · × Assume I is the disjoint union of dyadic cubes Q j : j N . Then n { ∈ } Y X I = (bj aj) = σ(Q j). | | j=1 − j N ∈ PROOF. We argue in the case d = 1 and leave the higher dimensions as an exercise. We need to prove that whenever I is an interval and Q j : j N is a collection of pairwise disjoint dyadic intervals whose union is I, we have{ ∈ } X sup I inf I = σ(Q j). − j N ∈ Let a = inf I. Write Q j = [uj, vj) for definiteness. Let " > 0 be given and set x j = uj j j − "2− , yj = vj +"2− , R j = (x j, yj). Then R j is an open cover of the compact set [a, a+b "]. { } − It thus admits a finite subcover, which we relabel (x j, yj), j = 1, . . . , N with x j increasingly { } ordered and such that if R j Rk = , then neither R j Rk nor the opposite inclusion holds. ∩ 6 ; ⊂ It must be that x1 < a, yN > b ", and x j+1 < yj so that − N 1 N X− X X b " a yN x1 = yN xN + x j+1 x j + yj x j σ(Q j) + 2" − − ≤ − − j=1 − ≤ j=1 − ≤ j N ∈ which by taking " 0 yields σ Q P σ Q . To prove the reverse inequality, by taking ( ) j N ( j) limits, it suffices to→ show that ≤ ∈ N X σ(Q j) σ(Q) = 1 j=1 ≤ 1. OUTER MEASURES, MEASURES AND THE LEBESGUE MEASURE 5

Q j N Q b Q a We can rearrange j : = 1, . . . , so that sup j = j min j+1 = j+1. Then { } N 1 N ≤ N X X X b a b a − a a b a Q 1 N 1 = N N + j+1 j j j = σ( j) ≥ − − j=1 − ≥ j=1 − j=1 as claimed. d 1.3.2 PROPOSITION (CONSISTENCY). Let µ denote the Lebesgue outer measure on R above deﬁned. Then d µ(Q) = `(Q) for all dyadic cubes Q .

∈ D d PROOF. First of all, since Q is a covering of Q, we have µ(Q) `(Q) = σ(Q). Assume now that Q j is a cover of Q{by} dyadic cubes such that ≤ { } X∞ (1.3) σ(Q j) σ(Q). j=1 ≤ We prove that equality must hold in (1.3), whence µ(Q) = σ(Q). Notice that (1.3) entails that Q j Q for all j (otherwise the opposite inequality would hold). Let jk be the indices of those⊆ elements of Q j which are maximal with respect to the order relation induced by inclusion. Then it must{ hold} that [ [ Q Q Q, k ` Q Q jk = j = = = jk j` = k j 6 ⇒ ∩ ; The ﬁrst of these assertions follows by observing that every point of Q is contained in a maximal Q j, since `(Q j) is bounded above by `(Q). The second follows because if the intersection Q Q were nontrivial, one of the inclusions would hold, contradicting maximality. jk j` But then ∩ X X ∞ Q ∞ Q Q σ( j) σ( jk ) = σ( ) j=1 ≥ k=1 using Lemma 1.3.1 for the second equality. Hence equality must hold in (1.3), and the proof is complete. We now extend the above consistency to open intervals. The main tool (in fact, a more general version which will be useful in the sequel) is the following covering lemma.

d 1.3.3 WHITNEYCOVERINGLEMMA. Let O R be an open set. Then there exists a subcollection I of dyadic cubes such that ⊂ [ (1) O = I. I I (2) the cubes∈ of I are pairwise disjoint (3) the distance of the center of I, cI from the complement of O is comparable to `(I). PROOF (SKETCH). Let I be the collection of maximal dyadic cubes with the property that 1 c (1.4) I O, `(I) 10 dist(cI , O ). ⊂ ≤ These are pairwise disjoint by maximality. Let now x O. Since O is open, the collection of 1 ∈ c those dyadic cubes containing x with `(I) 10 dist(cI , O ) is nonempty. Thus there exists a ≤ 6 1. OUTER MEASURES, MEASURES AND THE LEBESGUE MEASURE

maximal one, which belongs to I. This proves that O is covered by the union of I I. Now, (1) 1 1 ∈ c we must have that I the parent of I violates (1.4). In either case, `(I ) 10 dist(cI(1) , O ), so that ≥ c c dist(cI , O ) 2`(I) + dist(cI(1) , O ) 30`(I), ≤ ≤ and comparability follows. 1.3.4 COROLLARIES. Let µ denote the Lebesgue outer measure generated by the dyadic cubes and denote Euclidean volume. | · | d (1) if I R is any axis parallel cube then µ(I) = I , the Euclidean volume of I; d (2) for all⊂ A R , | | ⊂ X µ(A) = inf I I I | | ∈ d where I ranges over all countable collections of cubes of R whose union covers A; (3) the Lebesgue outer measure is translation invariant and respects dilations, in the sense d d that, for all t R , s > 0, A R ∈ ⊂ d µ(A + t) = µ(A), µ(sA) = s µ(A). PROOF. Exercise. We now move onto the approximation properties of the Lebesgue outer measure µ. From now on, in view of Proposition 1.3.2, we omit the distinction on µ and σ on dyadic cubes.

d 1.3.5 LEMMA. Let F R be any set. Then ⊂ µ F inf µ O ( ) = O F ( ) O open⊃ PROOF. It sufﬁces to assume that µ(F) < , otherwise there is nothing to prove, and show that for all " > 0 there exists an open∞ set O with µ(O) µ(F) + 2". To do so, let ≤ E0 = Q j : j N be a cover of F by dyadic cubes such that { ∈ } X (1.5) µ(Q j) µ(F) + ". j N ≤ ∈ j Let Qfj be the open cube with the same center as Q j and volume µ(Q j)+"2− . Then O = jQfj is an open set containing F. Furthermore, in view of the preceding corollary and of the above∪ display,

X∞ X j µ(O) µ(Qfj) = (µ(Q j) + "2− ) µ(F) + 2" ≤ j=1 j N ≤ ∈ as claimed. 1.4 σ-algebras and measures. A subcollection (X ) such that (1) (nontrivial) , F ⊂ P c (2) (closed under; ∈ complement) F F = F , (3f) (closed under ﬁnite union) F, G∈ F ⇒= F∈ FG hold is said to be an algebra of subsets of∈X F. If (1),⇒ (2)∪ and∈ F [ (3) (closed under countable union) Fn : n N = F = Fn { ∈ } ⊂ F ⇒ n N ∈ F ∈ 1. OUTER MEASURES, MEASURES AND THE LEBESGUE MEASURE 7

hold instead, is called a σ-algebra of subsets of X . We refer to a pair (X , ) as a measurable space. F F

1.4.1 REMARK. Firstly, it follows from the definition that if α : α A is any collection of {F ∈ } σ-algebras on X , then = α α is also a σ-algebra. Given any collection of sets E (X ) we denote by σ(E) theF intersection∩ F of all the σ-algebras containing E. Such an intersection⊂ P is nonvoid, since (X ) is a σ-algebra containing E. It is useful to observe that, by definition P E0 σ(E) = σ(E0) σ(E). ⊂ ⇒ ⊂ Let now (X , ) be a measurable space. A positive measure is a set function F µ : [0, ], F → ∞ with the countable additivity property X (1.6) µ(F) = µ(Fn) n N ∈ whenever Fn : n N are pairwise disjoint sets whose union is F. Then (X , , µ) is called measure{ space∈ and} ⊂ the F elements of are called measurable (or -measurable)F sets. F F 1.4.2 PROPOSITION. (basic properties of measure) Let (X , , µ) be a measure space and throughout Ej be measurable sets. There holds F (P1) µ( ) = 0; ; (P2) (monotonicity) if E1 E2 then µ(E1) µ(E2); (P3) µ is countably subadditive;⊂ ≤ [ (P4) If Ej Ej 1 for all j, then µ Ej = lim µ(Ej) + \ If E⊂ and E E for all j, then E E , and the assumption (P5) µ( 1) < j j+1 µ j = lim µ( j) ∞ ⊃ µ(E1) < is necessary; ∞ (P6) (lower semicontinuity) µ lim inf Ej lim inf µ(Ej); ≤ S (P7) (upper semicontinuity) lim sup µ(Ej) µ lim sup Ej provided µ Ej < . ≤ ∞ 1.4.3 REMARK. In fact, for properties (P1) and (P2), the finite additivity property (1.7) F, G , F G = = µ(F) + µ(G) = µ(F G) ∈ F ∩ ; ⇒ ∪ would suffice. PROOFOF PROPOSITION 1.4.2. Exercise. The following lemma provides a partial converse to (P3) 1.4.4 LEMMA. Let be a σ-algebra and µ : [0, ] be a countably subadditive set function satisfying theF finite additivity property (1.7)F →. Then∞µ is a measure on . F PROOF. We need to show that µ is countably additive. Let Fn : n N be pairwise disjoint sets whose union is F. We need to prove (1.6). Relying{ on countable∈ } ⊂ F subadditivity, we only need to prove the inequality in (1.6). To do so we may assume that the left hand side µ(F) is finite. This actually≥ implies that the right hand side is finite as well. Indeed, using finite additivity and monotonicity N N X [ µ(Fn) = µ Fn µ (F) . n=1 n=1 ≤ 8 1. OUTER MEASURES, MEASURES AND THE LEBESGUE MEASURE

and the in (1.6) follows by taking the limit as N in the above left hand side. ≥ → ∞ 1.5 Complete measures. Carathéodory’s theorem. For a measure space (X , , µ), define F := F : µ(F) = 0 , 0 := N X : F with N F . N { ∈ F } N { ⊂ ∃ ∈ N ⊂ } Note that and 0 are both closed under countable unions. The measure space (X , , µ) N N F is said to be complete if 0 . N ⊂ F 1.5.1 LEMMA. Let (X , , µ) be a measure space. Then F := E N : E , N 0 F { ∪ ∈ F ∈ N } is a σ-algebra and there exists a unique measure µ on such that (X , , µ) is a complete measure space, with µ(F) = µ(F) for all F . F F ∈ F PROOF. Since and 0 are both closed under countable unions, so is . We show closure under complement.F N Let A = E N , with E, F and N F. ReplacingF possibly c c c c c F by F E , N by N E , we can assume∪ E∈ FF = . This∈ way F (E N⊂) = (E F) (F N ). c c But (F ∩ N ) F and∩ (E F) . This proves∩ the; closure under∪ complement.∪ ∪ ∩ ∩ ⊂ ∪ ∈ F Now define µ(E N) := µ(E). This is well defined, since if E N = F M, E, E0 , ∪ ∪ ∪ ∈ F N, N 0 0 subsets respectively of F, F 0 one has E E0 F 0, whence ∈ N ∈ N ⊂ ∪ µ(E) µ(E0 F 0) µ(E0) + µ(F 0) µ(E0) ≤ ∪ ≤ ≤ and by symmetry, equality must hold. Clearly this extends µ on . The verification that µ is a measure is easy and we omit it. To show that it is unique, let νFbe another measure with

the same properties, E , N 0, F such that µ(F) = 0 and N F. Then ∈ F ∈ N ∈ N ⊂ ν(E N) ν(E) + ν(N) ν(E) + ν(F) = µ(E) = µ(E N) ∪ ≤ ≤ ∪ and by symmetry we conclude µ = ν To show that it is complete, let E , N 0, µ(E N) = 0 and F E N. By a similar argument as above we can assume∈ FE N∈= N . ∪ ⊂ ∪ ∩ ; Then µ(E) = 0, and F E E, so that F E 0. We also have N F 0. So actually ∩ ⊂ ∩ ∈ N ∩ ∈ N F = (F E) (F N) 0, and in particular F . Therefore µ is complete. ∩ ∪ ∩ ∈ N ∈ F Now, let µ be an outer measure on X . We say that F is a µ-measurable set if c (1.8) µ(G) = µ(F G) + µ(F G) G (X ). ∩ ∩ ∀ ∈ P 1.5.2 REMARK. Suppose µ is generated by a premeasure (σ, E). Then (1.8) holds for F if and only if c (1.9) µ(E) = µ(F E) + µ(F E) E E ∩ ∩ ∀ ∈ We leave this proof as an exercise.

1.5.3 CARATHÉODORY’S THEOREM. Let µ be an outer measure on X . Then the subcollection F of (X ) such that (1.8) holds is a σ-algebra • the restriction of µ ∈to F isP a complete measure. • F The proof is divided into steps. The ﬁrst, trivial one is to observe that contains and is closed under complement by virtue of the symmetry of (1.8). We beginF the actual work.; 1. OUTER MEASURES, MEASURES AND THE LEBESGUE MEASURE 9

STEP 1. is closed under finite union and µ is finitely additive on . Let A, B X be such that (1.8)F holds for F = A and F = B. Let G X be arbitrary. Then,F using (1.8)⊂ twice and monotonicity ⊂ c c c µ(G) = µ(G A B) + µ(G A B) + µ(G A B ) + µ(G (A B) ) ∩ ∩ ∩ ∩ c ∩ ∩ ∩ ∪ µ(G A B) + µ(G (A B) ) ≥ ∩ ∪ ∩ ∪ which is the nontrivial half of (1.8) for A B. Therefore A B . Now, if A and B are pairwise disjoint and belong to , the fact∪ that µ(A B) = µ∪(A) +∈µ F(B) is a triviality. F ∪ STEP 2. We show that is closed under countable disjoint union, which, together with closure with respect to complementF and finite union, suffices by the usual argument. Let Fn : n N be pairwise disjoint, A be their union, and AN = F1 FN . By the previous{ ∈ step,} ⊂AN Fbelongs to . We then have for an arbitrary G, ∪ · · · ∪ F N c c X c µ(G) = µ(G AN ) + µ(G (AN ) ) µ(G AN ) + µ(G A ) = µ(G Fn) + µ(G A ) ∩ ∩ ≥ ∩ ∩ n=1 ∩ ∩ and letting N and using subadditivity of outer measure → ∞ X∞ c [ c c µ(G) µ(G Fn) + µ(G A ) µ G Fn + µ(G A ) = µ(G A) + µ(G A ), ≥ n=1 ∩ ∩ ≥ n N ∩ ∩ ∩ ∩ ∈ which is the nontrivial part of (1.8) for A. So is a σ-algebra, and using the finite additivity and Lemma 1.4.4, we conclude µ restricted toF is a measure. F STEP 3. We show that µ is complete. Let F be a set with µ(F) = 0 and N F. Then µ(N) = 0. In fact we have using subadditivity∈ F and monotonicity that ⊂ c µ(G) µ(G N) + µ(G N ) µ(N) + µ(G) µ(G) ≤ ∩ ∩ ≤ ≤ 1.6 The Lebesgue and Borel σ-algebra. We keep denoting the Lebesgue outer measure d d on R by µ. Let be the σ-algebra of subsets of R such that (1.8) holds. We call the Lebesgue σ-algebra.L We already know that µ restricted to is a complete translationL invariant, dilation-respecting measure. As we have seen, givenL any topological space X , we denote the σ-algebra generated by the collection of open subsets of X by (X ) and call it the Borel σ-algebra. By the Whitney covering lemma,O each open set can beB written as a d countable union of dyadic cubes. Therefore (R ) = σ( ). Let us explore the relationship d between (R ), or simply , and . B D B B L 1.6.1 LEMMA. The following statements hold: (1) . (2) D ⊂ L . B L PROOF. To derive the second statement from the first, Let us observe that, by virtue for ( instance of Whitney’s covering Lemma 1.3.3, σ( ), therefore σ( ). On the other hand, it is easy to write the unit half closedO cube ⊂ asD the countableB intersection ⊂ D of open rectangles. Thus σ( ) as well. Hence, the first statement also implies . To disprove equality,B there ⊃ areD two possibilities which are explored in later exercises.B ⊂ L 10 1. OUTER MEASURES, MEASURES AND THE LEBESGUE MEASURE

We now move onto the proof of (1). Let Q be a dyadic cube. By Remark 1.5.2, to show that Q it sufﬁces to show that

∈ L c µ(Q0) µ(Q0 Q) + µ(Q0 Q ) Q0 . ≥ ∩ ∩ ∀ ∈ D If either Q0 Q = , or Q0 Q, there is nothing to prove. The last remaining case is when ∩ ; ⊂ Q Q0. Let then R j be a cover of Q0 by disjoint dyadic cubes such that { } X ( σ(R j) µ(Q0) + ". j ≤

Clearly, there holds `(R j) > `(Q) for ﬁnitely many j. By breaking each such R j with into subcubes of sidelength `(Q) we can construct another countable disjoint cover of Q0, say P P Sj , with j σ(R j) = j σ(Sj) and such that either Sj Q = or Sj Q. Then, Sj : Sj Q { } c ∩ ; ⊂ { ⊂ } covers Q and Sj : Sj Q = covers Q0 Q whence { ∩ ;} ∩ c X X X µ(Q) + µ(Q0 Q ) σ(Sj) + σ(Sj) = σ(R j) µ(Q0) + ", ∩ ≤ Sj Q sj Q= j ≤ ⊂ ∩ ; and we are done.

1.6.2 PROPOSITION. (continuities) Denote by µ the Lebesgue (outer) measure. Then (1) if A then µ(A) = inf µ(O) : A O, O open (2) if A ∈ L then µ(A) = sup{ µ(K) : K⊂ A, K compact} (3) A ∈ Lif and only if for all{ " > 0 there⊂ exists an open} set O such that ∈ L A O, µ(O A) < "; ⊂ \ (4) A if and only if for all " > 0 there exists an open set O and a closed set C such that∈ L C A O, µ(O C) < ", ⊂ ⊂ \ and C can be chosen to be compact if µ(A) < ; d ∞ (5) if A then there exist a set G δ(R ) (countable intersection of open sets) and ∈ L d ∈ G a set F σ(R ) (countable union of closed sets) such that ∈ F F A G, µ(G A) = µ(A F) = 0. ⊂ ⊂ \ \ PROOF. The property (1) is a simple consequence of Lemma 1.3.5. The properties (2)- (5) are left as an exercise. d 1.6.3 COROLLARY. The measure space (R , , µ) is the completion of the measure space d (R , , µ) L B d PROOF. The measure space (R , , µ) is complete, it thus suffices to show that it is con- d tained in the abstract completion ofL(R , , µ). Let A . We can use (4) of Proposition 1.6.2 to find a Borel set F with M := A FBof null Lebesgue∈ L measure. Then applying again the same property we find a Borel set N\ M with µ(N M) = 0. It follows that µ(N) = 0 as well. So A = F M with F Borel and M⊃ subset of a Borel\ null set N. Thus A belongs to the completion of ∪ . which finishes the proof. B 1. OUTER MEASURES, MEASURES AND THE LEBESGUE MEASURE 11

1.7 Non-Lebesgue measurable sets. We present hereby an extended version of the classical example of a subset of R which is not Lebesgue measurable. Coupled with completeness, that is if A is Lebesgue measurable and µ(A) = 0, then all subsets of A are Lebesgue measurable, this yields a necessary and sufficient condition for the existence of nonmeasurable subsets of a set A ∈ L 1.7.1 PROPOSITION. Let A be a Lebesgue measurable subset such that B (X ) : B A . Then µ(A) = 0. { ∈ P ⊂ } ⊂ L PROOF. The equivalence relation x y x y Q ∼ ⇐⇒ − ∈ partitions R into (uncountably many) equivalence classes. Using the axiom of choice, we may select from each a representative. Let E R be the set whose members are the selected representatives. ⊂ Let qn be an enumeration of the rationals and En = E + qn. The definition leads to the identities{ } [ En Em = (m = n), En = R. ∩ ; 6 Define An = A En. Let K be any compact subset of An. By assumption An, K are measurable. ∩ We claim that µ(K) = 0, which implies µ(An) = 0 for all n. Since A = An by the second of the above relations, µ(A) = 0 as well, which was to be proved. ∪ We turn to the proof of the claim The set [ H = K + q q Q: q <1 ∈ | | is bounded, therefore µ(H) < . However, by the first of the above, the sets K +q are pairwise disjoint and measurable (being∞ translates of a measurable set), so that using translation invariance, X X > µ(H) = µ(K + q) = µ(K) ∞ q Q: q <1 q Q: q <1 ∈ | | ∈ | | and the claim follows. 1.7.2 REMARK. The two properties of the Lebesgue measure we have used in the above construction are translation invariance and the fact that bounded sets have finite measure. 12 1. OUTER MEASURES, MEASURES AND THE LEBESGUE MEASURE Chapter 1 Exercises

1.1. Let µ be an outer measure on X generated by the premeasure (σ, E). Let F X . Prove that F is µ-measurable, namely ⊂ c µ(G) = µ(F G) + µ(F G) G (X ) ∩ ∩ ∀ ∈ P if and only if c µ(E) = µ(F E) + µ(F E) E E. ∩ ∩ ∀ ∈ 1.2. Let be the dyadic grid on R and µ denote the Lebesgue outer measure on R, namely for A D R ¨ « ⊂ X∞ [∞ µ(A) = inf `(I j) : A I j, I j j . j=1 ⊂ j=1 ∈ D ∀ n #1 Let n Z. Prove that µ does not change if we restrict to intervals with `(I j) 2− , namely∈ ≤ ¨ « X (n) (n) ∞ [∞ n µ(A) = µ (A), µ (A) := inf `(I j) : A I j, I j j, `(I j) 2− . j=1 ⊂ j=1 ∈ D ∀ ≤ #2 Let t be of the form t PN k 2 n for suitable integers N, k ,..., k . R = n= N n − N N Prove∈ that µ is invariant under translations− by t, namely −

µ(A) = µ(A + t) A R. n ∀ ⊂ Hints. For #2, reduce to the case t = 2− for some n. Then use #1 and that m = I : `(I) = m n D { ∈ D 2− is invariant under translation by 2− whenever m n. } ≥ d d 1.3. Let be the collection of all open cubes I R and deﬁne the outer measure on R given by O ⊂ ¨ « X∞ [∞ ν(A) = inf Rn : A R j, R j j n=0 | | ⊂ n=0 ∈ O ∀ where R is the Euclidean volume of the cube R. #1| |Show that ν coincides with the Lebesgue outer measure µ generated by dyadic cubes. You can use the fact, proved in class, that µ(R) = R for all cubes R. #2 Use part (1) to obtain that the Lebesgue outer measure is| translation| invariant and d d respects dilations, in the sense that, for all t R , s > 0, A R ∈ d ⊂ µ(A + t) = µ(A), µ(sA) = s µ(A). Hint. For #1 there are two inequalities. One is just the "/2j enlargement trick. For the other one, assume ν(A) is ﬁnite. You can use the Whitney Lemma 1.3.3 from the notes to reduce a cover of A by open cubes to a disjoint dyadic cover with lesser total volume. 1.4. Prove Proposition 1.4.2. 1.5 Dynkin systems. A collection of subsets of X is called a Dynkin system if (1) (nontrivial) , F c (2) (closed under; ∈ complement) F F = F , ∈ F ⇒ ∈ F 1. OUTER MEASURES, MEASURES AND THE LEBESGUE MEASURE 13

(3) (closed under countable disjoint union) if Fn : n N are pairwise disjoint { ∈ } ⊂ F sets, namely n = k = Fn Fk = , then 6 ⇒ ∩ [; F := Fn . n N ∈ F ∈ Given (X ), we denote by δ( ) and by Σ( ) respectively the smallest Dynkin system and theG smallest ⊂ P Σ-algebra containingG . G G #1 Prove that δ( ) is well deﬁned and that δ( ) Σ( ). #2 Let be a DynkinG system with the furtherG property⊂ G ⊂ thatG F (π) F, G = F G ∈ F ⇒ ∩ ∈ F (stable under ﬁnite intersection). Prove that is a σ-algebra. #3 Prove that if has property (π) so does δ( F). G G d 1.6 Uniqueness of Lebesgue measure. Let Q0 = [0, 1) . In this problem, we prove that d the Lebesgue measure on R is the unique complete measure µ such that µ(Q0) = 1 which is translation invariant. Its solution is divided into three parts. For simplicity we work in d = 1. For a dyadic cube Q in R we indicate by (Q) = ∈ D D Q0 : Q0 Q and by (Q) the σ-algebra of Borel subsets of Q. Recall that (Q) = Σ{ ( ∈(Q D)) for instance.⊂ } B B D #1 Let µ, ν be two measures on (Q0, (Q0)) with the property that µ(Q) = ν(Q) for B all Q (Q0), and µ(Q0) = 1. Prove that µ = ν. #2 Let µ,∈ν Dbe two measures on (R, ) with the property that µ(Q) = ν(Q) < for all Q . Prove that µ = ν. B ∞ ∈ D #3 Let ν be a measure on (R, ) such that ν(Q0) = 1 and B n A , n Z = ν(A + 2 ) = ν(A) ∈ B ∈ ⇒ Prove that the completion of ν is the Lebesgue measure.

Hints. For 1. show that A (Q0) : µ(A) = ν(A) is a Dynkin system and then a σ-algebra via Exercise 1.5. For 2. use 1.{ and∈ B exhaustion. For part} 3. verify the assumptions of part 2. by using dyadic translation invariance.

d d 1.7. #1 Let µ denote Lebesgue outer measure on R . Show that for all A R there exists a Lebesgue measurable set L with A L and µ(A) = µ(L). ⊂ ⊂ 1.8. Prove Proposition 1.6.2.

d d 1.9 (R ) = (R ), I. Deﬁne the n-th truncated Cantor set as B 6 L n n [ X n n X n Cn = an3− , 3− + an3− . n (a1,...,an) 0,2 j=1 j=1 T ∈{ } Then C = n 0 Cn is called the Cantor middle thirds set. We have seen in class that C is compact and uncountable≥ 1. #1 Show that C contains no intervals: thus C is called nowhere dense or meager.

1Moreover, every point of C is an accumulation point for C, but you don’t need to write the proof of this fact here. 14 1. OUTER MEASURES, MEASURES AND THE LEBESGUE MEASURE

#2 Show that C has zero Lebesgue measure. Sketch the construction a set C 1 still 2 1 satisfying #1 but whose Lebesgue measure is 2 . #3 Use point 1. and a cardinality argument to prove that (R) = (R). You can use without proof that (R) has the same cardinality as RB. See Folland,6 L Prop 1.23, p. 43 for a proof. B

1.10. Let µ be the Lebesgue outer measure on R. Show that µ(F) = sup µ(O) O F O open⊂ fails in general, even if F is Borel. Hint. Modify the construction of the Cantor set to obtain a set of positive Lebesgue measure containing no intervals.

d d 1.11 (R ) = (R ), II (d 2). #1 Let E R be a non Lebesgue measurable set, E˜ = 2 2 2 E 0B R .6 ExplainL why E˜ ≥ (R ) (R )⊂. You can use without proof that × { } ⊂ 2 ∈ L \B A (R ) x R : (x, y) A (R) y R. ∈ B ⇐⇒ { ∈ ∈ } ∈ B ∀ ∈ CHAPTER 2

Abstract integration and the Lebesgue integral

Throughout this chapter, unless otherwise specified, (X , ) is a measurable space and Y is a topological space. We will make use of the Borel σ-algebraF of Y , denoted by (Y ), which is the smallest σ-algebra containing the open sets in Y . B 2.1 Measurable functions. Let now Y be a topological space. A function f : X Y is said 1 → -measurable (or briefly, measurable) -measurable if f − (O) for all O Y open sets. F F ∈ F ⊂ 2.1.1 REMARK. Assume (X , ) and (X , 0) are measurable spaces with 0. Clearly F F F ⊂ F f : X Y is -measurable implies f : X Y is 0-measurable as well. → F → F 2.1.2 REMARK. Assume now that X is also a topological space and that f : X Y is con- 1 →1 tinuous, or equivalently f − (O) is open in X , whenever O is open in Y . Thus f − (O) = (X ) whenever O is open in Y , that is to say f : X Y is (X )-measurable (or∈ Borel F measurable).B → B The next lemma will lead us to a useful equivalent definition of measurable function.

2.1.3 LEMMA. Let Y be a nonempty set and f : X Y . The collection

1 → (2.1) := A Y : f − (A) (Y ) G { ⊂ ∈ F } ⊂ P is a σ-algebra, referred to as the push-forward of under f 1. Furthermore, if Y is a topological space F f : X Y measurable (Y ) . → ⇐⇒ B ⊂ G 1 PROOF. We ﬁrst prove that is a σ-algebra. Since f − (Y ) = X , we have Y . 1 G ∈ F ∈ G Let A . Then f − (A) = B . Now ∈ G c ∈ F c 1 c x B = f (x) A = f (x) A = x f − (A ) ∈ ⇒ 6∈ ⇒ 1 ∈c ⇒c ∈ c and all the implications are reversible. Thus f − (A ) = B , and hence A . The closure under countable union is similar. ∈ F ∈ G We now prove the equivalence. By deﬁnition we have that f : X Y is measurable if and only if contains all open sets. But then by minimality of (Y ),→ (Y ) . G B B ⊂ G In other words f : X Y is measurable if all Borel sets belong to the push-forward (2.1) of (X , ) via f . A further→ consequence is that whenever is a collection of subsets of Y such thatF Σ( ) = (Y ), then C C B 1 f : X Y measurable f − (C) C . → ⇐⇒ ∈ F ∀ ∈ C 1Sometimes the notation f is used for ∗F G 15 16 2. ABSTRACT INTEGRATION AND THE LEBESGUE INTEGRAL

2.1.4 REMARK (TOPOLOGYINTHEEXTENDEDREALLINE). We recall that the open sets in the extended real line [ , ] are declared to be the closure under arbitrary union of the collection −∞ ∞ [ , a), (a, b), (a, ] : a, b R Since each of the three type of−∞ intervals above can∞ be written∈ as a countable union of intervals of the type (a, ] or their complements, for instance ∞ c [ 1 [ , a) = a n , −∞ n N − ∞ ∈ the Borel σ-algebra is generated by such intervals. Thus, f : X [ , ] is -measurable 1 → −∞ ∞ F if and only if f − ((a, ]) for all a R. ∞ ∈ F ∈ We now show that a measurable function f on a measure space (X , , µ) induces a measure on (Y, ), which is referred to as the pushforward of µ. F G 2.1.5 PUSH-FORWARD LEMMA. Let (X , , µ) be a measure space, Y be a topological space and F 1 f : X Y be a Borel-measurable function with := A Y : f − (A) its push-forward σ-algebra.→ Then G { ⊂ ∈ F } 1 A f µ(A) = µ(f − (A)) ∈ G 7→ ∗ deﬁnes a measure on (Y, ) called the pushforward of µ via f . G PROOF. We already know is a σ-algebra. To verify countable additivity of µ, let Aj be a sequence of pairwise disjointG sets whose union is A. We have ∈ G 1 [ 1 X 1 X f µ(A) = µ(f − (A)) = µ f − (Aj) = µ f − (Aj) = f µ Aj ∗ j j j ∗ 1 since the sets f − (Aj) are be pairwise disjoint. In the next proposition, we record some elementary properties of the notion of measurability.

2.1.6 PROPOSITION. Throughout, ﬁx a measurable space (X , ). F 1. Let E X and 1E : X R be the indicator function of E, Then 1E is -measurable if and⊂ only if E . → F 2. Let M, N be topological∈ F spaces, f : X M be -measurable, g : M N be Borel measurable. Then g f is -measurable.→ F → 2 3. Let f , g : X R be measurable◦ F functions. Let Φ : X R be deﬁned by → → Φ(x) = (f (x), g(x)), x X . ∈ Then Φ is measurable. 4. Let u, v : X R be -measurable. Then f := u + iv : X C is -measurable. 5. Let f , g : X → C be F -measurable. Then → F → F u = Ref , v = Imf , f , f + g, f g | | are -measurable. Furthermore there exists α : X C -measurable such that F → F α(x) = 1 x C, f = α f . | | ∀ ∈ | | 2. ABSTRACT INTEGRATION AND THE LEBESGUE INTEGRAL 17

6. Let f : X [ , ] be measurable. Then +→ −∞ ∞ + f := max f , 0 , f − := min f , 0 , f = f + f − { } − { } | | are measurable functions. 7. Let fn : X [ , ] be a sequence of -measurable functions. Then → −∞ ∞ F g = sup fn, h = lim sup fn n are -measurable functions. F PROOF. Assertion 1. is simply verified by inspection. 1 To prove 2. let A N be a Borel set and h = g f . Then g− (A) is a Borel set in M by Borel ⊂ 1 ◦ 1 1 1 measurability of g. But f − (B) for all B M Borel. Thus h− (A) = f − (g− (A)) . 2 To prove 3. let us recall that open∈ F sets in R⊂ are countable unions of products of∈ open F 1 intervals I J. Thus it suffices to prove that Φ− (I J) is measurable. But this follows since 1 × 1 1 × Φ− (I J) = f − (I) g− (J). 2 To× prove 4. recall∩ that the map Ψ : R C, (u, v) u + iv is continuous. So f (x) = u(x)+iv(x) = Ψ Φ(x), with Φ as in previous→ point is measurable7→ by part 2. To prove 4. use 2 2. and the the fact◦ that max , 0 : R [0, ), and (u, v) u+iv : R C are continuous functions (and thus Borel measurable).{· } → ∞ 7→ → To prove the first part of 5. use also 2. and continuity of the corresponding maps. We 1 prove the last statement which is the most involved. Observe that the z φ(z) = z z− is a 1 7→ | | continuous map of C 0 into C. Let A = f − ( 0 ) which is clearly measurable. We define \{ } { } α = φ(f + 1A)

which clearly satisfies αf = f . By the first part, f +1A is measurable, and φ is continuous, so that α is also measurable.| We| leave the very similar proof of 6. as an exercise. To prove the first part of 7. simply use that 1 [ 1 g− ((a, ]) = fn− ((a, ]) ∞ n ∞ and Remark 2.1.4. The remaining claim is immediate and left as an exercise. 2.2 Invariances of the Lebesgue measure under GL(d, R). We exploit the above definitions of measurability to establish two further invariance properties of the Lebesgue measure d d d µ on R . Let x, y denote the standard bilinear scalar product on R and let us endow R 〈 〉 p d d with the Euclidean norm x = x, x . A linear map T : R R is said to be orthogonal d d if x, y = T x, T y for allk xk, y R〈 . Notice〉 in particular that→T x = x for all x R . 〈 The〉 pushforward〈 〉 of the Lebesgue∈ measure via an orthogonalk transformationk k k is the∈ Lebes- gue measure itself.

d d d 2.2.1 LEMMA. Let T : R R be an orthogonal transformation. Then A (R ) if and 1 d → ∈ B only if T − A (R ). Furthermore T µ = µ. ∗ ∈ B 1 PROOF. Since T is a linear orthogonal map, both T and T − are continuous and thus d Borel measurable, which is the ﬁrst claim. For the second claim, observe that for all x R and Borel sets A ∈ 1 1 1 1 T µ(x + A) := µ(T − (x + A)) = µ(T − x + T − A) = µ(T − A) = T µ(A) ∗ ∗ 18 2. ABSTRACT INTEGRATION AND THE LEBESGUE INTEGRAL

using Lebesgue translation invariance in the third step. In other words T µ is a translation invariant Borel measure. It follows from Exercise 1.6 that T µ(A) = κµ(A)∗for some constant ∗ 1 κ and all Borel sets A. Taking A to be the closed unit ball B, which is such that T − B = B, it follows that κ = 1. d d d 2.2.2 LEMMA. Let T : R R be a linear invertible transformation. Then A (R ) if and 1 d → ∈ B only if T − A (R ). Furthermore T µ = detT µ. ∈ B ∗ | | 1 d 1 PROOF. Exercise. It can be used freely that µ(T − ([0, 1) )) = detT − . | | 2.3 Simple functions. Let (X , ) be a measurable space. A function s : X C is said to be simple if it is of the form F → N X (2.2) s x a 1 x ( ) = j Aj ( ) j=1

where aj, j = 1, . . . , N are complex coefﬁcients and Aj, j = 1, . . . , N are measurable sets. 2.3.1 LEMMA. Let s be a simple function as in (2.2). Then 1. s is measurable; 2. s can be uniquely written as M M X [ (2.3) s b 1 , B j 1, . . . , M, B , B B when j k, B X = j Bj j = j = j k = = j = j=1 ∈ F ∀ 6 ; ∩ ; 6 j=1

with values bj = bk whenever k = j, which we call standard form of s. 6 6 PROOF. The measurability of s follows by points 1. and 5. of Proposition 2.1.6 and induction. To prove the second part, notice that the range of s is contained in the finite set a a : j ,..., j 1, . . . , N , n 0, . . . , N . j1 + + jn 1 n ··· ∈ { } ∈ { } Thus if b1,..., bM is an enumeration of the range of s, the unique representation above is { } 1 obtained by setting Bj = s− (bj). It remains to prove that each Bj is measurable. This is left as an exercise. 2.3.2 LEMMA. Let f : X [0, ] be a measurable function. Then there exists a sequence of → ∞ simple functions sn : X [0, ) such that → ∞s x s x x X n n( ) n+1( ) , N, (2.4) ≤ ∀ ∈ ∈ f (x) = lim sn(x) x X . n ∀ ∈ 2 →∞ PROOF. By replacing f by π arctan f , we can assume f takes values in [0, 1]. Let n N and define ∈ 1 n n n 1 n Aj,n = f − (j 1)2− , j2− , j = 1, . . . 2 1, A2n,n = f − [1 2− , 1] − − − The sets Aj,n are measurable, as they are preimages of Borel sets via a measurable function. Define n X s x j 1 2 n1 x . n( ) = ( ) − Aj,n ( ) j=1 − n namely sn(x) is the left endpoint of the dyadic subinterval of [0, 1] of length 2− containing f (x). This immediately entails both properties of (2.4). 2. ABSTRACT INTEGRATION AND THE LEBESGUE INTEGRAL 19

2.3.3 REMARK. Inspection of the above proof entails immediately that, if supX f (x) < has bounded range, ∞ lim sup fn(x) s(x) = 0 n x X − →∞ ∈ that is sn f uniformly on X . → 2.4 Integration of simple functions. Let (X , , µ) be a measure space. We deﬁne the µ-integral of a simple function s written in standardF form as in (2.3) by M Z X (2.5) s dµ = bjµ(Bj). j=1

Here, we use the convention that, if µ(Bj) = for some j, then bjµ(Bj) = if bj > 0, ∞ ∞ while bjµ(Bj) = 0 if bj = 0. 2.4.1 PROPOSITION. Let s, u be simple functions and α R. Then Z ∈ 1. s 0 = s dµ 0 ≥ ⇒ ≥ Z Z Z 2. s + αu dµ = s dµ + α u dµ Z Z 3. s u = s dµ u dµ ≥ ⇒ ≥ N N X Z X 4. s a 1 s d a A . = k Ak = µ = kµ( k) j=1 ⇒ j=1 PROOF. The first assertion is immediate from the definitions. The third follows by cou- pling 1. and 2. with α = 1. The fourth is an immediate consequence of 2. and of the fact that − Z 1A dµ = µ(A), which is immediate from the definition. It remains to prove the second assertion. To do so, write n m X X s b 1 , u c 1 = j Bj = j Cj j=1 j=1 in standard form. The range of s + αu is given by the finite set

R = ξ : ξ = (bj + αck), j = 1, . . . , n, k = 1, . . . , m . { } Let ξ` : ` = 1, . . . , L be an enumeration of R (so that the values ξ` are all distinct). Let I` { } be the set of those ordered pairs (j, k) such that ξ` = bj +αck. Defining the measurable sets [ A` = Bj Ck (j,k) I` ∩ ∈ then s αu has standard form + r X s αu ξ 1 + = ` A` `=1 20 2. ABSTRACT INTEGRATION AND THE LEBESGUE INTEGRAL is also in standard form. Using definitions and, in the second step the fact that Bj { ∩ Ck, (j, k) I` partition A` ∈ } L L Z X X X s + αu dµ = ξ`µ(A`) = (bj + αck)µ(Bj Ck) `=1 `=1 (j,k) I` ∩ n m m n ∈ X X X X = bj µ(Bj Ck) + α ck µ(Bj Ck) j=1 k=1 ∩ k=1 j=1 ∩ n m X X Z Z = bjµ(Bj) + α ckµ(Ck) = s dµ + α u dµ j=1 k=1 and in the last step the fact that Bj Ck : k = 1, . . . , m partition Bj and similarly for the other summand. { ∩ } 2.4.2 PROPOSITION. Let s be a nonnegative simple function. Then Z Z A µs(A) := s dµ := s1A dµ ∈ F 7→ A defines a measure µs on . F PROOF. Immediate from the definitions. 2.5 Integration of positive functions. Let (X , , µ) be a measure space. For f : X [0, ] measurable, define F → ∞ Z Z f dµ = sup s dµ : s simple, 0 s f [0, ]. ≤ ≤ ∈ ∞ From property 3. of Proposition 2.5.4, if s is simple, the two definitions coincide (and the abuse of notation is thus justified). The following two important properties are immediate consequence of the definition.

2.5.1 MONOTONICITY LEMMA. Let f , g : X [0, ] be measurable. Then →Z ∞ Z g f = gdµ f dµ. ≤ ⇒ ≤

2.5.2 MONOTONECONVERGENCETHEOREM. Let fn : X [0, ] be an increasing sequence, namely such that f f for all n. Assume each f is measurable→ ∞ and let f X be n n+1 n : [0, ] the pointwise limit of≤ the sequence fn . Then → ∞ {Z } Z f dµ = lim fndµ.

PROOF. Note that f is measurable by Proposition 2.1.6, fourth point. Of course fn f ≤ for all n, so that using Lemma 2.5.1 with g = fn and passing to the limit on the left hand sides, we get Z Z B := lim fndµ A := f dµ. ≤ 2. ABSTRACT INTEGRATION AND THE LEBESGUE INTEGRAL 21

Let now s be a simple function with 0 s f and 0 c < 1 be arbitrary. The sets E f cs ≤ ≤ f ≤cs E E n = n 0 are measurable (since the functions n are), moreover n n+1, and their{ union− is≥X (since} fn increase to f and f > cs). Therefore− ⊂ Z Z Z f d f 1 d c s1 d c E n µ n En µ En µ = µs( n) ≥ ≥ The second inequality is monotonicity, and third inequality follows by deﬁnition since cs1 En is simple and cs1 f 1 . Passing to the limit on both sides, and using Proposition 2.4.2 En n En ≤ Z B cµs(X ) = c sdµ. ≥ But since 0 c < 1 and s are arbitrary, B A as well. ≤ ≥ 2.5.3 FATOU’S LEMMA. Let fn : X [0, ] be a sequence of measurable functions. Then Z → ∞ Z lim inf fn dµ lim inf fndµ. n n →∞ ≤ →∞ PROOF. Let

gn inf fn, g lim inf fn sup gn = k n = n = ≥ →∞ We know that gn, g are measurable functions. Moreover gn increases to g. Therefore, by the monotone convergence Theorem 2.5.2 Z Z Z Z gdµ lim gndµ lim inf gndµ lim inf fndµ = n = n n →∞ →∞ ≤ →∞ R R where the right inequality follows from gn fn, by lim inf on both sides, since gn ≤ ≤ fn. We now use the monotone convergence theorem to extend the remaining point of Propo- sition 2.5.4, as well as Proposition 2.4.2 to the cone of positive functions.

2.5.4 PROPOSITION. Let f , g : X [0, ] be measurable functions. Then Z Z → Z ∞ 1. f + αg dµ = f dµ + α g dµ if α 0; ≥ 2. The set function Z Z A µf (A) := f dµ := f 1A dµ ∈ F 7→ A 2 deﬁnes a measure µf on and FZ Z g dµf = g f dµ.

PROOF. Exercise. Essentially a combinations of the simple function versions, Theorem 2.5.2 and Lemma 2.3.2. 2 We say that f is the Radon-Nykodym derivative of µf with respect to µ. 22 2. ABSTRACT INTEGRATION AND THE LEBESGUE INTEGRAL

2.5.5 MONOTONECONVERGENCETHEOREMFORSERIES. Let fn : X [0, ] be a sequence of measurable functions. Then → ∞ Z X X Z fn dµ = fndµ. n n

PROOF. Let Fn = f1 + + fn. Then Fn increases to F, and ··· N Z X Z FN dµ = fndµ. n=1 Passing to the limit on both sides and using the monotone convergence theorem the result follows. We provide a probabilistic example of application of the monotone convergence theorem.

2.5.6 BOREL-CANTELLIPRINCIPLE, I. Let En be a sequence of measurable sets and E = lim sup En. Then X µ(En) < = µ(E) = 0. n ∞ ⇒ P PROOF. Let F 1 . It is easy to see that E x X : f x . From the = En = ( ) = assumption and the monotone convergence theorem for{ series∈ ∞} Z X F dµ = µ(En) < . n ∞ whence µ(E) = 0. 2.6 Integration of complex functions. The space 1. Throughout this paragraph, we keep the measure space (X , , µ) fixed unless otherwiseL noted. We define F Z 1 1 = (X , , µ) := f = u + iv : X C measurable : f dµ < . L L F → | | ∞ Notice that, from Proposition 2.1.6, f is measurable whenever f is, so that the integral in the above definition makes sense. We| then| define Z Z Z Z Z + + f dµ = u dµ u−dµ + i v dµ i v−dµ. − −

From the same proposition, u±, v are all positive measurable functions, dominated by f so that the four summands in the above± definition are well defined and finite. Notice also| that| if f : X [0, ) the above definition coincides with the one given for positive functions. → ∞ 1 2.6.1 LEMMA. Let f , g , α C. 1 ∈ L ∈ 1 is a linear space, namely f + αg ; •Lintegration is a linear bounded functional∈ L on 1, namely • Z Z Z L Z Z

f + αg dµ = f dµ + α g dµ, f dµ f dµ. ≤ | | 2. ABSTRACT INTEGRATION AND THE LEBESGUE INTEGRAL 23

PROOF. The proof of the ﬁrst part is straightforward, since f + αg is measurable and f +αg f +α g . For the second part, linearity follows from the deﬁnitions and the cor- |responding| ≤ | result| | for| positive functions. We leave details as an exercise. For the inequality, we can assume the left hand side is strictly positive, otherwise there is nothing to prove. Let α be a complex unimodular such that the left hand side is equal to α R f dµ. Then Z Z Z Z Z Z Z

f dµ = α f dµ = αf dµ = Re αf dµ = Re(αf )dµ Re(αf ) dµ f dµ. ≤ | | ≤ | | We have used linearity in the second equality, the fact that the first left hand side (and thus all members) are real in the third, the definition of integral in the fourth equality and fifth inequality and finally the fact that Re(αf ) αf = f . | | ≤ | | | | 2.6.2 DOMINATED CONVERGENCE THEOREM. Let fn : X C be a sequence of measurable functions such that →

f x lim fn x ( ) = n ( ) →∞ exists for all x X . Suppose that there exists g 1 such that ∈ ∈ L fn(x) g(x) x X . | | ≤ ∀ ∈ Then Z 1 f , lim fn f dµ 0 n = ∈ L →∞ | − | 2.6.3 REMARK. In particular, if the assumptions of the theorem are satisﬁed, Z Z lim fn dµ f dµ. n = →∞ 1 PROOFOF THEOREM 2.6.2. Since f g it is clear that f . The function hn = 2g f fn are nonnegative and their| | ≤ pointwise limit is 2g.∈ Therefore, L using Fatou’s lemma− | and− linearity| in the second step Z Z Z Z 2g dµ lim inf hn dµ = 2g dµ lim sup f fn dµ. ≤ − | − | Subtracting the ﬁnite number R 2g dµ, we obtain the last inequality Z Z 0 lim inf f fn dµ lim sup f fn dµ 0 ≤ | − | ≤ | − | ≤ which yields the conclusion of the theorem.

2.7 Change of variable formula. Let (X , ), (X 0, 0) be measurable spaces and φ : X F F → X 0 be a function with the property that 1 φ− (E0) E0 0, ∈ F ∀ ∈ F in other words the pushforward of via φ contains 0. Let ν = φ µ be the pushforward of µ, in other words F F ∗ 1 ν(E0) = µ(φ− (E0)) E0 0. ∀ ∈ F 24 2. ABSTRACT INTEGRATION AND THE LEBESGUE INTEGRAL

2.7.1 LEMMA. Let f : X 0 [0, ] be measurable. Then f φ : X [0, ] is measurable and → ∞ ◦ → ∞ Z Z f dν = f φ dµ. ◦ 1 1 The same equality holds whenever f (X 0, 0, ν), in which case f φ (X , , µ). ∈ L F ◦ ∈ L F PROOF. Exercise. 2.8 The role of sets of measure zero. Let (X , , µ) be a measure space and P be a property that x X may or may not have. For instance,F given f : X [0, ], P can be “f (x) < at the point∈ x. We say that P holds almost everywhere with→ respect∞ to µ, a.e.[µ] if ∞ A = x X : P fails at x , µ(A) = 0. { ∈ } ∈ F Letting Y be a topological space, we can define an equivalence relation on the class of measurable functions f : X Y by saying → f g f = g a.e.[µ]. ∼ ⇐⇒ 2.8.1 LEMMA. Let f , g : X [0, ] be measurable. Then → ∞ Z Z f = g a.e.[µ] = f dµ = g dµ E . ⇒ E E ∀ ∈ F 1 The same statement holds for f , g (X , , µ). ∈ L F PROOF. We only prove the first part. Let N = x X : f (x) = g(x) . Note that this is c a measurable set. Then µ(N ) = 0, which we use{ in∈ the first and last step} of the chain of equalities Z Z Z Z f dµ = f dµ = gdµ = g dµ. E E N E N E ∩ ∩ The proof is complete. The above lemma shows that, within the scope of µ-integration, functions which are equal a.e.[µ] are undistinguishable. This suggests that the notion of measurability can be suitably enlarged, at least when the measure space (X , , µ) is complete. We say that Fc 1 f : X Y is measurable if there exists X0 with µ(X0) = 0 such that f − (A) X0 is → ∈ F c ∩ measurable for all open sets A Y . If so, we can redefine f to be constant on X0 (for in- c ⊂ stance, setting f = 0 on X0 if Y = [0, ]) obtaining a new function which is equal to f almost everywhere and is measurable in∞ the previous sense. Indeed, with this redefinition, 1 c c f − (A) X0 is either empty or X0 for all (not just open) sets A Y . We∩ dwell more explicitly on some examples of properties defined⊂ almost everywhere. We say f : X [ , ], measurable is finite a.e. if • → ∞ ∞1 1 µ(f − ( ) = µ(f − ( ) = 0. {∞} {−∞} We say f : X [ , ], measurable is essentially bounded if there exists M 0 • such that → ∞ ∞ ≥ 1 1 µ(f − ((M, ]) = µ(f − ([ , M)) = 0. ∞ −∞ − 2. ABSTRACT INTEGRATION AND THE LEBESGUE INTEGRAL 25

We say f : X [ , ] measurable is continuous almost everywhere if the set • of discontinuity→ points∞ ∞ of f has zero measure. Notice that there are functions f that are continuous a.e. but for which there exist no continuous function g with g f . Viceversa, f may be equal a.e. to some continuous function g, but not almost∼ everywhere continuous.

2.8.2 REMARK. Both the monotone and dominated convergence theorems can be reformulated by requiring monotonicity and domination a.e.[µ]. An example of this is the next corollary.

2.8.3 COROLLARY. Let fn : X C be a sequence of measurable functions such that → Z X∞ fn dµ < . n=1 | | ∞ Then the series X∞ f (x) := fn(x) n=1 converges a.e., f 1 and ∈ L Z Z X∞ f dµ = fn dµ. n=1 PROOF. Let X∞ F(x) = fn(x) . n=0 | | By the monotone convergence theorem for series, we have Z Z X∞ F dµ = fn dµ < , n=1 | | ∞ P so in particular F is ﬁnite almost everywhere. It means that the series f = fn converges absolutely almost everywhere. Since f1 + ... + fN F almost everywhere, by the (almost everywhere) dominated convergence| theorem, f | ≤ 1, and N ∈ L N Z Z X X Z f dµ lim fn dµ lim fn dµ = N = N →∞ n=1 →∞ n=1 which was the ﬁnal claim to be proved. 2.8.4 REMARK. However, if f = g a.e. it is not necessarily true that f φ = g φ a.e.; hence, when dealing with composition, the choice of the representative in◦ the equivalence◦ classes does matter.

1 p 2.9 The spaces L (X , , µ), L (X , , µ). The preceding section shows how, for integration purposes, we can identifyF functionsF which are equal almost everywhere. The following lemma gives us a further reason to do so.

2.9.1 LEMMA. Let (X , , µ) be a measure space. F 26 2. ABSTRACT INTEGRATION AND THE LEBESGUE INTEGRAL

1. If f : X [0, ] is measurable, then → ∞ Z f dµ = 0 f = 0 a.e.[µ]. ⇐⇒ 2. If f 1, ∈ L Z f dµ = 0 f = 0 a.e.[µ]. | | ⇐⇒ 3 If f 1, ∈ L Z Z

f dµ = f dµ α C : f = α f a.e.[µ] | | ⇐⇒ ∃ ∈ | | 1 PROOF. Let us prove the first part. Let E = x : f (x) > 0 . The sets Ek = x : f (x) > k− increase to E. Moreover { } { } Z Z 1 0 k− µ(Ek) f dµ f dµ = 0 ≤ ≤ Ek ≤ thus µ(Ek) = 0 for all k. It follows µ(E) = 0 as well. The second part for f real is simply the first part applied to positive and negative parts of f . For f complex, we omit the easy details. To prove the third claim, we notice that if the left hand side condition holds, the proof of the second part of Lemma 2.6.1 yields for some unimodular α the equality Z f Re αf dµ = 0 | | − but since f Reαf is nonnegative, it must be equal to zero almost everywhere by 1st part. It means that| |− f = αf = Reαf a.e. which is the same as f = α f a.e., which is what we had to prove. | | | | | | 1 1 We now define L (X , µ, ) as the space of equivalence classes of functions in (X , µ, ) 1 with respect to the equivalenceF relation f = g a.e.[µ]. For f L (X , µ, ), weL introduceF the 1-norm ∈ F Z

f 1 f f L (X ,µ, ) = 1 = dµ k k F k k | | 1 Notice that we omit, in the notation, the distinction between f L (X , µ, ) (appearing in the first two members) and its representative (appearing in the∈ right handF side), since the right hand side is clearly independent from the representative chosen. 1 It follows from Lemma 2.6.1 that L (X , µ, ) is a linear space. Moreover, 1 is a norm 1 on L (X , µ, ), in the sense that the followingF properties are satisfied: k·k F f 1 = 0 = f = 0; •k k ⇒1 α C, f L (X , µ, ) = αf 1 = α f 1 • ∈ 1∈ F ⇒ k k | |k k f , g L (X , µ, ) = f + g 1 f 1 + g 1 • ∈ F ⇒ k k ≤ k k k k The first property, which has to be understood in the a.e.[µ] sense, in particular, is a consequence of Lemma 2.9.1, second part. The second property is linearity of the integral, and the third property is linearity and monotonicity of the integral. 2. ABSTRACT INTEGRATION AND THE LEBESGUE INTEGRAL 27

2.10 The relationship between in measure, pointwise, uniform and L1 convergence. Let (X , , µ) be a measure space and fn, f : X [ , ] be measurable functions. We say thatF → −∞ ∞

fn f in measure if " > 0 • → ∀ lim µ x X : fn x f x > " 0. n ( ( ) ( ) ) = →∞ { ∈ | − | } fn f pointwise a.e.[µ] if • → µ ( x X : lim sup fn(x) f (x) > 0 ) = 0 1 { ∈ | − | } fn f in L if • → lim fn f 1 0 n = →∞ k − k fn f almost uniformly if for all " > 0 there exists a set X" with • → ∈ F µ X X" < ", lim sup fn x f x 0. ( ) n ( ) ( ) = \ x X" | − | →∞ ∈ fn f uniformly a.e.[µ] if there exists X 0 with µ(X X 0) = 0 and • → ∈ F \ lim sup fn(x) f (x) = 0. n x X | − | →∞ ∈ 0 The exercises explore some of the implications and forbidden implications between the four types of convergence. The next lemma, whose easy proof is left as an exercise, tells us that all the notions of convergence are stable under ﬁnite linear combination.

2.10.1 LEMMA. Let fn, gn : X C be measurable functions. Assume that → fn f , gn g → → according to any one of the above convergence modes. Then, for all α C, ∈ fn + αgn f + αg. → Pointwise convergence a.e.[µ] implies almost uniform convergence on sets of ﬁnite measure.

2.10.2 EGOROV’STHEOREM. Let (X , , µ) be a measure space and fn : X C be measurable F → functions such that fn f pointwise a.e.[µ]. Let A have µ(A) < . Then fn f almost uniformly on A, in the→ sense that fn1A f 1A almost∈ uniformly.F ∞ → → c PROOF. We can assume without loss of generality that fn, f are equal to zero on A . For 1 each k 1 let Ck,m = x A : fm(x) f (x) > k− . Then ≥§ { ∈ | − |ª } 1 \ [ \ [ Bk = x A : lim sup fn(x) f (x) > k− = Ck,m = Bk,n, Bk,n := Ck,m. ∈ n | − | n m n n m n →∞ ≥ ≥ Since fn f a.e., we know that µ(Bk) = 0. Since µ(Bk,1) µ(A) < , and Bk,n decreases in n, one→ has ≤ ∞

lim µ Bk,n µ Bk 0. n ( ) = ( ) = Now let 0 be given and for each→∞ k choose n large enough so that B 2 k. " > k µ( k,nk ) < " − S c c Letting Y ∞ B , we set X Y , so that µ X µ Y < ". Furthermore for all k = k=1 k,nk " = ( " ) = ( ) 1 n nk = sup fn(x) f (x) < k− ≥ ⇒ x X" | − | ∈ 28 2. ABSTRACT INTEGRATION AND THE LEBESGUE INTEGRAL

which is uniform convergence on X". The proof is complete. 1 2.11 Approximation of L functions. Let (X , , µ) be a measure space and , be a class of complex-valued measurable functions. WeF say that the class is dense inY theZ class , with respect to one of the above modes of convergence if for anyYf , there exists a sequenceZ fn such that fn f in that mode of convergence. It has∈ to Z be noted that, in each of the{ modes} ⊂ Y considered, being→ dense is a transitive property. Furthermore, by Lemma 2.10.1, if is dense in and is closed under finite linear combinations, then is also dense in theY class of finiteZ linearY combinations of functions from . We leave theY simple proof of these relevant properties as an exercise. Z 1 Almost by construction, simple functions are dense in L (X , , µ). F 1 2.11.1 PROPOSITION. Let f L (X , , µ). Then there is a sequence of simple functions fn : 1 ∈ F X C with fn f in L . → → PROOF. Let f = u + iv. By the linearity Lemma 2.10.1, it suffices to construct four sequences of simple functions converging respectively to u±, v±. This allows us to reduce to the case where f 0. Lemma 2.3.2 yields a sequence fn of simple functions whose pointwise increasing limit≥ is f . By the monotone convergence theorem (for instance) Z Z Z 0 f fn 1 = f fn dµ = f dµ fn dµ 0 ≥ k − k − − → and the lemma is proved. d The next approximation theorem is specialized to R with the Lebesgue measure. We d d denote by 0(R ) the linear space of continuous, complex-valued functions on R with compact supportC 3.

1 d d 2.11.2 THEOREM. Let f L (R ) (R with Lebesgue measure). Then there is a sequence d ∈ 1 fn 0(R ) with fn f in L . ∈ C → 1 d d PROOF. The statement is equivalent to: for all " > 0, f L (R ), there exists g (R ) with f g 1 < ". We use the strategy of reducing this statement∈ to functions f in∈ smaller C and smallerk − k classes. Firstly, by Proposition 2.11.1, we can reduce to f being a nonnegative simple function 1 in L . Secondly, by transitivity under finite linear combination, we can reduce to f = 1A, where A is a Lebesgue measurable set of finite measure. Thirdly, we can reduce to open sets. Suppose that for every δ > 0 and open set O R of finite measure, there exists a d ⊂ d function gO,δ 0(R ) with gO,δ 1O 1 < δ. Let then A R be a set of finite measure. By regularity of∈ C Lebesgue measure,k − therek exists O A open⊂ with µ(O A) < "/2. Then ⊃ " \ gO,"/2 1A 1 gO,"/2 1O 1 + 1O 1A 1 2 + µ(O A) < " k − k ≤ k − k k − k ≤ \ and the reduction is complete. Subsequently, we can reduce to sets B which are finite union of disjoint dyadic cubes. Let O be an open set of finite measure. By the Whitney lemma,

3Recall that the support of f : X Y , where X is a topological space and Y is a linear space, is the closure of the set x X : f (x) = 0 . → { ∈ 6 } 2. ABSTRACT INTEGRATION AND THE LEBESGUE INTEGRAL 29 there is a sequence of disjoint dyadic cubes Qn with {X} 1 1 . O = Qn n P In particular, µ(O) = n µ(Qn). Given " > 0 we can choose N large enough so that N X X X µ Q 1 1 1 < " ( n) = Qn + O Qn n>N n>N 1 − n=1 1

The first equality is simply the dominated convergence theorem for series. With B = Q1 QN , we are done. By closure under linear combination, we have reduced to the case of∪ · · · ∪ f = 1Q, where Q is a dyadic cube. By translation and dilation invariance, it suffices to treat d d Q = [0, 1) . The proof is finished by the construction of g 0(R ) with ∈ C g d 1[0,1) 1 < " k − k which is is left as an exercise. d In particular, the above theorem tells us that 0([0, 1] ), for instance, is not complete with respect to the metric induced by the L1-norm.C We present one more approximation result for measurable functions, where L1 plays a role in the proof.

d 2.11.3 LUSIN’STHEOREM. Let f : R C be a measurable function and " > 0. Then there → d c exists a measurable set E such that f 1E is a continuous function on E, and µ([0, 1] E ) < ". \ Before the proof, we remark that the continuity of the above theorem has to be understood in the sense of the induced topology on E. This means that

xn, x E, xn x = f (xn) f (x). ∈ → ⇒ → This does not mean that f is continuous at points in E in the usual sense. For instance, f 1 is nowhere continuous, while if one deﬁnes E , µ Ec and f 1 : E is = Q = R Q ( ) = E identically zero, and thus a continuous function in the above\ sense. ; →

PROOF. Filling in the details of the proof are left as an exercise. It suffices to prove the theorem for f 0. Furthermore, by using the arctan trick, we can reduce to the case where f : d 0, 1≥. Let B be the unit ball of radius N. Now f 1 L1. thus there exists a R [ ) N BN → ∈ N 2n sequence fn,N of continuous functions with compact support such that f fN,n < "2− − . It follows by Chebychev’s inequality that k − k n n N Fn,N := x BN : f (x) fn,N (x) > 2− , µ(Fn,N ) "2− − S S { ∈ | − | } c ≤ Define F ∞ ∞ F . Then µ F < ". Now set E F . You should see why f is a = N=1 n=1 n,N ( ) = continuous function on E, first argue for E BN . ∩ 30 2. ABSTRACT INTEGRATION AND THE LEBESGUE INTEGRAL Chapter 2 Exercises

2.1. We have seen that f : R R is Lebesgue-measurable if and only if the preimage of → 1 every Borel set is Lebesgue measurable, that is f − (B) whenever B . Show that this is false if “Borel set” is replaced by “Lebesgue measurable∈ L set”. Namely,∈ B construct a 1 Lebesgue measurable function f : [0, 1] R and A such that f − (A) . Hints. In this exercise, we refer to the (countable)→ collection∈ L \BD of endpoints of dyadic6∈ L subintervals of [0, 1] as dyadic points. Note that the set dyadic points are a σ set. Let N = [0, 1] D and note that each x N can be written uniquely as F \ ∈ X ∞ n N x = an(x)2− , an(x) 0, 1 , n=0 { } ∈ { } and the map x an(x) is a bijection. Prove that 7→ { }  0 x D f : 0, 1 0, 1 , f x X n ∈ [ ] [ ] ( ) = 2an(x)3 x N →  − n N ∈ ∈ is measurable by writing it as a pointwise (increasing) limit of measurable (in fact, simple) functions. What is the range of f ? Note that f is a bijection from N onto its range. Recall that N has measure 1 and thus possesses non-Lebesgue measurable subsets. This should be enough to conclude.

2.2. Prove the following statement. Let (X , , µ) be a measure space. Given f : X [0, ], deﬁne the measure F → ∞ Z ν(A) = f 1A dµ, A . ∈ F (there is no need to prove that this is a measure). Prove that for all g : X [0, ] Z Z → ∞ g dν = g f dµ.

2.3. Find a sequence of integrable functions fn : [0, 1] R whose pointwise limit on [0, 1] is f and → Z Z Z lim fn = f , lim fn f = | − | ∞

2 Answer. Take fn x n 1 1 1 1 1 1 ( ) = ( 2n , n ) (1 n ,1 2n ) − − − 2.4. This counterexample shows that pointwise domination by an integrable function is only a sufﬁcient condition for convergence in L1. Find a sequence of integrable functions fn : [0, 1] R whose pointwise limit on [0, 1] is f and such that → Z lim fn(x) f (x) dx = 0 • | − | 1 there is no g L such that fn g a. e. on [0, 1] for all n. • 1 ∈ | | ≤ Answer. Take fn x 1 n 1 1,n 1 x . ( ) = x (( + )− − ]( ) 2. ABSTRACT INTEGRATION AND THE LEBESGUE INTEGRAL 31

2.5. Compute 1 Z e nt 1 t n lim − ( ) dt. n − t − →∞ 0 You can freely use that the Lebesgue integral coincides with the Riemann integral when both exist. Hint. You want to use the equality Z t nt n n(t τ) n 1 e− (1 t) = nτe− − (1 τ) − dτ. − − 0 −

2.6 Countable additivity. Let En : n N be a measurable partition of X (that is, En = X and the sets En are pairwise disjoint).{ ∈ Prove} that ∪ X Z f 1 f dµ < ,

∈ L ⇐⇒ n En | | ∞ in which case Z X Z f dµ = f dµ < , n En ∞ 2.7. Using the previous exercise, prove that, for µ Lebesgue measure,

sin x 1 f (x) = (R, , µ). x 6∈ L L 2.8 Absolute continuity of the integral. Let (X , , µ) be a measure space. Prove that F Z 1 f L (X , , µ) " > 0 δ = δ(", f ) > 0 s.t. µ(A) < δ = f dµ < ". ∈ F ⇐⇒ ∀ ∃ ⇒ A | | 2.9 Modes of convergence. Prove or disprove with a counterexample:

#1 fn f pointwise a.e.[µ] = fn f in measure → 1 ⇒ → #2 fn f in L = fn f in measure → ⇒ → 1 #3 fn f uniformly = fn f in L → 1 ⇒ → #4 fn f in L = fn f pointwise a.e.[µ] P→ ⇒ → #5 fn f 1 < = fn f almost uniformly k − k 1 ∞ ⇒ → #6 fn f in L = fn f almost uniformly, up to a subsequence → ⇒ → 2.10 Sharpness of Egorov’s theorem. Construct a sequence fn : R R of measurable functions, and f : R R measurable such that → → 1 fn f pointwise everywhere and in L • → fn does not converge to f almost uniformly on R. •

CHAPTER 3

Elementary theory of Lp spaces

3.1 Convexity. Jensen, Hölder, Minkowski inequality. We begin with two definitions. A d set A R is said to be convex if, for any x0, x1 A ⊂ ∈ xα = x0 + α(x1 x0) A α [0, 1]. − ∈ ∀ ∈ With this definition, the empty set and singletons are convex. If d = 1, A is convex if and only if A is an interval (possibly unbounded). Thus all convex subsets of R are measurable. d It is not hard to see that if A R is convex, then all of its projections onto the first d 1 coordinates ⊂ −

At = x 0(x1,..., xd 1) : (x 0, t) A , t R { − ∈d } ∈ are also convex. By using inductively that A (R ) is Borel measurable if and only if d 1 ∈ B At (R − ) for all t R, all convex sets are Borel sets. d ∈Let BA R be a convex∈ set. A function f : A R is said to be convex if ⊂ → f (xα) f (x0) + α(f (x1) f (x0)) α [0, 1]. ≤ − ∀ d∈ 1 Equivalently the set (x, y) A R : y f (x) is convex in R + . Also equivalently, for all { ∈ × ≥ } α (0, 1), x0 = x1 ∈ 6 f (xα) f (x0) f (x1) f (xα) (3.1) − − xα x0 ≤ x1 xα d − − 3.1.1 LEMMA. Let A R be open and convex set, and f : A R be convex. Then f is a d continuous function. Furthermore,⊂ for all v R the limits → ∈ f (x + tv) f (x) (3.2) D± f (x) := lim v t 0 t − → ± + exist ﬁnite for all x A and Dv− f (x) Dv f (x). ∈ ≤ PROOF. Let x A. To show that f is continuous at x, it sufﬁces to show that for all d v R the function∈ ∈ t Ix,v gx,v(t) = f (x + tv), ∈ 7→ where Ix,v = t R : x + tv A is an open interval containing zero, is continuous at 0 R. We omit the{ subscript∈ in the∈ notation} from now on. It is easy to see that g is a convex∈ function. Let a, s, t, b I with a < s < 0 < t < c < d. From (3.1), we learn that ∈ g(0) g(a) g(s) g(a) + (s a) − , ≤ − 0 a and similarly for the other pair. This yields − g(0) g(a) g(0) g(s) g(t) g(0) g(b) g(0) L(a) := − L(s) := − R(t) := − R(b) := − 0 a ≤ 0 s ≤ t 0 ≤ b 0 − − 33 − − 34 3. ELEMENTARY THEORY OF Lp SPACES

Hence s L(s) is increasing and bounded above, t R(t) is increasing and bounded below, therefore→ the limits below exist ﬁnite and → lim L s lim R t , s 0 ( ) t 0 ( ) ↑ ≤ ↓ which establishes (3.2), and in particular, implies that g is continuous at zero. d A consequence of the above Lemma is that convex functions on R are measurable. The next inequality is a very instructive and useful application of convexity.

3.1.2 JENSEN’SINEQUALITY. Let (X , , µ) be a measure space. Assume that µ(X ) < . Let 1 f L (X , , µ) be real-valued and ΦF: R R be a convex function. Then ∞ ∈ F → 1 Z 1 Z Φ f dµ Φ f dµ. µ(X ) ≤ µ(X ) ◦ PROOF. Deﬁne 1 Z t = f dµ. µ(X )

Let β = D−Φ(t). From the proof of Lemma 3.1.1 one has Φ(s) Φ(t) β(s t), s R. − ≥ − ∀ ∈ It follows that Φ(f (x)) Φ(t) + β(f (x) t) ≥ − and integrating both sides with respect to x in dµ yields the claimed inequality, since the integral of the second summand on the right hand side is zero. 1 3.1.3 REMARK. We do not know whether Φ f L (X , , µ). However, the proof shows 1 ◦ +∈ 1 F that (Φ f )− L (X , , µ). Thus either (Φ f ) L (X , , µ) or we may deﬁne the right hand side◦ to be∈ , andF the inequality holds◦ trivially∈ F ∞ t 3.1.4 EXAMPLE -YOUNG’SINEQUALITY. Take for instance Φ(t) = e . Then, if µ(X ) = 1, the particular case of Jensen’s inequality reads Z Z exp f dµ ef dµ ≤

Let us consider X = x1,..., xn with = (X ) and µ deﬁned via { } F P n X µ( x j ) = αj [0, 1], j = 1, . . . , n, µ(X ) = αj = 1 { } ∈ j=1

Let yj be positive numbers f : X R be deﬁned by f (x j) = log yj. The above inequality becomes the familiar relationship→ between the arithmetic and the geometric mean, known as Young’s inequality. n n Y αj X (3.3) yj αj yj. j=1 ≤ j=1 3. ELEMENTARY THEORY OF Lp SPACES 35

2 A pair (p, p0) [1, ] is called conjugate if there exists 0 α 1 such that ∈ ∞ 1 1 ≤ ≤ (3.4) p = α , p0 = 1 α − 1 When we write α = p and p = , we mean that α = 0. ∞ 3.1.5 HOLDER’SINEQUALITYFORMEASURABLEFUNCTIONS. Let f , g : X [0, ] be mea- → ∞ surable functions and (p, p0) be a conjugate pair with 1 < p < . Then 1 ∞1 Z Z p Z p p p 0 f g dµ f dµ g 0 dµ . ≤ PROOF. Call A and B respectively each factor in the right hand side. We may assume that A, B are both nonzero and ﬁnite, otherwise there is nothing to prove in each of the other cases. Writing F = f /A, G = g/B, note that Z Z p p F dµ = G 0 dµ = 1

Now, by Young’s inequality applied with α1 = α, α2 = 1 α, y1 = F(x), y2 = G(x) we obtain that − p p FG αF + (1 α)G 0 so that ≤ − Z Z Z Z 1 p p f gdµ = FG dµ α F dµ + (1 α) G 0 dµ = 1 AB ≤ − which is what we had to prove. 3.1.6 MINKOWSKI’SINEQUALITYFORMEASURABLEFUNCTIONS. Let f , g : X [0, ] be measurable and 1 p < . Then → ∞

≤ ∞ 1 1 1 Z p Z p Z p p p p (f + g) dµ f dµ + g dµ . ≤ PROOF. We already know the case p = 1. So assume p > 1. We can assume both summands on the right hand side are finite and nonzero. Otherwise the inequality holds trivially. In this case, since t t p is convex, 7→ p p 1 p p (f + g) 2 − (f + g ) ≤ and the left hand side is finite and nonzero as well. Write p p 1 p 1 (f + g) = f (f + g) − + g(f + g) − . Integrating both sides and applying Hölder inequality to both summands on the right hand side, we obtain 1 1 ! 1 Z Z p Z p Z p p p p p 0 (f + g) dµ f dµ + g dµ (f + g) dµ , ≤ and dividing by the second factor on the right hand side of the last display, which is finite and nonzero, we obtain the claimed inequality. 36 3. ELEMENTARY THEORY OF Lp SPACES

p 3.2 L spaces. Let (X , , µ) be a measure space which we assume fixed throughout unless otherwise mentioned, andF let 0 < p < . We define p ∞ p 1 L (X , , µ) = f : X C measurable : f L (X , , µ) . p F → | | ∈ F If f L (X , , µ), the quantity ∈ F 1 Z p p (3.5) f p := f dµ k k | | 1 is finite, and moreover f p = 0 f = 0 . We extend the definition of (3.6) to p = by k k ⇐⇒ ∞ (3.6) f = inf M > 0 : µ x X : f (x) > M = 0 . k k∞ { ∈ | | } and define

L∞(X , , µ) = f : X C measurable : f < . F { → k k∞ ∞} A nontrivial remark is that, if f < , it must be µ x X : f (x) > f ) = 0 since k k∞ ∞ { ∈ | | k k∞}1 x X : f (x) > f is the countable union of x X : f (x) > f + n− , each of {which∈ must| have| measurek k∞} zero. Consequently, f { =∈ M if| and| onlyk ifk there∞ exists} g = f a.e.[µ] with k k∞ sup g(x) = M. x X | | We record the fundamental Hölder’s inequality,∈ which is easily derived from (3.1.6). The missing case p = 1 is certainly the easiest and is left as an exercise

3.2.1 HÖLDER’SINEQUALITY. Let (p, p0) be a pair of conjugate exponents with 1 p . p p 1 Let f L (X , , µ), g L 0 (X , , µ). Then the product f g L (X , , µ) and ≤ ≤ ∞ ∈ F ∈ F ∈ F f g 1 f p g p . k k ≤ k k k k 0 3.2.2 PROPOSITION. Let (X , , µ) be a measure space. Fp 1. for all 0 < p ,L (X , , µ) is a linear space, in the sense that ≤ ∞p F p f , g L (X , , µ), α C = αf + g L (X , , µ). ∈ F ∈ p⇒ ∈ F 2. For 1 p < , p is a norm on L (X , , µ), that is ≤ ∞ k · k F 2a. f p = 0 = f = 0; k k ⇒p 2b. α C, f L (X , µ, ) = αf p = α f p ∈ p∈ F ⇒ k k | |k k 2c. f , g L (X , µ, ) = f + g p f p + g p ∈ F ⇒ k k ≤ kp k k k 3. For 0 p < 1, p is a quasinorm on L (X , , µ), that is properties 2.a and 2.b above≤ continue tok · hold k and 2c. is replaced by F

1 p 1 (3.7) f + g p 2 − f p + g p k k ≤ k k k k 4. For 0 < p < , there holds Chebychev’s inequality

∞ 1 p sup "µ( x X : f (x) > " ) f p. ">0 { ∈ | | } ≤ k k 1 Here, and everywhere else, we identify functions which are equal a.e.[µ] without explicit mention 3. ELEMENTARY THEORY OF Lp SPACES 37

p 5. The spaces L (X , , µ) are complete, in the sense that if fn is a Cauchy sequence in p L (X , , µ), i.e. F F fn fm p 0, n, m , k − k → →p ∞ p L then there exists f L (X , , µ) such that fn f , i.e. fn f p 0 as n . p 6. Simple functions with∈ compactF support are dense→ in L (Xk , −, µk), 0→< p < →. ∞ F ∞ PROOF. We leave the proof of points 1. to 4. as an easy exercise. In particular, part 3. uses the inequalities p p p (a + b) a + b ; 1 ≤ 1 1 1 p p 1 p p (a + b) 2 − a + b . ≤ for a, b 0, 0 < p 1, which are to be proved in Exercise 3.3. We turn to the proof of completeness.≥ There≤ is nothing to prove in the case p = . We turn to the case 0 < p < . p Let fn be a Cauchy sequence in L . Then, from Chebychev’s∞ inequality, fn is Cauchy∞ in measure and furthermore fn p is a bounded sequence. The main tool we will use is the following k k

3.2.3 SUBLEMMA. Let fn be a sequence of measurable functions which is Cauchy in measure, in the sense that for all " > 0

µ( x X : fn(x) fm(x) > " ) 0, m, n . { ∈ | − | } → → ∞ Then there exists a subsequence f and f : X such that f f pointwise. Moreover, nj C nj → → fn f in measure as well. → We postpone the proof of the Sublemma to the exercises. We obtain f : X C such that f f pointwise. By Fatou’s lemma → nj → Z Z p p f dµ lim inf fn dµ < j j | | ≤ →∞ | | ∞ so that f p L1. At this point, f f p 22p f p for j large enough, thus nj | | ∈ | − | ≤ | | f f p 0 nj p k − k → by the dominated convergence theorem. It follows that f f in Lp. But then nj → f f C f f C f f n p p n nj p + p nj p k − k ≤ k − k k − k and both terms on the right hand side go to zero as n, nj , which implies that fn f in Lp, concluding the proof of completeness. Density of simple→ ∞ functions with is proved→ in a totally analogous way to the case p = 1. p 3.2.4 REMARK. Point 5. of Proposition 3.2.2 can be reformulated as completeness of L (X , , µ) with respect to the metric F min p,1 dp(f , g) = f g p { }, 0 < p . k − k ≤ ∞ The fact that the above is a metric is immediate when p 1. In the exercises, the numerical inequality ≥ p p p a b a + b 0 < p 1 | − | ≤ | | | | ∀ ≤ 38 3. ELEMENTARY THEORY OF Lp SPACES

is proved. This may be relied upon to infer that p p p f g p f h p + h g p k − k ≤ k − k k − k and conclude that dp is a metric when 0 < p < 1 as well. We conclude this introductory subsection with an approximation theorem. The proof is identical to the case p = 1 when 1 p < , and very similar in the case 0 < p < 1, where the quasi-triangle inequality replaces≤ the∞ triangle inequality.

p d d 3.2.5 THEOREM. Let 0 < p < . and f L (R ) (R with Lebesgue measure). Then there d ∞ ∈p is a sequence fn 0(R ) with fn f in L . ∈ C → 3.3 Linear bounded functionals. Let (X , , µ) be a measure space and 1 p .A p linear bounded, linear continuous or simply linearF functional on L (X , , µ) is≤ a map≤ ∞ p F Λ : L (X , , µ) C F → satisfying p (linearity) f , g L , α C = Λ(f + αg) = Λ(f ) + αΛ(g) • ∈ ∈ ⇒ p (boundedness) C 0 such that Λ(f ) C f p for all f L . The least• C > 0 such that∃ the≥ above holds| for all| ≤f kLp kis denoted∈ by Λ . It is easy to see that ∈ k k Λ(f ) Λ = sup | | = sup Λ(f ) p f k k f L :f =0 p f p=1 | | ∈ 6 k k k k 3.3.1 REMARK (CONTINUITY). There is no difference between boundedness and (Lipschitz) continuity for linear functionals. In other words a linear bounded functional is uniformly Lipschitz continuous. This simply follows from

Λ(f ) Λ(g) = Λ(f g) Λ f g p. | − | | − | ≤ k kk − k p 1 p A pivotal example is as follows. Let g L 0 (X , , µ). Since f g¯ L whenever f L we can deﬁne the map ∈ F ∈ ∈ Z p (3.8) Λg : L (X , , µ) C, Λg (f ) = f g¯ dµ. F → It is easy to see that this is a linear map, bounded by virtue of Hölder inequality which yields

Λg (f ) f g 1 g p f p. | | ≤ k k ≤ k k 0 k k Therefore Λg p g p . In fact, there holds k k ≤ k k 0 3.3.2 LEMMA. Let 1 p and deﬁne Λg as in (3.8). Then ≤ ≤ ∞ Λg = g p k k k k 0 PROOF. Exercise 3.9. A natural question which arises from Lemma 3.3.2 is whether all linear bounded func- p p tionals on L (X , , µ) arise from g L 0 as in (3.8). This is true if 1 p < and false for p = . The ﬁrstF half of this statement∈ goes under the name of Riesz≤ representation∞ theorem p for the∞ dual space of L (X , , µ). We will prove both statements in Chapter 5 in a general F 3. ELEMENTARY THEORY OF Lp SPACES 39

setting. In the next section, we devote ourselves to the special case of this statement for `p spaces. 3.4 The `p spaces and their Riesz representation. In this paragraph, we consider the spaces p p ` = L (N, (N), ν) P ~ where ν is the counting measure, namely, the space of sequences f = (f1,..., f j,...) such that ( 1 P p p ∞j 1 f j 0 < p < , > f `p = = sup | f | p ∞ ∞ k k j N j = ∈ | | ∞ The sequences k k k k 0 j = k b = (b1,..., bj ,...), bj = 6 1 j = k form an unconditional basis of `p, in the sense that if f `p, the sequence n ∈ n X k f = fk b k=1 converges to f in `p.

p 3.4.1 THEOREM. Let 1 p < and Λ : ` C be a linear bounded functional. Then there p 0 ≤ ∞ → exists a unique g ` such that Λ = Λg as in (3.8), namely ∈ X∞ p Λ(f ) = Λg (f ) = f j g j f ` , j=1 ∀ ∈

and furthermore Λ = g `p . k k k k 0 PROOF. We deal with 1 < p < . The case p = 1 follows with minor changes. The candidate g is obtained by setting ∞

j g j = Λ(b ), j N. ∈ Let f `p. By linearity, for all n ∈ n n n X k X k n Λ(f ) = Λ fk b = fkΛ b = Λg (f ). k=1 k=1 It follows that for n m ≥ n X n m n m f j g j = Λ(f ) Λ(f ) Λ f f `p , j=m+1 | − | ≤ k kk − k which since f n is Cauchy, implies that the series

X∞ Λg (f ) = f j g j j=1 40 3. ELEMENTARY THEORY OF Lp SPACES

p converges to Λ(f ). By Lemma 3.3.2, to finish the proof it suffices to prove that g ` 0 and g p Λ . Define ∈ 0 k k ≤ k k 0 g j = 0 h hj : j , hj = N = p0 2 { ∈ } g j − g j g j = 0 | | 6 and let again hn Pn h b j, g n Pn g b j be the truncation to the first n components. = j=1 j = j=1 j Then hn `p and p 1 ∈ n n 0 h p = g p − 0 Furthermore, a computation reveals thatk k k k

n p0 n n n p0 1 g p = Λg (h ) Λ h p = Λ g p − , 0 0 n k k ≤ k kk k k kk k and rearranging g p Λ . The monotone convergence theorem ﬁnally yields g p 0 0 Λ as well. k k ≤ k k k k ≤ k k 3. ELEMENTARY THEORY OF Lp SPACES 41 Chapter 3 Exercises

3.1. Prove Sublemma 3.2.3.

3.2 Converse of Jensen’s inequality. In this problem, X = [0, 1] with the Lebesgue measure. Suppose φ : R R is a function with the property that → Z Z φ f dµ φ f dµ [0,1] ≤ [0,1] ◦ 1 for all f L (X ). Prove that φ is a convex function. ∈ 3.3. Let 0 < α 1, a, b 0 Prove the two numerical inequalities ≤ ≥ α α α (a + b) a + b ; 1 ≤ 1 1 1 α α 1 α α (a + b) 2 − a + b . ≤ 3.4 Log-convexity of Lp-norms and endpoint spaces. Let f Lp Lq with 1 p < q . a. Prove that ∈ ∩ ≤ ≤ ∞

θ 1 θ 1 θ (1 θ) f r f p f q− θ [0, 1], r = p + −q . k k ≤ kp k k k ∀ ∈ b. Prove that if f L L for p large enough, lim f p f . You should use ∞ p = ∈ ∩ k k k k∞ a. for part of the statement. →∞ p c. Assume in addition that µ(X ) = 1. Let f L for all 0 < p < and such that ∈ ∞ f p sup k k 17. 1 p< p ≤ ≤ ∞ f 1 f Prove that e| | L and e| | 1 1 + 17e. r d. Assume again that∈ µ(X )k = 1.k Let≤ f L for some 0 < r . Prove that ∈Z ≤ ∞ lim f p exp log f dµ p 0 = ↓ k k | | with the convention that exp( ) = 0. −∞ 3.5 Converses of Hölder’s inequality. In this exercise, (X , , µ) is a measure space, µ(X ) = 1, and 1 < q is ﬁxed. F ≤ ∞ q a. Let f : X C be a measurable function with the following property: for all g L , the product→ f g L1 and ∈ ∈ Z f g dµ g q. | | ≤ k k p Prove that f L , with p = q0, and that f p 1. ∈ k k ≤ 2 b. Let f : X C be a measurable function with the following property : for all q → 1 p g L , the product f g L . Prove that f L , with p = q0. ∈ ∈ ∈ 2For this part, any solution using tools which do not fall into the scope of the ﬁrst three chapters will not be considered. 42 3. ELEMENTARY THEORY OF Lp SPACES

3.6. Here X = R with Lebesgue measure. Find a function F : X [0, ) such that q → ∞ q (0, ) : F L = 7 . { ∈ ∞ ∈ } {q } Can there be a function G such that q (0, ) : F L = e, π ? p { ∈ ∞ ∈ } { } k 3.7. Let 0 < p < , f L (X , , µ). Deﬁne for k Z, Fk = x X : f (x) > 2 and ∞ ∈ F ∈ 1 { ∈ | | } p X kp [f ]p := 2 µ(Fk) . k Z ∈ Prove that there exist constants cp < Cp such that

cp[f ]p f p Cp[f ]p. ≤ k k ≤ Do the case with µ(X ) < ﬁrst. ∞ 3.8. Let Φ : [0, ) [0, ) be a convex, strictly increasing function with Φ(0) = 0. This exercise is about∞ ﬁnding→ a∞ norm for the space Φ Φ 1 L = L (X , , µ) = f : X C measurable : x Φ( f (x) ) L (X , , µ) . F → 7→ | | ∈ F 1. Prove that if f : X C → Z f (x) Iλ(f ) := Φ | λ | dµ(x), λ > 0

is a decreasing function of λ and

lim Iλ(f ) = λ 0+ → ∞ unless f = 0 a.e., in which case Iλ(f ) = 0 for all λ > 0. 2. Prove that if f LΦ ∈ (3.9) lim Iλ f 0 λ ( ) = →∞ and therefore

f Φ := inf λ > 0 : Iλ(f ) 1 [0, ) k k { ≤ } ∈ ∞ is well deﬁned. Furthermore prove that f Φ = 0 if and only if f = 0 a.e. and that k k f LΦ, f 0 I f 1. = = f Φ ( ) = k k Φ ∈ 6 ⇒ 3. Assume that f , g L , α C. Prove that ∈ ∈ αf Φ = α f Φ, f + g Φ f Φ + g Φ k k | |k k k Φ k ≤ k k k k and conclude that Φ is a norm on L (in fact, arguing in the same way we used for Lp, LΦ is completek · k with respect to this norm).

p p 3.4.2 REMARK. In fact, using the above for Φ(t) = t proves the triangle inequality for L spaces without any appeal to Hölder’s inequality. There is actually a way to derive Hölder’s inequality in this framework, using the concept

of conjugate function. Given Φ : [0, ) [0, ) convex, one deﬁne the conjugate Φ∗ : [0, ) [0, ) by ∞ → ∞ ∞ → ∞ Φ∗(s) = sup ts Φ(t) : t 0 . { − ≥ } 3. ELEMENTARY THEORY OF Lp SPACES 43

p It turns out that Φ∗ is also convex and (Φ∗)∗ = Φ. For instance, if Φ(t) = t with 1 < p < p 0 ∞ then Φ∗(s) = s . Using the deﬁnition of conjugate functions, it should be easy to prove the Φ Φ 1 following: if f L , g L ∗ then f g L and ∈ ∈ ∈ f g 1 2 f Φ g Φ . k k ≤ k k k k ∗ 3.9. Prove Lemma 3.3.2.

Hint.

CHAPTER 4

Elementary Hilbert space theory

4.1 Hilbert space. Let H be a linear space over C. An inner product on H is a map ( , ) : H H C with the following properties: for all x, y, z H and α C · · × → ∈ ∈ (skew-symmetry) (y, x) = (x, y); • (linearity) (x + αy, z) = (x, z) + α(y, z); • (positive definite) (x, x) > 0 for all x = 0. A consequence• of the definitions is that 6 (x, 0) = (0, x) = 0 x H. ∀ ∈ We will use the notation Æ x H = (x, x) sometimes omitting the subscript. It isk easyk to see that x 0 with equality if and only if k k ≥ x = 0, and αx α x whenever α C, x H. We now show that x H is a norm on H by proving thek trianglek| |k k inequality. The∈ first∈ step is the Hilbert spacek analoguek of Hölder’s inequality

4.1.1 CAUCHY-SCHWARZ INEQUALITY. Let ( , ) be an inner product on H. Then · · (x, y) x y . | | ≤ k kk k PROOF. If either x = 0 or y = 0 there is nothing to prove. We may thus assume that the right hand side does not vanish. By replacing x with x/ x and y with y/ y it sufﬁces to prove that k k k k x = y = 1 = (x, y) 1. k k k k ⇒ | | ≤ Let α be a unimodular such that α(x, y) = (x, y) and t be a real parameter. We compute 2 | | 2 2 0 x αt y = (x, x) t α(x, y) + α(x, y) + t (y, y) = 1 2t (x, y) + t ≤ k − k − − | | Taking t = (x, y) and rearranging yields the result. | | From Cauchy-Schwarz, it follows that 2 2 2 2 2 2 x + y = x + y + (x, y) + (y, x) x + y + 2 x y = ( x + y ) k k k k k k ≤ k k k k k kk k k k k k which is the triangle inequality. We also record the useful parallelogram identity x y 2 x y 2 (4.1) + 1 x 2 1 y 2. + − = 2 + 2 2 2 k k k k Now that we have seen that an inner product induces a norm on H, we say that H is a

Hilbert space if it is complete with respect to the norm H induced by the inner product. A consequence of the Cauchy-Schwartz inequality is thatk · k

xn x 0 = (xn, y) (x, y) y H k − k → ⇒ → ∀ ∈ 45 46 4. ELEMENTARY HILBERT SPACE THEORY

namely, the map x (x, y) is continuous in the topology induced by the Hilbert norm of H. 7→

4.1.2 EXAMPLE. Let (X , , µ) be a measure space. We can easily verify that F Z (f , g) = f g dµ,

2 which is well defined for f , g L (X , , µ) by virtue of Hölder’s inequality, defines an inner 2 2 product on L (X , , µ) whose∈ inducedF norm coincides with the L -norm, with respect to 2 2 which L (X , , µ)Fis complete. Thus L (X , , µ) is a Hilbert space. Taking X = 1, . . . , d d 2d with the countingF measure, we recover the EuclideanF space C R . { } ∼ 4.2 Subspaces, orthogonal complement and projection theorem. Recall that a subspace M of a linear space H is a subset of H which is invariant under respect to linear combination. For any subset A of a Hilbert space, we define the orthogonal complement

A⊥ = x H : (x, y) = 0 y M . { ∈ ∀ ∈ } Notice that A⊥ is always a subspace. Furthermore A is closed with respect to the topology induced by the Hilbert norm of H. Indeed \ A⊥ = y ⊥ y A{ } ∈ and each y ⊥ is closed, being the preimage of 0 with respect to the continuous map x (x, y). { } 7→ 4.2.1 REMARK. We give an example of subspaces of Hilbert spaces H which are not closed. 2 Let H = ` . We can define 2 2 `0 = x ` : x j = 0 for all but finitely many j N . 2 { ∈ N6 N N ∈ } N 2 For x ` , recall the definition of x as x j = x j if j N, x j = 0 otherwise. Then x `0 N ∈ 2 2 2 ≤ ∈ x x in ` , thus `0 is dense in ` . A whole variety of dense, nontrivial subspaces can be obtained→ as follows. Let a(k) be a sequence with 1 inf a(k) lim sup a(k) = , and define ≤ ≤ ∞ ¨ « 2 2 X∞ 2 à = x ` : a(k) xk < . ∈ k=1 | | ∞ 4.2.2 MINIMAL DISTANCE LEMMA. Let A be a convex closed nonempty subset of H. Then there exists a unique y A of minimal norm, namely ∈ z A, z y = z = y. ∈ k k ≤ k k ⇒ PROOF. Denote by δ = infz A z . Let us prove uniqueness. Suppose y, y0 H are such ∈ k k ∈ that y = y0 = δ. Then by convexity z = (y + y0)/2 A. By the parallelogram equality k k k k1 2 1 2 1 2 2 1∈ 2 1 2 2 (4.2) 4 y y0 = 2 y + 2 y z 2 y + 2 y δ k − k k k k k − k k ≤ k k k k − which is = 0, so it must be y = y0. Let now yn be a sequence of points of A such that yn δ. Writing (4.2) for y = yn, y0 = ym, the right hand side will tend to 0 as n, m . k k → → ∞ Thus yn is Cauchy, and yn y in A. By continuity, it must be δ = y , which proves existence. → k k 4. ELEMENTARY HILBERT SPACE THEORY 47

4.2.3 PROJECTIONTHEOREM. Let M be a closed subspace of a Hilbert space H. Then for all h H, there exist a unique x = Ph M and y = Qh M ⊥ such that ∈ ∈ ∈ h = Ph + Qh, each satisfying z M, z h x h = x = z ∈ k − k ≤ k − k ⇒ w M ⊥, w h y h = w = y ∈ k − k ≤ k − k ⇒ Furthermore the maps h Ph and h Qh are linear and 7→ 7→2 2 2 h = Ph + Qh k k k k k k PROOF. Let us prove uniqueness. It is easy to see that M M ⊥ = 0 . Suppose there ∩ { } exist x, x 0 M, y, y0 M ⊥ such that h = x + y = x 0 + y0. Then x x 0 = y0 y belongs to ∈ ∈ − − M M ⊥, which means x = x 0, y = y0. ∩To prove existence, note that the set y H : y h M is convex and closed. Thus there exists an element y of minimal norm.{ ∈ Call Qh−:= y∈, Ph}= h Qh. Let us prove that − Qh M ⊥. For any z M of unit norm and α C, Qh h αz M as well, so by deﬁnition of Qh∈ ∈ ∈ − − ∈ 2 2 2 0 Qh αz Qh = α(Qh, z) α(Qh, z) + α ≤ k − k − k k 2 − − | | which for α = (Qh, z) yields 0 (Qh, z) . So Qh M ⊥. ≤ −| | ∈ PROOF. Denote by M the closed linear span of A. It is easy to see that A (A⊥)⊥. Since ⊂ the latter is closed, it must be M (A⊥)⊥. To prove the reverse inclusion, suppose there ⊂ exists x = 0 such that x (A⊥)⊥ and x M. Then x = Ph M by deﬁnition and if z M, by orthogonality6 ∈ 6∈ ∈ ∈ 2 2 2 z h = z Ph Qh = z Ph + Qh k − k k − − k k − k k k which is obviously minimized when z = Ph. For linearity, let α C, h, k H. Then ∈ ∈ P(h + αk) Ph αPk = Q(h + αk) Qh αQk − − − − so they are both zero, which means P,Q are linear. This completes the proof.

A corollary of the projection theorem is that if M is a proper subspace of H, then M ⊥ is not equal to 0 . To see this, notice that if M is a proper subspace, there exists h H which { } ∈ does not belong to M. Thus Qh = 0 is a nontrivial element of M ⊥. Another corollary is the following.6 We deﬁne closed linear span of A H to be the the intersection of all closed subspaces M containing A. The closed linear span⊂ of A is a closed subspace.

4.2.4 LEMMA. (A⊥)⊥ is the closed linear span of A. PROOF. Exercise 4.3 Linear functionals. Let H be a Hilbert space. A linear continuous functional is a linear map Λ : H C such that → Λ(h) Λ := sup | | < . k k h H: h =1 h ∞ ∈ k k k k 48 4. ELEMENTARY HILBERT SPACE THEORY

It is easy to deﬁne plenty of nontrivial linear continuous functionals on H. Namely, if h H is nonzero the map ∈

Λh : H C, Λh(k) = (k, h) → is linear, by Cauchy-Schwarz inequality

k = 1 = Λh(k) h k k ⇒ | | ≤ k k with equality when k = h/ h , whence Λh = h . In fact, all linear functionals arise in this way. k k k k k k

4.3.1 RIESZ REPRESENTATION THEOREM. Let H be a Hilbert space and Λ : H C be a linear continuous functional. Then there exists a unique vector h, with h = Λ , such→ that k k k k Λ = Λh.

PROOF. Let us ﬁrst prove uniqueness. If Λ = Λh = Λk one has (x, k) = (x, h) for all 2 x H. Taking x = k h it follows k h = 0, thus h = k. ∈Let us prove existence.− If Λ 0,k simply− k take h = 0. If not, the subspace ≡ M = k H : Λ(k) = 0 { ∈ } is a proper subspace of H. Therefore there exists z M ⊥ of unit norm. Notice that for all x, we have ∈ Λ(x) Λ x z = 0, − Λ(z) Λ(x) which means that x Λ z z M, which means that − ( ) ∈ Λ(x) Λ(x) 0 = x z, z = (x, z) . − Λ(z) − Λ(z) It follows that for all x H ∈ z (x, h) = Λ(x), h := Λ(z)

whence Λ = Λh. The proof is complete. We state the following important corollary which answers the question of the preceding chapter at least for p = 2.

4.3.2 COROLLARY. Let (X , , µ) be a measure space and Λ be a linear bounded functional on 2 F 2 L (X , , µ). Then Λ = Λg for some g L , namely, F ∈ Z 2 Λ(f ) = f g dµ f L (X , , µ). ∀ ∈ F 4.4 The Riesz theorem and the Dirichlet problem. For this paragraph, D will stand for 2 the space 0 (0, 1) of real-valued, inﬁnitely differentiable functions with compact support in C 4. ELEMENTARY HILBERT SPACE THEORY 49

(0, 1). We want to make D a (real) inner product space with respect to two different inner products1: Z 1 Z 1 (4.3) (f , g)H = f (x)g(x) dx, (f , g)V = f 0(x)g0(x) dx, 0 0 and denote by H , V the corresponding norms. For now the subscripts H, V are mere k · k k · k notation. While we already know that ( , )H is an inner product, this is not the case for · · ( , )V : namely, while linearity, symmetry, and nonnegativity are clear, we must observe that · · f V = 0 = f = 0. The lemma that follows tells us this and much more. k k ⇒ 4.4.1 POINCARÉINEQUALITY. Let f D. Then ∈ (4.4) f H f V . k k ≤ k k In particular, f V = 0 if and only if f = 0. k k PROOF. Since f is compactly supported we have for all 0 < x < 1, by Hölder,

1 Z x Z x 2 2 f (x) f 0(t) dt px f 0(t) dt f V . | | ≤ 0 | | ≤ 0 | | ≤ k k Square and integrate over (0, 1). 2 Since D is dense in L , its completion with respect to H is the Hilbert space H = 2 k · k L (0, 1). It can be seen that D is not complete with respect to V (exercise). In the next lemma we characterize its completion V . k · k

4.4.2 LEMMA. Let v V . Then v H. Moreover there exists w H with the property that ∈ ∈ ∈ (4.5) (w, φ) = (v, φ0) φ D. − ∀ ∈ Furthermore v is a continuous function on (0, 1) amd (4.6) lim v(x) = lim v(x) = 0 x 0+ x 1 → → − The function w is often called the weak derivative of v. Further ahead in the course we will see that v(y) v(x) (4.7) v V = v0(x) = lim − exists a.e, v0 = w a.e. ∈ ⇒ y x y x → − The sense of (4.16) is that v extends continuously (to zero) at x = 0, 1.

PROOFOF LEMMA 4.4.2. Let vn D be a Cauchy sequence with respect to V . This ∈ k · k means vn0 vm0 H 0 so vn0 w H by completeness of H. Now by Lemma 4.4.2, it must be vn k vm−H k 0→ as well, so→vn ∈ v H. Let φ . An integration by part yields k − k → → ∈ ∈ D (vn0 , φ) = (vn, φ0) − and the lhs tends to (w, φ), while the right hand side tends to (v, φ) by continuity of the maps u (u, φ), u (u, φ0) in H. The second part is left as an exercise. 7→ 7→ 1The integral notation is for the Lebesgue integral, which coincides with the Riemann integral whenever the integrand are Riemann-integrable. This is of course the case of functions in D. 50 4. ELEMENTARY HILBERT SPACE THEORY

The most important boundary value problem in Partial Differential equations is the Dirichlet problem. In this paragraph, we study the simple one-dimensional version. Let f : (0, 1) R be a given datum. We say that u : [0, 1] R is a classical solution to the Dirichlet problem→ on [0, 1] with datum f if → u00(x) = f (x) 0 < x < 1 (4.8) − u(0) = u(1) = 0. For instance, if the segment [0, 1] represents a metal rod whose extremities are kept at a constant reference temperature, Q = Q(x) is an external source of heat per unit length, and f = Q0, u describes the temperature of the rod once a stationary state is reached. Now, if f is a continuous function with compact support in (0, 1), then u D. Let φ D. Multiply and integrate the equation on (0, 1), integrating by parts one gets∈ that ∈

(u, φ)V = (f , φ)H φ D. ∀ ∈ Let now v V , and φn v in V , passing to the limit (u, φn)V (u, v)V and similarly for the right hand∈ side. It means→ that →

(4.9) (u, v)V = (f , v)H v V ∀ ∈ We can observe that the right hand side of (4.9) makes sense if f H. More precisely, since by Lemma 4.4.1 ∈

(f , v)H f H v H f H v V | | ≤ k k k k ≤ k k k k the left hand side deﬁnes a linear continuous functional with norm f H . By the Riesz representation theorem ≤ k k

(4.10) !u = u(f ) V : (u, v)V = (f , v)H v V, u V f H , ∃ ∈ ∀ ∈ k k ≤ k k Such an u is called a weak solution to the Dirichlet problem with datum f H. ∈ 4.5 Orthonormal sets. A set B H is termed an orthonormal system if ⊂ (h, k) = 0 h, k B, h = k, h = 1 h B ∀ ∈ 6 k k ∀ ∈ An orthonormal system B is an orthonormal basis if

(x, h) = 0 h = x = 0, ∀ ⇒ that is B⊥ = 0 . { } 4.5.1 REMARK. Let α1,..., αn be complex numbers. It is easy to see that 2 n n X X 2 αjhj = αj . j=1 j=1 | |

4.5.2 EXAMPLE. The canonical vectors b j form an orthonormal basis of the Hilbert space `2.

4.5.3 THEOREM. Every Hilbert space H admits an orthonormal basis B.

PROOF. We only give a proof of the above theorem under the further assumption that H 2 d is separable, that is, there exists a countable dense subset N H. For instance, L (R ) with the Lebesgue measure is separable, since ﬁnite linear combinations⊂ of indicator functions 4. ELEMENTARY HILBERT SPACE THEORY 51

2 d of dyadic cubes are dense in L (R ). This proof is actually constructive, while the general case is not. First of all, if N is countable and dense in H, we can construct a sequence yn : n N N J y y J { J ∈ } ⊂ such that, if n is the linear span of 1,..., n , n is a proper subspace of n+1 and for all x H and " > 0 there exists n large{ enough and} y Jn with x y < ". We leave the ∈ ∈ k − k details as an exercise. We will now construct a sequence hj : j N which is orthonormal, { ∈ } such that h1,..., hn spans Jn. Then hj : j N must be an orthonormal basis. To see this, let x H {be such that} { ∈ } ∈ (x, hj) = 0 j. ∀ Now for any " > 0, we can ﬁnd n large enough and coefﬁcients α1,..., αn such that n X x xn < ", xn = αjhj. k − k j=1

However, it must be (x, x xn) = 0, therefore − (x, x) = (x, x xn) x xn x " x − ≤ k − kk k ≤ k k since " is arbitrary, it must be x = 0. We leave the inductive construction of h1,..., hn as an exercise. { } The proof actually provides the stronger conclusion that n X (4.11) x lim x, hj hj = n ( ) →∞ j=1

4.5.4 BESSELAND PARSEVAL LEMMATA. If hj is an orthonormal set (ﬁnite or countably inﬁ- nite), then { }

X∞ 2 2 (x, hj) x j=1 | | ≤ k k and equality holds if and only if hj is an orthonormal basis. { } PROOF. Let M be the closed linear span of hj and P x be the projection of x on M. The same argument as in the proof of Theorem 4.5.3{ } yields that n X P x lim x, hj hj = n ( ) →∞ j=1 so in particular by Remark 4.5.1

2 2 X∞ 2 x = P x = (x, hj) . k k k k j=1 | | since P x = x for all x if and only if M = H, the second claim follows. 4.6 Orthonormal systems and bases on the real line. Let us consider the Hilbert space 2 L (0, 1) (Lebesgue measure). An orthonormal system is given by the complex exponentials 2πikx ek(x) := e , k Z. ∈ 52 4. ELEMENTARY HILBERT SPACE THEORY

Finite linear combinations N X P(x) = αkek(x) k= N − are called trigonometric polynomials. We recall the following result of Weierstrass: the proof is omitted for now. It can be found in Rudin, p.89.

4.6.1 THEOREM. Trigonometric polynomials are dense in ([0, 1]). Namely, for each continuous functions f : [0, 1] C and " > 0 there exists a trigonometricC polynomial P with → max P(x) f (x) < ". x [0,1] ∈ | − | We leave for exercise, using Weierstrass’ theorem above, the proof that ek : k Z is an 2 orthonormal basis of L (0, 1). An immediate consequence is the following{ theorem.∈ } 2 4.6.2 THEOREM. Let f L (0, 1). Then the partial Fourier sums ∈ N X fN (x) = (f , ek)ek(x) k= N − converge to f in L2 as N and there holds the Parseval identity → ∞ 2 X∞ 2 f 2 = (f , ek) . k k k= | | −∞ Since convergence in L2 is stronger than convergence in L1, the above theorem yields

that fN f pointwise a.e. up to a subsequence. → 2 4.6.3 REMARK. The fact that if f L (0, 1) then the full sequence fN converges to f pointwise a.e. is MUCH HARDER and was∈ proved by Lennart Carleson in 19662.

For each dyadic interval I denote by I`, Ir respectively the left and right dyadic child of I. The I-th Haar function is∈ defined D as 1 x 1 x I` ( ) Ir ( ) (4.12) hI (x) = − p I | | The collection hI : I is an orthonormal system: this is easily verified. In the exercises, { ∈ D} 2 we will prove that hI : I is an orthonormal basis of L (R). The first step, which yields as a corollary that{ whenever∈ D}J is a dyadic interval

1 2 (4.13) B(J) := hI : I , I J J − 1J { ∈ D ⊂ } ∪ {| | } 2 is an orthonormal basis of L (J), is summarized in the following lemma.

4.6.4 LEMMA. Let I J be a dyadic interval. Then 1I belongs to the linear span of B(J). ⊂ 2At the time when Carleson proved his theorem, it was not even known whether pointwise a.e. convergence held for f continuous (and several prominent analysts, like Lusin and Zygmund, thought it was false). We will get back to Fourier series of continuous functions later in the class 4. ELEMENTARY HILBERT SPACE THEORY 53

n PROOF. Let n 0 be such that I = 2− J . We prove this by induction on n. The case ≥ | | | | ˜ n = 0 is trivial. Suppose n 1. Then the dyadic parent of I, call it I is such that 1˜I belongs to the linear span of B(J) ≥by induction assumption. For deﬁniteness, suppose I is the left child of ˜I. Then Æ ˜ I h˜I + 1˜I = 21I | | so that 1I belongs to linear span of B(J) as well. This completes the proof. 54 4. ELEMENTARY HILBERT SPACE THEORY Chapter 4 Exercises

4.1 Polarization identity. Let ( , ) be an inner product on H and be the associated norm. Prove that, for all x, y H· · k · k ∈ 1 X x, y a x ay 2. ( ) = 4 + a 1, 1,i, i k k ∈{ − − } 4.2 Gram-Schmidt orthonormalization. This problem completes the proof of Theorem 3 4.5.3. Let H be an inner product space and yj : j N be a sequence of vectors of H such that { ∈ } J y y J n := span 1,..., n n+1 { } Construct an orthonormal system hj : j N such that Jn = span h1,..., hn . { ∈ } ( { } 4.3. Let H be a Hilbert space and M be a subspace of H. Prove that

(M ⊥)⊥ = M

where M denotes the closure of M with respect to the norm H . Use this result and the k · k previous problem to prove that if M is ﬁnite dimensional, that is M = span y1,..., yn , then M is closed. { }

2 4.4. Let D be the space of real valued functions in (0, 1) with compact support in (0, 1) and V be the closure of D with respect to the innerC product (see Section 4.4) Z 1 (f , g)V = f 0(x)g0(x) dx. 0 Let v V . Prove that ∈ v is a continuous function on (0, 1) • lim v(x) = lim v(x) = 0. x 0+ x 1 • → → − 4.5 Lax-Milgram Theorem. This problem consists of two parts. 1. Let V be a Hilbert space over C and B : V V C be a map with the properties × → the map x B(x, y) is linear for all y V ; • 7→ ∈ the map y B(x, y) is linear for all x V ; • there exist 7→M δ > 0 such that for all ∈x, y V

• ≥ ∈ 2 B(x, y) M x V y V , B(x, x) δ x V . | | ≤ k k k k ≥ k k Let λ be a linear bounded functional on V . Prove that there exists a unique u V such that ∈ 1 B(v, u) = λ(v) v V, u V λ . ∀ ∈ k k ≤ δk k Note: there is a totally analogous result when V is a Hilbert space over R by removing the conjugate in the second requirement.

3 For the sake of deﬁniteness, we deal with the case where the sequence yj is countably inﬁnite. { } 4. ELEMENTARY HILBERT SPACE THEORY 55

2. Let V and H be the same spaces introduced in Section 4.4. Let β 0 and consider the boundary value problem ≥ u00(x) + βu(x) = f (x) 0 < x < 1 (4.14) − u(0) = u(1) = 0. Write the weak formulation of (4.14) as: find u V such ∈ B(u, v) = (f , v)H v V. ∀ ∈ for a suitable B. Verify that B and the right hand side fulfill the assumptions of Part 1 and conclude that there exists a unique weak solution u V . Give an upper bound on u V . ∈ k k Hint. For Part 1, observe that for all u V the map ∈ v B(v, u) 7→ is a linear continuous functional. Fix u and apply the Riesz theorem to get the existence of z = z(u) V such that ∈ B(v, u) = v, z(u) v V. ∀ ∈ Prove that the map u z(u) is linear and that the image of V under such map 7→ Z = z V : z = z(u) for some u V { ∈ ∈ } is a closed linear subspace of V and Z⊥ = 0 . Thus Z = V . This can be readily used to conclude. { } 4.6. Prove that ek : k Z , where { ∈ } 2πikx ek(x) = e 2 is an orthonormal basis of L (0, 1). You can use Weierstrass’ theorem. 4 4.7. Let I R be a dyadic interval . Define ⊂ Z 2 M = f L (J) : f dµ = 0 . ∈ J 2 1. Show that M is a closed subspace of L (J). 2. Determine M ⊥. 2 Hint. M is the nullspace of a certain linear functional on L (J). 2 4.8 The Haar basis of L (R). In this problem, we complete the proof that the Haar system 2 hJ : J defined in (4.12) is an orthonormal basis of L (R). { 1.∈ D}Use Lemma 4.6.4 to prove that 1 2 B(J) := hI : I , I J J − 1J { 2 ∈ D ⊂ } ∪ {| | } is an orthonormal basis of L (J). 2. Use the first part and Problem 4.7 to prove that 2 f L (R), (f , hI ) = 0 I = f 0 ∈ ∀ ∈ D ⇒ ≡ 2 thus completing the proof that the Haar basis is an orthonormal basis of L (R).

4 2 2 Here L (J) denotes the L (J, (J), µ) space, (J) are the Lebesgue measurable subsets of J and µ is Lebesgue measure L L 56 4. ELEMENTARY HILBERT SPACE THEORY

Solutions to selected Exercises of Chapter 4.

4.2. We proceed by induction. The base case is J1 = y1 . Take h1 = y1/ y1 . Suppose now h1,... hn is an orthonormal system that spans Jn.{ Deﬁne} k k { } n X rn+1 = yn+1 (yn+1, hj)hj. − j=1 r y J It must be n+1 = 0, otherwise n+1 n, which contradicts our assumption. Then set h r r 6 h h ∈ J n+1 = n+1/ n+1 . The set 1,..., n+1 is orthonormal and spans n+1. Inductive step is thus complete.k k { }

4.3. Preliminarily notice that (M)⊥ = M ⊥. Indeed, if v M, x M ⊥, and vn M, vn v, ∈ ∈ ∈ → it must be 0 = (vn, x) (v, x), whence (M)⊥ M ⊥ and the other inclusion is obvious. So by the projection theorem,→ the projections PM ,⊃PM satisfy ⊥

x = PM x + PM x x H. ⊥ ∀ ∈ We come to the actual proof. Since M (M ⊥)⊥ and the latter is closed it must be ⊂ M (M ⊥)⊥. Suppose that the opposite inclusion does not hold, thus there is v (M ⊥)⊥ M. ⊂ ∈ \ By the above display PM v = 0. However since v (M ⊥)⊥, 0 = (PM v, v) = (PM v, PM v) = 2 ⊥ ⊥ ⊥ ⊥ PM v . Contradiction. 6 ∈ k ⊥ k 4.4. Let v V and vn be a sequence of functions of D with vn v in V . We can assume ∈ → vn V 2 v V . Let 0 x < y 1. Then by Cauchy-Schwarz and vn(0) = vn(1) = 0, we havek k ≤ k k ≤ ≤ Z y 1 1 2 2 (4.15) vn(x) vn(y) vn0 (t) dt x y vn V 2 v V x y . | − | ≤ x | | ≤ | − | k k ≤ k k | − | 1 That is, the sequence vn is uniformly equicontinuous on [0, 1]; in fact, it is uniformly 2 - Hölder continuous. By Ascoli-Arzelà, up to a subsequence, vn w to some w ([0, 1]) → ∈ C uniformly, and it must be w(0) = w(1) = 0. But we know vn v pointwise almost everywhere up to a subsequence. Repeating the argument for such→ a subsequence, it must be w = v almost everywhere. Redefining v on a null set, we obtain that v ([0, 1]) and v(0) = v(1) = 1, which is the second part of the statement. The proof yields∈ C the stronger 1 result that v 2 ([0, 1]). ∈ C 4.5. We pick up where the hint stops for the first part. Let u1, u2 V and α C Using the definitions and the conjugate linearity in the second component ∈ ∈ v, z(αu1 + u2) = B(v, αu1 + u2) = αB(v, u1) + B(v, u2) = v, αz(u1) + v, z(u2) = v, αz(u1) + z(u2)

for all v V . Thus rearranging, αz(u1) + z(u2) z(αu1 + u2) ⊥ = V , and we conclude that the map∈u z(u) is linear. It follows{ that Z is a− linear subspace} of V . Let us show that it is closed. We7→ need two estimates. Let u = 0. We have by the third property 6 B(u, u) (u, z(u)) δ u = M z(u) k k ≤ u u ≤ k k k k k k 4. ELEMENTARY HILBERT SPACE THEORY 57

Furthermore 2 z(u) = z(u), z(u) = B(z(u), u) M z(u) u . k k ≤ k kk k If z(u) = 0, we may divide and obtain z(u) M u . In fact, the inequality trivially holds true when6 z(u) = 0 as well. Summarizingk thek ≤ lastk twok conclusions (4.16) δ u z(u) M u k k ≤ k k ≤ k k It follows that the map u z(u) is invertible with continuous inverse, and Z is the preimage of the closed set V under7→ the inverse z u(z). Thus Z is closed. If you don’t like this 7→ explanation, you can show using (4.16) that if zn = z(un) V converges to ξ H, then un ∈ ∈ must converge to some u V and it must be ξ = z(u). Now let us show that Z ⊥ = 0. This is obvious, since ∈

2 (v, z) = 0 z Z = 0 = (v, z(v)) = B(v, v) δ v . ∀ ∈ ⇒ ≥ k k To conclude the proof, given any λ linear continuous functional we may ﬁnd zλ V such ∈ that (v, zλ) = λ(v) for all v. But then there exists uλ such that z(uλ) = zλ, namely

B(v, uλ) = (v, zλ) = λ(v) v V ∀ ∈ 1 1 and furthermore by (4.16) uλ δ− zλ = δ− λ . This completes the proof of part 1. Part 2 is a straightforwardk k application ≤ k k of Partk 1.k If u, Φ D we can multiply by φ the equation and integrate on (0, 1) to obtain ∈ Z 1 Z 1 Z 1 φ0(x)u0(x) dx + β φ(x)u(x) dx = φ(x)f (x) dx 0 0 0 Namely

B(φ, u) = (φ, f )H , B(φ, u) = (φ, u)V + β(φ, u)H . Since both sides are continuous with respect to φ in the topology of V , we obtain that the above holds if and only if

B(v, u) = (v, f )H v V. ∀ ∈ This is the weak formulation of the problem. It is straightforward to prove that B and the right hand side fulﬁll the assumption of part 1.

4.6. Weierstrass’theorem yields that ﬁnite linear combinations of ek : k Z are dense in ([0, 1]) in the norm. We are on a bounded interval, thus { norm∈ dominates} the C k·k∞ k·k∞ 2 norm and therefore ﬁnite linear combinations of ek : k Z are dense in ([0, 1]) in k · k 2 { ∈ } C the 2 norm as well. But ([0, 1]) is dense in L (0, 1). This completes the exercise. k · k C 4.7. Consider the linear functional Z 2 f L (J) Λ(f ) = f dµ ∈ 7→ J By Hölder, 1 f J 2 f 2 Λ( ) L (J), | | ≤ | | k k 58 4. ELEMENTARY HILBERT SPACE THEORY

f 2 M namely Λ is a linear continuous functional on L (J). It thus follows that , the nullspace 2 k k of Λ, is a closed linear subspace of L (J). Thus f = PM f + PM f for all f f L2(J). Since for such f ⊥ ∈ k k Z Z 1J 1J f = f f dµ + f dµ − J J J J | | 1J R | | and the left summand belongs to M, PM f f dµ. In other words, M span 1J . ⊥ = J J ⊥ = | | { } 4.8. A perusal of the proof of Theorem 2.11.2 yields that the linear span of indicator func- 2 tions of dyadic subintervals of J is dense in L (J). I omit the details (but make sure you can ﬁll in). Lemma 4.6.4 then implies 2 span B(J) = span 1I : I , I J = L (J). { ∈ D ⊂ } It follows, by applying the previous problem, that (4.17) hI : I , I J ⊥ = span 1J . { ∈ D ⊂ } { } Now onto the second part. We will succeed if we show that 2 f L (R), (f , hI ) = 0 I = f = 0. ∈ ∀ ∈ D ⇒ It will sufﬁce to show that the left side of the above implies that f is constant a.e. on both ( , 0) and [0, ), since then both constants need to be zero. But this follows since (4.17) N yields−∞ that f is constant∞ a.e. on every dyadic interval, and dyadic intervals [0, 2 ), N N exhaust [0, ) (say). ∈ ∞ CHAPTER 5

Complex measures and the Radon-Nykodym theorem

5.1 Motivation. The Vitali-Cantor function. This chapter is the first step towards a differ- d entiation theory for (at the very least) the Lebesgue integral on R and R . Let us start with a simple example. If f is a continuous function on [0, 1] we may define its integral function by Z x Z F x f t t f ( ) = ( ) d = 1[0,x) dµ. 0 respectively in Riemann and Lebesgue integral notation: Riemann and Lebesgue integral coincide for continuous functions. The first fundamental theorem of Calculus tells us that F(x + h) F(x) F 0(x) := lim = f (x) x (0, 1). h 0 h − → ∀ ∈ Conversely, if F is continuously differentiable on (0, 1), then Z F x F F ( ) (0) = 01[0,x) dµ. − 1 Both statements fail badly when the assumption of f continuous is replaced by f L (0, 1). ∈ Here is a telling example. For n = 0, 1, . . . let En be the n-th iteration of the Cantor set, namely, 2n [ En = In,j j=1 n where the union is disjoint and In,j is a suitable closed interval of length 2− . Notice that I E I I f n,j n+1 = n+1,2j 1 n+1,2j. Define the Riemann integrable function n : [0, 1] C ∩ − ∪ → 2n x 3n+1 X Z f 1 , F x f t dt. n = 2 In,j n( ) = n( ) j=1 0

It is easy to see that Fn is a continuous function, Fn(0) = 0, Fn(1) = 1 and Fn is constant on the complement of En. The easiest way to see that Fn converges uniformly on [0, 1] to F g f f a continuous function on [0, 1] is the following. Notice that n = n+1 n is zero unless x En, and there holds that − ∈ n 1 2 + f x f x 1 x 1 1 x 1 x 2 1 x 3 n+1( ) n( ) In,j ( ) = 3 In+1,2j 1 ( ) + In+1,2j 1 ( ) 3 Jn,j ( ) = − − − − where Jn,j is the middle third of In,j. In particular, since the above has mean zero on each

In,j, Fn+1 = Fn outside En. Using the above display, one has 1 sup Fn 1 x Fn x + ( ) ( ) n+1 x [0,1] | − | ≤ 3 2 ∈ · 59 60 5. COMPLEX MEASURES AND THE RADON-NYKODYM THEOREM

whence, by standard arguments, Fn is uniformly Cauchy and converges uniformly to a continuous function F : [0, 1] R with F(0) = 0, F(1) = 1. Moreover, F is constant on the → S complement of the Cantor set E = En. It follows that F(x + h) F(x) F 0(x) = lim = 0 h 0 h − → for all x E (and therefore, F’=0 almost everywhere). Thus, the second fundamental theorem of6∈ Calculus fails for F. 1 Here, what happens is that the limit of fn should not be understood as an L function 1 (fn 0 pointwise a.e. and in L ), but rather as a measure which is concentrated on the Cantor→ set, and thus is singular with respect to Lebesgue measure. The theory developed in

the upcoming sections will allow us to correctly characterize the limit of fn. 5.2 Complex measures. Let be a σ-algebra on a set X . If E , and F ∈ F [∞ E = Ej, Ej j, j = k = Ej Ek = j=1 ∈ F ∀ 6 ⇒ ∩ ;

then Ej : j N is called a measurable partition of the set E.A complex measure is a map { ∈ } ν : C with the property that, whenever E and Ej : j N is a measurable partitionF 7→ of E, ∈ F { ∈ }

X∞ (5.1) ν(E) = ν(Ej). j=1 Notice that the convergence of the series (of complex numbers) is a part of the requirement, unlike the case of (positive) measure where ν is allowed to take the value .

1 ∞ 5.2.1 EXAMPLE. Let (X , , µ) be a measure space and f L (X , , µ). Let F Z ∈ F νf (E) = f dµ, A . E ∈ F

Exercise 6 from Chapter 3 yields exactly that νf is a complex measure on . We observe F for further use that the same statement yields ν f is a ﬁnite positive measure and νf (E) | | | | ≤ ν f (E) for all E . | | ∈ F The second part of the above example suggests that, given a complex measure ν on , one should look for a positive ﬁnite measure λ on such that ν(E) λ(E) for all A F . Such a measure would need to satisfy F | | ≤ ∈ F

X∞ X∞ λ(E) = λ(Ej) ν(Ej) j=1 ≥ j=1 | |

whenever E and Ej : j N is a measurable partition of E. Thus, we deﬁne the map ∈ F { ∈ } X∞ (5.2) ν : [0, ], ν (E) := sup ν(Ej) | | F → ∞ | | Ej meas. part. of E j 1 | | { } = which is the least possible such measure λ and is termed the total variation of ν. It should be pretty obvious that if ν is a complex measure taking nonnegative values, then ν = ν. | | 5. COMPLEX MEASURES AND THE RADON-NYKODYM THEOREM 61

5.2.2 LEMMA. Let ν be a complex measure on . Then the total variation ν of (5.2) is a finite positive measure on . F | | F PROOF. Let us first prove that ν is a measure. Let Ej , F` be any two measurable | | { } { } partitions of E. Then F` Ej : j N is a measurable partition of F` and, viceversa F` { ∩ ∈ } { ∩ Ej : ` N is a measurable partition of Ej. Thus, with the first equality justified by the convergence∈ } of the sums, there holds

X∞ X∞ X∞ X∞ X∞ X∞ X∞ X∞ ν(F`) = ν(F` Ej) ν(F` Ej) = ν(F` Ej) ν (Ej). `=1 | | `=1 j=1 ∩ ≤ `=1 j=1 | ∩ | j=1 `=1 | ∩ | ≤ j=1 | |

Taking the supremum over F` in the left hand side we obtain { } X∞ (5.3) ν (E) ν (Ej). | | ≤ j=1 | |

For the opposite inequality, let Aj,` : ` N be a measurable partition of each Ej such that { ∈ } X∞ j ν (Ej) "2− + ν(Aj,`) . | | ≤ `=1 | |

Clearly Aj,` : j, ` N is a measurable partition of E. Thus { ∈ } X∞ X∞ X∞ (5.4) ν(Ej) " + ν(Aj,`) " + ν (E) + " j=1 | | ≤ j=1 `=1 | | ≤ | | and " > 0 is arbitrary. Combining (5.3) with (5.4) yields countable additivity of ν . Let us move to ν being a ﬁnite measure. We will need the following Sublemma,| | whose proof is left as an exercise.| |

5.2.3 SUBLEMMA. Let z1,..., zN be complex numbers. Then there is a subset of indices S 1, . . . , N such that ⊂ { } N X 1 X z z . j π j j S ≥ j=1 | | ∈ Suppose ﬁrst that there exists a measurable set E with ν (E) = . Then there must | | ∞ exist a partition Ej : j T of E and a ﬁnite N such that { ∈ } N X ν(Ej) > π(1 + ν(E) ). j=1 | | | | Using the sublemma, choose a subcollection of indices S 1, . . . , N such that ⊂ { }

X [ (1 + ν(E) ) ν(Ej) = ν(A), A = Ej. | | ≤ j S | j S ∈ ∈ c Let B = E A . It must be by ﬁnite additivity of ν that ∩ ν(B) = ν(E) ν(A) ν(E) ν(A) 1 | | | − | ≥ | | − | | ≥ 62 5. COMPLEX MEASURES AND THE RADON-NYKODYM THEOREM

So we have found two disjoint measurable sets A, B such that ν(A) , ν(B) 1 and A B = E, whence, without loss of generality ν (A) = . | | | | ≥ ∪ We now show that if ν (X ) =| | we reach∞ a contradiction. By repeating the above procedure, replacing E with| | A each∞ time, we construct a sequence Bj of pairwise disjoint sets with ν(Bj) 1. But this contradicts the convergence of the series | | ≥ [∞ X∞ ν Bj = ν(Bj), j=1 j=1 which completes the proof. 5.2.4 REMARK. It is easy to see, and we leave the proof as an exercise, that the space

M(X , ) : ν : C complex measures F { F → } is a linear space and ν X = ν(X ) k k 1 is a norm on M(X , ). Furthermore, if µ is a positive measure on and f L (X , , µ), F F ∈ F then νf = ν f and νf = f 1. | | | | k k k k Let ν be a complex measure on taking real values. We can decompose F ν ν ν ν+ ν , ν : , = + − ± = | |2 ± such that the measures ν± are ﬁnite positive measures on . This decomposition is referred to as the Jordan decomposition of ν. F

5.3 Absolute continuity. Let (X , , µ) be a measure space and ν be an arbitrary (complex or positive) measure on . We sayF that ν is absolutely continuous with respect to µ, and write ν Î µ, if F (5.5) E , µ(E) = 0 = ν(E) = 0. ∈ F ⇒ It is immediate to see that if ν is a complex measure then ν Î ν . A measure ν (positive or complex) on is said to be concentrated on the set A if | | F ∈ F ν(A B) = ν(B) B . ∩ ∀ ∈ F If ν1, ν2 are two measures on such that νj is concentrated on Aj and A1, A2 partition X , then we say that ν1, ν2 are mutuallyF singular and write ν1 ν2. It is not difﬁcult to ﬁnd examples for each property, thus⊥ we leave these as an exercise, as well as the following properties.

5.3.1 PROPOSITION. Let µ be a positive measure, ν, ν1, ν2 be either positive or complex measure on the same σ-algebra . Then if ν is concentratedF on A, then ν is concentrated on A as well;

• if ν1 ν2 then ν1 ν2 ; | | • ⊥ | | ⊥ | | if ν1 µ, ν2 µ then ν1 + ν2 µ; • ⊥ ⊥ ⊥ if ν1 Î µ, ν2 Î µ then ν1 + ν2 Î µ; • if ν1 Î µ and ν2 µ then ν1 ν2; • ν Î µ, ν µ if and⊥ only if ν =⊥0. • ⊥ 5. COMPLEX MEASURES AND THE RADON-NYKODYM THEOREM 63

PROOF. Exercise. We will need the following absolute continuity lemma for σ-ﬁnite measures.

5.3.2 LEMMA. Let (X , , µ) be a measure space and µ be a σ-finite measure. Then there exists a finite positive measureF ν on such that F ν Î µ, µ Î ν and ν takes the form Z ν(A) = w dµ, A A ∀ ∈ F 1 for some w L (X , , µ) with 0 < w < 1. ∈ F S PROOF. By σ-finiteness, X X with µ X < . The function = n N n ( n) ∈ ∞ X 2 n w x ∞ − 1 ( ) = Xn 1 µ X n n=1 + ( ) has the required properties. 5.4 The Lebesgue-Radon-Nykodym theorem. The following is the most important theorem in measure theory. We present two versions, the second of which will be obtained as a corollary of the first.

5.4.1 LEBESGUE-RADON-NYKODYM THEOREMFORCOMPLEXMEASURES. Let (X , , µ) be a measure space and assume that µ is a σ-ﬁnite measure. Let ν : C be a complexF measure. Then there exists a unique decomposition F →

ν = νa + νs, νa Î µ, νs µ ⊥ and if ν is a positive ﬁnite measure, so are νa, νs. Furthermore, there exists a unique f 1 L (X , , µ) such that ∈ F Z νa(A) = f dµ, A and f is termed the Radon-Nykodym derivative of ν with respect to µ.

5.4.2 LEBESGUE-RADON-NYKODYM THEOREMFORPOSITIVEMEASURES. Let (X , , µ) be a measure space and assume that µ is a σ-ﬁnite measure. Let ν : C be a σ-ﬁniteF positive measure. Then there exists a unique decomposition F →

ν = νa + νs, νa Î µ, νs µ ⊥ Furthermore, there exists a unique positive measurable f such that Z νa(A) = f dµ, A and f is termed the Radon-Nykodym derivative of ν with respect to µ. 64 5. COMPLEX MEASURES AND THE RADON-NYKODYM THEOREM

Before the proof, we observe that the second theorem may fail in case ν is not assumed to be σ-ﬁnite. Indeed, let µ be the Lebesgue measure on (0, 1) and ν be the counting measure on the Lebesgue σ-algebra. Assuming that ν = νa + νs is a Lebesgue decomposition of ν with respect to µ leads to a contradiction. Suppose that (0, 1) = A B with A B = and ∪ ∩ ; νs(E) = νs(E B) = . It follows that for all E ﬁnite ∩ ; νs(E B) ν(E) = νa(E) + νs(E B), ∩ ≤ ∩ but νa(E) = . Then B = (0, 1), which clearly clashes with µ( ) = µ( A). We are now ready for the; proofs. · · ∩

PROOF THAT THEOREM 5.4.1 = THEOREM 5.4.2. Let X n be a pairwise disjoint sequence ⇒ n of measurable subsets whose union is X and such that the restriction ν ( ) = ν( X n) is a ﬁnite measure. Applying Theorem 5.4.1 for each n we obtain the decomposition· · ∩ Z n n n n ν(A X n) = ν (A) = νa(A) + νs (A), νa(A) = fn dµ. ∩ A for some nonnegative f L1 . We can replace f with f 1 without perturbing the last n (µ) n n Xn equality. Then ∈

X∞ n νs(A) := νs (A) n=1 is a positive measure which is singular with respect to µ. To see this, observe that νn is S s concentrated on Bn X n and µ(Bn) = 0, so νs is concentrated on B = Bn which has ⊂ n µ(B) = 0. Similarly µ is concentrated on An X n and ν (An) = 0, which yields the claim. Deﬁne now ∩

X∞ f = fn n=1 Then by the monotone convergence theorem Z Z X n X νa(A) := νa(A) = fn dµ = f dµ n n A Xn A ∩ and it is obvious that νa Î µ. This completes the proof.

PROOFOF THEOREM 5.4.1. We first reduce to the case where ν is a positive finite measure. If ν is complex, write ν = ν1 +iν2 with νj real and apply the statement to each, taking advantage of Proposition 5.3.1. If ν is real, apply the statement to ν± and take advantage of Proposition 5.3.1. The core case is when ν is a finite positive measure. Let us also associate to µ a function 0 < w < 1 as in Lemma 5.3.2, such that wdµ is a finite measure. Let us define the further finite measure dφ = dν + wdµ. The functional Z 2 f L (φ) I(f ) := f dν ∈ 7→ 5. COMPLEX MEASURES AND THE RADON-NYKODYM THEOREM 65 is (clearly) linear and by Hölder

1 Z Z Z 2 Æ 2 I(f ) f dν f dφ φ(X ) f dφ . | | ≤ | | ≤ | | ≤ | | 2 By the Riesz representation, there exists g L (φ) such that Z ∈ Z (5.6) f dν = f g dφ

2 2 for all f L (φ). Let now E be a measurable set with φ(E) > 0, then 1E L∞(φ) L (φ) and first∈ equality in (5.6) above reads ∈ ⊂ Z Z ν(E) 1 1 (5.7) 0 = 1E dν = g dφ 1 ≤ φ(E) φ(E) φ(E) E ≤ Then it must be 0 g 1 a.e. φ. We can redefine g on a null set of φ (and thus of µ so that 0 g 1 everywhere.≤ ≤ Define ≤ ≤ A = x X : 0 g(x) < 1 , B = x X : g(x) = 1 , νa( ) = ν( A), νs( ) = ν( B) { ∈ ≤ } { ∈ } · · ∩ · · ∩ which yields a decomposition ν = νa + νs. Now (5.6) is the exact same as Z Z 2 (5.8) f (1 g) dν = f g wdµ f L (φ) − ∀ ∈ Now νs is concentrated on B, while for f = 1B the above reads 0 = (wdµ)(B), that is n µ(B) = 0. Let fn = (1 + g + ... + g ) L∞(φ) and test now (5.8) for f = fn1E, it yields Z ∈ Z Z n (1 g ) dν = fn gw dµ = fn gw1A dµ E − E E n where the second equality is because µ(B) = 0 But (1 g ) increases monotonically to 1A, while the integrand on the right hand side increases to− the nonnegative function g f : w1 = 1 g A By passing to the limit on both sides − Z νa(E) = ν(E A) = f , dµ ∩ E 1 It is then obvious that νa Î µ and since νa(X ) < , f L (µ). This completes the proof. ∞ ∈ 5.5 Corollaries of the Radon-Nykodym theorem.

5.5.1 POLARDECOMPOSITIONLEMMA. Let ν be a complex measure on (X ). Then there exists a measurable function α : X C such that F ⊂ P → α(x) = 1 x X , dν = αd ν . | | ∀ ∈ | | We leave the proof as an exercise.

1 5.5.2 COROLLARY. Let (X , , µ) be a measure space and f L . Then νf = ν f . F ∈ | | | | 66 5. COMPLEX MEASURES AND THE RADON-NYKODYM THEOREM

PROOF. By the polar decomposition lemma, we find a unimodular measurable function α such that αd νf = dνf = f dµ = d νf = αf dµ | | ⇒ | | but then αf 0 a.e, whence αf = f a.e, whence νf = ν f . ≥ | | | | | | 5.5.3 HAHNDECOMPOSITIONTHEOREM. Let ν be a real valued complex measure on . Then, F there exist measurable sets A± such that + + A A− = X , A A− = , µ±( ) = µ( A±). ∪ ∩ ; · · ∩ PROOF. Let αdν = d ν be the polar decomposition of ν. Since ν is real valued, also α is. Noticing | | 1 α dν± = ± d ν 2 | | 1 α and that ± 1A , where A x : α x 1 , we are done. 2 = ± ± = ( ) = p { ± } 5.5.4 THE L RIESZ REPRESENTATION THEOREM. Let (X , , µ) be a measure space and µ be a p σ-finite measure. Let 1 p < . For all linear boundedF functionals Λ on L (µ) there exists p 0 ≤ ∞ a unique g L (µ) such that Λ = Λg . ∈ PROOF. We prove the case where µ(X ) < and leave the σ-finite case as an exercise. p Fix a linear bounded functional Λ on L (µ). We∞ can normalize Λ = 1 and also µ(X ) = 1. Define the set function k k A ν(A) = Λ(1A) 1 ∈ F 7→ p First of all ν(A) = Λ(1A) 1A p = µ(A) 1. So ν takes values in the unit disk. If | | | | ≤ k k ≤ A, B are disjoint measurable sets, it is evident that ν(A B) = ν(A) + ν(B). Now if Ej is a measurable partition of E, and A E E , 1∪ 1 in Lp by the dominated n = 1 n An E convergence theorem (dominant is 1X ). Thus∪ by· · · continuity∪ → and finite additivity n X X∞ ν E lim Λ 1A lim ν En ν En . ( ) = n ( n ) = n ( ) = ( ) →∞ →∞ j=1 j=1 Summarizing, ν is a complex measure on which is clearly absolutely continuous with 1 respect to µ. By the Radon-Nykodym theorem,F there exists g L (µ) such that for all ∈ E , if s = 1E ∈ F Z Z Λ(s) = ν(E) = g dµ = sg dµ E This equality carries to simple functions s by linearity, and by density we obtain Z p (5.9) Λ(f ) = f g dµ f L (µ). ∀ ∈ p Since Λ(f ) f p the fact that g L 0 (µ) and g = 1 follows from Exercise 3.5, part a. This concludes| | ≤ k thek proof of the theorem∈ in the casek k when µ(X ) < ∞ 5. COMPLEX MEASURES AND THE RADON-NYKODYM THEOREM 67 Chapter 5 Exercises

5.1. Prove Sublemma 5.2.3 5.2. Let ν be a complex measure on . Prove that F X ν (E) := sup ν(Ej) | | | | where the supremum is taken over all ﬁnite Ej : j = 1, . . . , N measurable partitions of E. { } 5.3. Let µ be a positive measure, ν be a complex measure. Prove that, referring to (5.10) ν Î µ if and only if

(5.10) " > 0 δ = δ(") > 0 : E , µ(E) < δ = ν(E) < ". ∀ ∃ ∀ ∈ F ⇒ | | Show that the above equivalence fails in general if ν is a positive measure. 5.4. Prove Lemma 5.5.1. 5.5. Complete the proof of Theorem 5.5.4 in the σ-finite case. Furthermore, show that if 1 < p < , the restriction that µ is σ-finite can be removed. ∞ Hint. For the second part, for E X measurable and σ-finite, consider the functional ΛE( ) = Λ( 1E). ⊂ p · · By the first part, there exists gE L 0 (E, E, µ) such that Z ∈ F | p Λ(f ) = f gE dµ f L (E, E, µ), f 1E = 0. ∀ ∈ F |

From the first part you should inherit that if E E0 then gE = gE0 1E Note that ⊂ sup gE p = t Λ , k k 0 ≤ k k supremum being taken over σ-finite sets. Prove that the supremum is attained on a σ-finite set F p and that for all E F σ-finite it must be gE F = 0. Now use that when f L , and p > 1, the set x : f (x) = 0 is σ⊃-finite. \ ∈ { 6 } 5.6 The conditional expectation. This exercise involves the concept of conditional expectation. Let (X , , µ) be a σ-finite measure space and 1 be a σ-algebra. Let 1 F F ⊂ F f L (X , , µ) and define a set function (X , 1) by ∈ F FZ ν : 1 C, ν(A) = f dµ, A 1. F → A ∀ ∈ F 1 It is easy to see, by countable additivity of the integral, and by f L (X , , µ), that ν is a complex measure on 1. It is also easy to see that ν µ, where∈ we keepF calling µ the F 1 1 restriction to 1. By the Radon-Nykodym theorem, there exists h L (X , 1, µ) such that F Z ∈ F ν(A) = h dµ, A 1. A ∀ ∈ F

We write h := E(f 1), the conditional expectation of f with respect to 1. |F F 1 Note in particular that h is 1-measurable! F 68 5. COMPLEX MEASURES AND THE RADON-NYKODYM THEOREM

1. Explain why Z Z E(f 1) dµ = f dµ A 1. A |F A ∀ ∈ F Explain why the map f E(f 1) is linear. 2. Prove that if 2 1 7→ |F F ⊂ F ⊂ F E E(f 2) 1) = E E(f 1) 2) = E(f 2) |F 1|F |F |F |F 2 3. Prove that if f , g, f g L (µ) and g is 1-measurable, then E(f g 1) = gE(f 1) ∈ F |F |F 5.7 Discrete martingales. Let be the collection of dyadic intervals of R and for n Z n D 3 ∈ n = I : `(I) = 2− . Consider the increasing sequence of σ-algebras D { ∈ D } n = Σ( n), n n+1. G D G ⊂ G 1 These are all sub-σ-algebras of the Lebesgue σ-algebra (R). Let now f L (R), and L ∈ deﬁne fn := E(f n). 1. Prove that|G lim fn(x) = 0 a.e. x R. n ∈ →−∞ f d f f 2. Find an explicit formula for n. Compute n+1 = n+1 n. What you ﬁnd should be useful in the next step. − 1 2 3. Assume now f L (R) L (R). Prove ∈ ∩ lim fn = 0, lim fn = f n n + →−∞ → ∞ 2 4 pointwise almost everywhere and L (R). 4. EXTRA CREDIT. Show that the convergence in the above display also holds in 1 L (R).

2Hint: simple functions. 3 It can be useful to observe that, since n is countable, A n if and only if A is a countable union of intervals of n. D ∈ D 4 D Hint. Your dn+1 should suggest an application of Plancherel. 5. COMPLEX MEASURES AND THE RADON-NYKODYM THEOREM 69

Solutions to selected Exercises of Chapter 5. 1 5.4. Since ν ν , there exists h L ( ν ) such that | | ∈ Z| | ν(A) = h d ν A . A | | ∀ ∈ F We need to show h = 1 almost everywhere (wrt ν ). First of all, for all sets B | | |Z | ∈ F ν(B) 1 1 | | = h(x)d ν ≥ ν (B) ν (B) B | | | | | | which implies that h 1 almost everywhere. Let now Br = x : h(x) < r . Let Aj be a measurable partition| | of ≤Br , we have { | | }

Z Z X X X ν(Aj) = h(x)d ν r d ν = rν(Br ). | | Aj | | ≤ Aj | |

Taking the supremum on the left hand side we get µ(Br ) rν(Br ) which can only be true if ≤ µ(Br ) = 0 when 0 < r < 1. Thus h 1 almost everywhere, which completes the solution. | | ≥ 5.5. (part of it is due to Ankan Ganguly) We first complete the proof of the σ-finite measure case. Let X n be a sequence of disjoint measurable sets of finite measure whose union is X . p p Given the functional Λ L (X , µ), denote its compression to L (X n, µ), namely Λn(f ) := f 1 for f Lp X , ∈ . This is a linear bounded functional on the finite measure space Λ( Xn ) ( n µ) ∈ p (X n, µ). We obtain gn L 0 (X n, µ) such that Z∈ f g d f f 1 f Lp X , . n µ = Λn( ) = Λ( Xn ) ( n µ) ∀ ∈ Now notice that if f Lp X , , P f 1 converges pointwise a.e. to f and in Lp X , by ( µ) n Xn ( µ) p ∈ P the L dominated convergence theorem (dominant is f ). Define g = n gn. Notice that due to the disjoint supports of gn, pointwise convergence of the above series holds trivially. p If p = 1, then g < . We prove the same bound for p > 1. For every f L we can find a measurablek k∞ sequence∞ of unimodular functions such that f 1 g ∈ f g . αn Xn αn n = n Notice that the same argument as above yields that P f 1 converges in Lp X , to| some|| | Xn αn ( µ) p F L . and F p = f p It follows by continuity that ∈ k k k k X X Z Z f F f 1 f g d f g d . Λ p Λ( ) = Λ( αn Xn ) = n µ = µ k kk k ≥ n n | || | | || |

p p0 By taking f = g 1 g N , and using the monotone convergence theorem, we obtain that p | | | |≤ g L 0 (µ) and g p Λ . The fact that Λ = Λg now follows trivially by a similar argument as∈ the one givenk above.k 0 ≤ k k Now suppose that 1 < p < and µ is not necessarily σ-finite. Suppose E X is ∞ ⊂ measurable and σ-finite. Then as we did in the σ-finite case, define ΛE(f ) = Λ(f 1E). By p the first part of the problem, there exists a unique gE L 0 (E, µ) such that ∈ Z ΛE(f ) = f gE dµ 70 5. COMPLEX MEASURES AND THE RADON-NYKODYM THEOREM

p Now, suppose E E0 where E0 is also measurable σ-ﬁnite set. Then for all f L (E0, µ), ⊂ ∈ Z Λ f 1E f gE 1E dµ ( ) = 0

So by uniqueness, gE E = gE. Therefore, gE p gE p . Now, notice that for all σ-ﬁnite E, 0 | k 0 k 0 ≥ k k 0

f gE dµ Λ(f ) Λ f p ≤ | | ≤ k kk k So, by problem 5(a) of problem set 3, (and this is where we use that 1 < p < ) ∞

gE p Λ k k 0 ≤ k k So, the supremum over σ-finite sets is bounded. Suppose sup gE p T. Then for any 0 = ε > 0, there exists a σ-finite E such that gE p > T ε. By takingk unions,k we can construct 0 k k − k S a sequence of sigma-finite sets Ek with gE p > T 2 and Ek Ek 1. Let F ∞ Ek. k 0 − + = k=1 Then F is σ-finite and gF p T. Supposek k G −F. Then as⊂ before, gG F gF . But, 0 = = gG p gF p . So, k k ⊃ | k k 0 ≤ k k 0 Z p gG1G F 0 dµ = 0 | \ | p So gG G F = 0 a.e. In other words gN = gF 1N a.e. for all sigma finite N. Now, if f L , the set | \ R ∈ R N = µ( x X : f (x) > 0 is σ-finite. Therefore Λ(f ) = Λ(f 1N ) = f gN dµ = f gF dµ, where uniqueness{ ∈ | is given| } by the σ-finite case. 5.6. The equality Z Z E(f 1) dµ = f dµ A 1 A |F A ∀ ∈ F is simply the defining property of the fact that E(f 1) is the Radon-Nykodym derivative of νf restricted to 1 with respect to µ restricted to|F 1. Linearity follows simply because F F νf +αg = νf + ανg , whence Z Z f g A A A f g E( + α 1) dµ = νf +αg ( ) = νf ( ) + ανg ( ) = E( 1) + αE( 1) dµ A |F A |F |F

Let h1 = Re E(f + αg 1), h2 = Re (E(f 1) + αE(g 1)), then A± = (h1 h2) > 0 is |F |F R h |Fh {± − } 1 measurable and, from the above equality A 1 2 dµ = 0. Arguing similarly for the F ± ± − imaginary parts it follows E(f + αg 1) = (E(f 1) + αE(g 1)) a.e. |F |F |F We have E(f 2) = E E(f 2) 1) since the former is 1-measurable. Now if A 2, A 1 as well and|F we have the|F equality|F F ∈ F ∈ FZ Z Z Z E(f 2) dµ = f dµ = E(f 1) dµ = E E(f 1) 2 dµ A 2 A |F A A |F A |F |F ∀ ∈ F whence E(f 2) = E E(f 1) 2 . |F |F |F 5. COMPLEX MEASURES AND THE RADON-NYKODYM THEOREM 71

First of all, we can assume f , g 0 by linearity. Now, if g = 1B with B 1 and A 1 Z Z ≥ Z Z ∈ F ∈ F f g dµ = f dµ = E(f 1) dµ = E(f 1)g dµ A A B A B |F A |F ∩ ∩ Ok, then E(f g 1) = gE(f 1) for indicator, and thus for simple functions by linearity. Now if sn is a sequence|F of 1|F-measurable functions increasing to g and A 1 Z FZ Z Z ∈ F f g dµ = lim f sn dµ = lim E(f sn 1) dµ = lim snE(f 1) dµ A A A |F A |F Z = lim gE(f 1) dµ A |F by two applications of the monotone convergence theorem. This completes the proof.

n 5.7. Since fn := E(f n) needs to be constant on the dyadic cubes with `(I) = 2− , we guess that |G X fn = f I 1I I n〈 〉 ∈D A R f R f It is trivial to verify that for all n, A = A n. Let us prove that fn 0 almost∈ G everywhere as n . We are free to assume that f and thus fn are nonnegative.→ Suppose the convergence→ −∞ does not hold, then there exists " > 0 such that µ fn > " > 0 for a subsequence nj . But then f I > " for at least ( j ) nj one I since f is constant on these intervals.→ It −∞ follows that 〈 〉 nj nj nj ∈ D Z Z f d f d I µ µ > " nj I ≤ nj | |

but the right hand side goes to as nj . Contradiction. Let ∞ → −∞ 1 h 1 1 I = ( I` Ir ) pI − An easy computation yields f 1 f 1 f 1 f , h h I` I` + Ir Ir I I = ( I ) I 〈 〉 〈 〉 − 〈 〉 whence X d f f f h h n+1 = n+1 n = ( , I ) I . − I n 2 2 ∈D Since f L , we have that (dn, dm) = dn 2δnm by orthonormality of hI . The series below converge∈ unconditionally and we havek thek equality

X 2 X X 2 2 dn 2 = (f , hI ) = f 2. n k k n I Dn | | k k ∈ From this, we obtain for n m ≤ 2 m m X X f f 2 d d 2. m n 2 = k = k 2 k − k k=n+1 2 k=n+1 k k 72 5. COMPLEX MEASURES AND THE RADON-NYKODYM THEOREM

2 Since the last sum is the tail of a convergent sum, fn is Cauchy and thus converges in L to some g L2. Fixing m, we get m ∈ X g = fm dk − k= ∞ m for all m. But the right hand side has zero inner product with all hI with `(I) < 2− . Letting m , we obtain that (g, hI ) = 0 for all I. Thus g = 0. Ok, now → −∞ m X fm = dk k= ∞ 2 and it is really the same argument to show that fm f in L as m . Pointwise convergence a.e. follows from the uniqueness of limit (for→ instance). → ∞ CHAPTER 6

Derivatives of measures and differentiation theorems

6.1 Motivation. Let µ denote the Lebesgue measure on R for this paragraph. Let ν be a complex measure on (R), and consider the integral function L F(x) = ν( , x]). 1 −∞ In the case where ν = νf for some f L (R) this is the usual integral function. Notice that ∈ νf µ In this chapter, we investigate the question of establishing for which x R the limit ∈ F(x + h) F(x) F 0(x) = lim h 0 h − → exists, and in this case whether F 0(x) = f (x) holds (for ν = νf ), or if more generally F 0(x) coincides with the Radon-Nykodym derivative of the absolutely continuous part of ν. A slightly more general formulation of this question is the following. Let µ be now d d d Lebesgue measure on R . Fix a point x R and let ν be a complex measure on (R ). Does there exist A C such that ∈ L ∈ ν I (6.1) lim ( n) A n = µ In →∞ ( ) d for all sequences of boxes In R such that x In for all n and `(In) 0 as n . Such ⊂ d ∈ → → ∞ a sequence In is said to shrink to x R . To recover{ } the previous question,∈ just take any sequence hn 0 and let In to be the → intervals (x, x + hn] for hn > 0, or alternatively [x + hn, x) when hn < 0. 6.2 The dyadic Lebesgue differentiation theorem. In this paragraph, we deal with the version of (6.1) where the intervals In are restricted to coming from , the standard dyadic d d grid on R . Throughout this section, µ will be the Lebesgue measureD on R . For a complex d measure ν be on (R ), we deﬁne L ν(Q) d ν Q = , Q cube of R . 〈 〉 µ(Q) ∀

If ν = νf this is simply the average of f on the cube Q; and the same should be thought of the complex measure case. Let us recall the notation Q n for the collection of dyadic d n ∈ D d cubes of R with sidelength `(Q) = 2− . Recall also that the cubes of n partition R . Deﬁne X D En(ν)(x) = ν Q1Q(x), Q n〈 〉 ∈D 1 d When ν = νf this coincides with the conditional expectation of f L (R ) with respect ∈ to the σ-algebra n = Σ( n), see Exercises 5.6 and 5.7. It is clear that En(ν) is Lebesgue G D 73 74 6. DERIVATIVES OF MEASURES AND DIFFERENTIATION THEOREMS

measurable. It follows from the conditional expectation property or from a simple computation that Z d En(ν) dµ = ν(R ).

Let us deﬁne the dyadic maximal operator

M ν = sup En( ν ) = sup ν Q. D n Z | | Q x 〈| |〉 ∈ Q3 ∈D For the remainder of the paragraph, we keep writing M in place of M . Then M ν is a measurable function (even when ν is a complex measure). Let us stressD that theD total variation ν (or simply f when ν = νf ) appear in the deﬁnition. We have the following | | | | 6.2.1 DYADIC MAXIMAL THEOREM. There holds d sup µ x R : M ν(x) > λ ν . λ>0 ∈ D ≤ k k d PROOF. Fix λ > 0 and call Aλ = x R : M ν(x) > λ . If x Aλ, there must be a D dyadic cube Q x x such that { ∈ } ∈ 3 (6.2) Q Q . ν Q x > λ λµ( x ) < ν ( x ) 〈| |〉 ≡ | | Let = Q x : x K . Notice that by (6.2) Q { ∈ } 1 1 d sup `(Q) λ− ν (X ) , Q ≤ | | ∈Q so that, if of elements which are maximal with respect to set inclusion Q ⊂ Q [ [ Aλ = Q = Q Q Q ∈Q ∈Q and the last union is disjoint. It follows that X X λµ(Aλ) = λµ(Q) ν (Q) ν (X ), Q ≤ Q | | ≤ | | ∈Q ∈Q and we are done.

We now deal separately with the absolutely continuous part of ν which is νf for f = dνa 1 dµ L (µ), and with the singular part. ∈ 1 d 6.2.2 THE DYADIC LEBESGUEDIFFERENTIATIONTHEOREM. Let f L (R ). Then for almost d every x R ∈ ∈ lim f f x I 0 n ( ) n = →∞〈| − |〉 d whenever In is a sequence of dyadic cubes of R which shrinks to x as n . In particular, → ∞ f lim n f = n E ( ) →∞ d 1 d almost everywhere in x R and in L (R ). ∈ 6. DERIVATIVES OF MEASURES AND DIFFERENTIATION THEOREMS 75

PROOF. Let us first observe how the second part follows from the first. Since fn 1 = f 1 it is actually enough to prove pointwise convergence, as the L1 convergencek followsk fromk k Exercise 3 of the supplementary problems of October 9. Let x be a point for which the first d 1 part holds: almost every x R is such a point and Q x be a dyadic cube. Then ∈ 3 f Q f (x) = f f (x) Q f f (x) Q |〈 〉 − | |〈 − 〉 | ≤ 〈| − |〉 and the right hand side goes to zero when Q shrinks to x. Let us prove the first part. Define

T f (x) = lim sup f f (x) I . I ,I x,`(I) 0〈| − |〉 ∈D 3 → We prove that T f = 0 almost everywhere. Fix λ > 0. Let " > 0 be given. By density of d 1 d d (R ) in L (R ) there exists g (R ) with f g 1 < "λ. Call h = f g. Then one has C ∈ C k − k − f f x g g x h h x h g g x h h x ( ) I = ( ( )) + ( )) Q + ( ) I + I + ( ) 〈| − |〉 − − | |〉 ≤ 〈| − |〉 〈| |〉 | | g g(x) I + Mh + h ≤ 〈| − |〉 | | and taking lim sup on left hand side, lim inf on right hand side and since T g = 0, T f Mh + h , ≤ | | whence, by the maximal theorem h µ( T f > 2λ ) µ( h > λ ) + µ( Mh > λ ) 2k k 2". { } ≤ {| | } {| | } ≤ λ ≤ Letting " 0 we get T f 2λ outside of a null set Eλ. But then T f = 0 outside the null set E .→ This completes≤ the proof of the ﬁrst part and in turn of the theorem. n N 1/n ∪ ∈ Let us now move to the singular part, which is easier.

d 6.2.3 THEOREM. Let ν be a complex measure on (R ). Suppose ν µ. Then for almost d every x R , L ⊥ ∈ lim 0 ν In = n 〈| |〉 →∞ d whenever In is a sequence of dyadic cubes of R which shrinks to x as n . In particular → ∞ lim n ν 0 n E ( ) = →∞ d almost everywhere in x R . ∈ PROOF. The second part is obtained from the first just like in the previous theorem. We can easily reduce to the case where ν is positive finite. Let A be a Lebesgue measurable set with µ(A) = 0 and ν concentrated on A. Fix λ > 0 and let " > 0 be given. Since ν is positive and finite, we may choose a compact set K such that ν(K) > ν "/λ. This can be proved d in the same way we used for the Lebesgue measure on R (Propositionk k − 1.6.2, point 2). Let us decompose c ν = ν1 + ν2, ν1(E) = ν(K E), ν2(E) = ν(K E). ∩ ∩ 1These are called dyadic Lebesgue points. 76 6. DERIVATIVES OF MEASURES AND DIFFERENTIATION THEOREMS

c Notice that ν2 "/λ. Choose a point x K which is not dyadic. Then if In shrinks to x, k k ≤ c ∈ it means that In K when `(In) is small enough, whence ⊂ ν(In) ν(In) ν2(In) Tν(x) := lim sup lim sup = lim sup Mν2(x). n µ I k k µ I k k µ I ( n) ≤ `(In) 2 ( n) ≤ `(In) 2 ( n) ≤ →∞ →∞ ≤ − →∞ ≤ − But then, by the maximal theorem

µ( Tν > λ ) µ(dyadic points) + µ(K) + µ( Mν2(x) > λ ) " { } ≤ { } ≤ Arguing as in the previous proof it follows that T(ν) = 0 almost everywhere, which is our claim. We summarize the ﬁndings of the previous theorems in the following

6.2.4 COROLLARY: THEDYADICFIRSTFUNDAMENTALTHEOREMOFCALCULUS. Let ν be a com- d plex measure on R such that Z ν = νf + νs, νs µ, νf = f dµ ⊥ 1 d d for some f L (R ). Then, for almost every x R and all sequences In of dyadic cubes shrinking to∈ x ∈

lim ν I f x 0. n n ( ) = →∞ |〈 〉 − | 1 d Furthermore En(ν) f pointwise almost everywhere. If νs = 0, then En(ν) f in L (R ). → → 6.2.5 REMARK. Throughout this paragraph, the only point where we used that the intervals over which we are averaging come from a dyadic grid, in particular the standard dyadic grid was in the covering argument within the proof of the maximal theorem. The same argumentD will go through for any choice of dyadic grid. 6.3 Shifted dyadic grids and the three grids lemma. This short paragraph is devoted to a technical tool which will be used to extend the differentiation theorems from the dyadic to the continuous case. We deﬁne three shifted dyadic grids on R, one of which is the standard dyadic grid

¦ n j n j © j = 2− k + 3 , 2− k + 1 + 3 : k, n Z , j 0, 1, 2 D ∈ ∈ { } In other words, j is the dyadic grid obtained by shifting each element I of the standard Dj d ∈~ D dyadic grid by 3 `(I). In dimension d 1, for each of the 3 choices j = (j1,..., jd ) d ≥ d ∈ 0, 1, 2 , we deﬁne the corresponding dyadic grid in R by { } Q I I : ` I ` I , I , ` 1, . . . , d . ~j = = 1 d ( 1) = ( d ) ` j` = D × · · · × ··· ∈ D We call d [ = ~j S ~j 0,1,2 d D ∈{ } d the standard dyadic net of R . 6. DERIVATIVES OF MEASURES AND DIFFERENTIATION THEOREMS 77

d d 6.3.1 THREEGRIDSLEMMA. Let A R be a closed cube. Then there exists a cube Q such that ⊂ ∈ S A Q, `(Q) 6`(A). ⊂ ≤ In fact Q may be chosen such that `(Q) is the least dyadic integer larger than 3`(A). PROOF. It sufﬁces to prove the d = 1 case. There exists a unique n Z such that n ∈ 3`(A) < 2− 6`(A). ≤1 n Then the left endpoints of the intervals Q : `(Q) = 2− form an arithmetic progres- n { ∈ S } sion of step 2− /3 > `(A). It follows that A contains at most one of these endpoints. If A contains exactly one endpoint of one of the above intervals Q0 j, it means that both of 1 ∈ D A’s endpoints are contained in Q = Q0 3 `(Q) j 1 mod 3. Select such Q. If A contains no endpoints, we may choose any of the above− intervals∈ D − intersecting A. 6.4 The Lebesgue differentiation theorem. In this paragraph, we extend the differentia- d tion results of Section 6.2 to arbitrary intervals. Let again ν : R C be a complex measure. We consider the continuous maximal operator →

Mν = sup ν A A x 〈| |〉 3 d where the supremum is taken over all cubes A R containing the point x. ⊂ 6.4.1 THEMAXIMALTHEOREM. There holds d Mν(x) 6 sup ν(x) ≤ ~j 0,1,2 d ∈{ } As a consequence d d sup µ x R : Mν(x) > λ (18) ν . λ>0 ∈ ≤ · k k PROOF. Let us show how the second claim follows from the first. We have [ d Mν > λ M~j 0,1,2 d > 6− λ { } ⊂ ~j 0,1,2 d { ∈{ } } ∈{ } d d 1 and the measure of each of the 3 sets on the right hand side is bounded by 6 λ− ν by the dyadic maximal theorem. k k d To prove the first claim it suffices to show that for each cube A R there exists Q such that ⊂ ∈ S d ν A 6 ν Q. Let Q be the output of the three grids〈| lemma|〉 ≤ corresponding〈| |〉 to A. Then ν A ν A ν A ν Q ν ( ) ( ) 6d ( ) 6d ( ) 6d ν . A = | | = | | d | | d | | = Q 〈| |〉 µ(A) `(A) ≤ `(Q) ≤ µ(Q) 〈| |〉 This completes the proof of the first claim. Now, the continuous versions of the results of Section 6.2 can be proved in the same exact fashion, but relying on the continuous maximal theorem. We give an alternative proof of the final statement, using the dyadic version and the three grids lemma. 78 6. DERIVATIVES OF MEASURES AND DIFFERENTIATION THEOREMS

d 6.4.2 THEFIRSTFUNDAMENTALTHEOREMOFCALCULUS. Let ν be a complex measure on R with Radon-Nykodym decomposition Z ν = νf + νs, νs µ, νf = f dµ ⊥ 1 d d for some f L (R ). Then, for almost every x R and all sequences In of cubes shrinking to x ∈ ∈

lim ν I f x . n n = ( ) →∞〈 〉 d PROOF. For almost every x R we can perform the following. Let In be a sequence of cubes shrinking to x. By the three∈ grids lemma we get a sequence of cubes Qn with ∈ S In Q and 3`(In) < `(Qn) 6`(In). Then Qn also shrinks to x. By the dyadic Theorem 6.2.2,⊂ ≤

(6.3) lim f f x Q 0, lim νs Q 0 n ( ) n = n n = →∞〈| − |〉 →∞〈| |〉 But 0 f f x 6d f f x , 0 6d ( ) In ( ) Qn νs In ν Qn ≤ |〈| − |〉 ≤ 〈| − |〉 ≤ 〈| |〉 ≤ 〈| |〉 whence (6.3) holds for In in place of Qn as well. Since f x f f x f f x ν In ( ) ( ) In + νs In ( ) In + νs In |〈 〉 − | ≤ |〈 − 〉 | |〈 〉 | ≤ 〈| − |〉 〈| |〉 the claim follows. This is a particular case of Theorem 6.4.2.

1 6.4.3 FIRST FUNDAMENTAL THEOREMOF CALCULUS. Let f L (R) and Z x ∈ F(x) = f (t) dµ(t).

−∞ Then F is differentiable at almost every x R and F 0 = f almost everywhere. ∈ 6.5 Absolutely continuous functions. We now tackle the converse (hard) direction of the fundamental theorem of calculus. Let F ([0, 1]) with F(0) = 0. We look for a necessary and sufﬁcient condition for ∈ C Z x F(x + h) F(x) (6.4) F 0(x) = lim a.e., F(x) = F 0(t)dµ(t). h 0 h − → ∃ 0 Of course, for such a formula to even be well deﬁned, in this context of Lebesgue inte- 1 gration, we need F 0 L . This may fail even if F is differentiable almost everywhere. ∈ XAMPLE F x x 2 x 2 x 6.5.1 E . Take ( ) = sin( − )1(0,1]( ).

We have already seen that the Vitali-Cantor function F has F 0 = 0 outside of a compact set of measure zero, so that (6.4) must fail. The issue with this F is that F(E) = [0, 1], namely F maps a measure zero set to a set of full measure. Let us derive a simple necessary condition for (6.4). 6. DERIVATIVES OF MEASURES AND DIFFERENTIATION THEOREMS 79

1 6.5.2 LEMMA. Let F ([0, 1]) with F(0) = 0 be such that (6.4) holds and f := F 0 L . Then for all " > 0 there∈ C exists δ > 0 such that ∈ X X (6.5) bj aj < δ = F(bj) F(aj) < " j − ⇒ | − |

whenever (aj, bj) is a countable collection of pairwise disjoint subintervals of (0, 1). { } We say that F is absolutely continuous, F AC([0, 1]) if (6.5) holds. ∈ PROOFOF LEMMA 6.5.2. Clearly νf is absolutely continuous with respect to µ. It follows by Exercise 5.3 that for all " > 0 there exists δ > 0 such that if µ(A) < δ then νf (A) < ". | | If A is the union of pairwise disjoint intervals (aj, bj) , by deﬁnition of νf we have X X { } | | F(bj) F(aj) = νf ((aj, bj)) νf (A) < " | − | | | ≤ | | and (6.5) follows. We now prove that condition (6.5) is actually sufﬁcient under the additional assumption that F is monotone.

6.5.3 LEMMA. Let F ([0, 1]) with F(0) = 0. Assume in addition that F is nondecreasing. The following are equivalent:∈ C 1. F AC([0, 1]), namely (6.5) holds 2. if µ∈(A) = 0 then µ(F(A)) = 0 1 3. formula (6.4) holds and F 0 = f L . ∈ PROOF. Lemma 6.5.2 already shows that 3. = 1. To prove 1. = 2., let A be a Lebesgue set of⇒ measure zero. Let " > 0 be arbitrary and associate δ to " as⇒ in (6.5). Pick O A open with µ(O) < δ. Then O is the countable union ⊃ of disjoint intervals (aj, bj) . It follows that, by monotonicity, { } [ X F(A) F(O) (F(aj), F(bj)), F(bj) F(aj) < " ⊂ ⊂ j j − and thus F(A) is covered by countably many intervals of total measure less than ". But " > 0 is arbitrary. So µ(F(A)) = 0. To prove 2. = 3., we may assume F is strictly increasing and therefore one to one. To do so, for instance,⇒ replace F(x) by G(x) = x + F(x): such a G is strictly increasing, and notice that 3. for G implies 3. for F. Our aim is to deﬁne a positive ﬁnite measure

ν(A) = µ(F(A)), A ([0, 1]). ∈ L In order for this deﬁnition to be well posed, ﬁrst of all, we need to ensure that F(A) ∈ ([0, 1]). This is done as follows. Recall that A = K B where K is a δ set and B has LebesgueL measure zero. In particular K is a countable∪ union of compactF sets, and F(K) is as well, being F continuous. Now by condition 2. µ(F(B)) = 0, which by completeness of the Lebesgue σ-algebra, implies that F(B) is measurable. It follows that F(A) is also measurable. Now, the fact that ν is countably additive follows from the fact that if A are a S n measurable partition of A, we have F(A) = F(An) and the union is disjoint, since F is one to one. 80 6. DERIVATIVES OF MEASURES AND DIFFERENTIATION THEOREMS

1 At this point, observe that condition 2. again yields ν µ. Thus there exists h L (µ) such that dν = hdµ. In particular ∈ Z x F(x) = µ((0, F(x)]) = µ(F((0, x]))) = ν([0, x)) = h dµ 0 that is, (6.4) holds. The fact that F is a.e. differentiable with F 0 = h follows from Corollary 6.4.3. This completes the proof of the theorem. We now extend the equivalences of Lemma 6.5.3 to all AC functions. There will be an additional definition needed. Let F ([0, 1]) with F(0) = 0. The total variation of F is defined as ∈ C N X VF (x) = sup F(t j) F(t j 1) , x [0, 1] − j=1 | − | ∈ the supremum being taken over all positive integers N and all finite sequences of points 0 = t0 < t1 < < tN = x. ··· 6.5.4 LEMMA. Let F ([0, 1]) with F(0) = 0 be such that (6.5) holds. Then VF , VF F are monotone nondecreasing∈ C functions and each belongs to AC([0, 1]). ±

PROOF. Fix 0 x < y 1. Take any finite sequence of points 0 = t0 < t1 < < tN = x. One has ≤ ≤ ··· N X VF (y) F(y) F(x) + F(t j) F(t j 1) , − ≥ | − | j=1 | − | Taking supremum over N, t0,..., tN yields VF (y) VF (x) F(y) F(x) , that is, for α 0, 1 − ≥ | − | ∈ { ± } (6.6) VF (y) + αF(y) VF (x) + αF(x) F(y) F(x) + α(F(y) F(x)) 0. − ≥ | − | − ≥ so VF + αF is nondecreasing for each α 0, 1 To prove that VF + αF satisfies (6.5), since F satisfies (6.5), it suffices to treat α =∈ {0.± Let}" > 0 be given, δ = δ(") associated to F P by (6.5), I j = (aj, bj) pairwise disjoint intervals with I j < δ. For all j we can choose t a t ... t b such that | | j,0 = j < j,1 < j,Nj = j

Nj j X VF (bj) VF (aj) "2− + F(t j,k) F(t j,k 1) , − − ≤ k=1 | − | Summing these relations it follows that

Nj X X X VF (bj) VF (aj) " + F(t j,k) F(t j,k 1) < 2" − j − ≤ j k=1 | − | where the last bound follows from the observation that

Nj X X X t j,k t j,k 1 = bj aj < δ − j k=1 − j − and by the use of (6.5) for F. The next to last display is (6.5) for VF , so that the proof is complete. 6. DERIVATIVES OF MEASURES AND DIFFERENTIATION THEOREMS 81

With Lemma 6.5.4 in hand, we may complete the proof of the converse part of the L1 fundamental theorem of Calculus.

6.5.5 SECOND FUNDAMENTAL THEOREMOF CALCULUS. Let F : [a, b] C. The following are equivalent. → 1. F AC([a, b]), that is (6.5) holds; ∈ 1 2. F is differentiable at almost every x (a, b),F 0 L (a, b) and Z x ∈ ∈ F(x) = F(a) + F 0(t) dµ(t), x [a, b]. a ∀ ∈ PROOF. The implication 2. = 1. is contained in Lemma 6.5.2. To prove the converse, it sufﬁces to assume F is real valued.⇒ Write F + VF VF F F = G H := − − 2 − 2 Lemma 6.5.4 yields that G, H are nonincreasing functions and belong to AC([a, b]). We may thus apply Lemma 6.5.3 and obtain that G, H are differentiable almost everywhere.

Then F also is differentiable almost everywhere with F 0 = G0 H0and there holds Z x − Z x Z x F(x) F(a) = G(x) G(a) (H(x) H(a)) = G0 dµ H0 dµ = F 0 dµ − − − − a − a a which is what was to be proved. 6.6 The change of variable formula for the Lebesgue integral. The content of the previous section allows us to state and prove, with no difﬁculties, a change of variable theorem for the Lebesgue integral. We devote most of our work to a version on the real line.

6.6.1 THEOREM. Let Φ : [a, b] [c, d] be a monotone function and assume Φ AC([a, b]). Then → ∈

If f : [c, d] [0, ] is Lebesgue measurable, then (f Φ) Φ0 is Lebesgue measurable • and → ∞ ◦ | | Z d Z b f (t) dµ(t) = f (Φ(t)) Φ0(t) dµ(t) c a | | 1 1 If f L (c, d) then (f Φ) Φ0 L (a, b) and the same formula holds. • ∈ ◦ | | ∈ 6.6.2 REMARK. We have previously observed that if f = g almost everywhere, it is not necessarily true that f Φ = g Φ almost everywhere. However, it is true that if Φ : [a, b] [c, d] be is a monotone◦ function◦ and Φ AC([a, b]), then → ∈ f = g a.e. = F = G a.e., F := (f Φ) Φ0 , G := (g Φ) Φ0 ⇒ ◦ | | ◦ | | A proof of this fact is sketched in the exercises. As a consequence, it sufﬁces to prove Theorem 6.6.1 for the case of f Borel measurable.

PROOFOF THEOREM 6.6.1. For simplicity of notation, we deal with the case of Φ nondecreasing and f : [c, d] [0, ]. We deﬁne the positive measure → ∞ Z ν(B) = Φ0(t) dµ(t), B ([a, b]). B ∈ B 82 6. DERIVATIVES OF MEASURES AND DIFFERENTIATION THEOREMS

We recall the setup of Lemma 2.7.1. We use that Φ : [a, b] [c, d] is such that 1 → Φ− (B0) ([a, b]) B0 ([c, d]). ∈ B ∀ ∈ B and recall that the pushforward measure of ν deﬁned by 1 λ(B0) = ν(Φ− (B0)) B0 ([c, d]) ∀ ∈ B satisﬁes, for all f : [c, d] [0, ] Borel measurable Z → ∞Z Z (6.7) f dλ = f Φ dν = f (Φ(t)) Φ0(t)dµ(t). ◦ 1 However, let B0 = [γ, δ] [c, d] be an interval. Then Φ− (B0) = [α, β] for some α < β [a, b] with Φ(α) = γ, Φ(β⊂) = δ. Then by the fundamental theorem of Calculus ∈ 1 λ(B0) = ν(Φ− (B0)) = Φ(β) Φ(α) = δ γ = µ(B0). − − The fact that this equality holds for intervals, which is a Dynkin system, implies that the

same equality holds for all B0 ([c, d]). In other words, λ is the Lebesgue measure, and the claim of the theorem follows∈ B by comparing with (6.7). d A similar version of this theorem holds in R . We do not provide a proof. Recall that d d if V R is an open set, the map T : V R is said to be differentiable at x V if there ⊂ d d → ∈ exists a linear operator Jx : R R such that → T(x + h) T(x) Jx h lim = 0. h 0 −h − | |→ d | | d 6.6.3 THEOREM. Let V R be an open set, U V be a measurable set. Assume T : V R is continuous on V , differentiable⊂ at every point⊂ of U and → x, y U, T(x) = T(y) = x = y ∈ ⇒ namely T is one to one on U. Also assume that µ(T(V U)) = 0. Then for all positive measur- 1 able functions f : V C (resp. f L (V, (V ), µ)) \ → Z ∈ ZL f dµ = f (T(x)) det Jx dµ(x). T(V ) V | | 6. DERIVATIVES OF MEASURES AND DIFFERENTIATION THEOREMS 83 Chapter 6 Exercises

1 p 6.1. Let 1 < p and f L (R) L (R). Prove that ≤ ∞ ∈ ∩ Z x F(x) = f (t) dt

∞ is Hölder continuous of exponent 1/p0. 6.2. Construct a Lipschitz continuous G : [0, 1] [0, 1] such that → a. G is strictly increasing on [0, 1] b. G0 = 0 on a set of positive measure. 6.3. Construct a Lipschitz continuous G : [0, 1] [0, 1] such that G has no intervals of monotonicity, namely, for all x < y such that G(→x) G(y) there exists x < z < y with G(z) < G(x) and for all x < y such that G(x) G(y≤) there exists x < z < y with G(x) < G(z). ≥ 6.4. Let Φ : [0, 1] [0, 1] be absolutely continuous and nondecreasing. → 1. For t [0, 1], deﬁne ∈ Ψ(t) = inf x [0, 1] : Φ(x) t . { ∈ ≥ } Prove that Ψ is a strictly increasing function which is continuous from the left, namely

t, tn 0, 1 , tn t, lim tn t lim Ψ tn Ψ t , [ ] n = = n ( ) = ( ) ∈ ≤ →∞ ⇒ →∞ and that Φ(Ψ(t)) = t for all t [0, 1]. 2. Deﬁne ∈

H = x [0, 1] : Φ0(x) exists, Φ0(x) > 0 . { ∈ } Prove that H Ψ([0, 1]), that Ψ is one to one on H, that Ψ(Φ(x)) = x for all x H and that Ψ is⊂ continuous on Φ(H). ∈ 3. Use part 2. to prove that 1 N [0, 1], µ(N) = 0 = µ(Φ− (N) H) = 0. ⊂ ⇒ ∩ 4. Use part 3. to prove that

f = g a.e. = F = G a.e., F := (f Φ) Φ0 , G := (g Φ) Φ0 . ⇒ 1 ◦ | | ◦ | | Hint. For part 3. deﬁne M = Φ− (N) H. Then prove that ∩ [∞ 1 1 1 M = Mk, Mk := x M : Φ(x) Φ(y) > k− x y , y (x k− , x + k− ) [0, 1] k=1 { ∈ | − | | − | ∀ ∈ − ∩ }

This should follow from Φ0(x) > 0 at all points of H. Now it sufﬁces to show that µ(Mk) = 0 for all 1 k. It also sufﬁces to show that given any interval I with I < k− we have µ(Mk I) = 0. This is useful since it forces | | ∩ 1 x, y Mk I, x y k− = x y < k u(x) u(y) . ∈ ∩ | − | ≤ ⇒ | − | | − | Now with this in hand and using that µ(Φ(Mk I)) = 0 it should be easy to prove that µ(Mk I) = 0. ∩ ∩ 84 6. DERIVATIVES OF MEASURES AND DIFFERENTIATION THEOREMS

1,2 6.5 The Sobolev space W0 (0, 1). We provide a stronger characterization of the Hilbert space V we have deﬁned in Section 4.4. Recall that V was deﬁned as the closure of 2 D = f (0, 1) : supp f is compact in (0, 1) { ∈ C } with respect to the norm induced by the real-valued inner product Z 1 (f , g)V = f 0(x)g0(x) dx. 0 1,2 Prove thar V = W0 (0, 1) where 1,2 2 W0 (0, 1) := F AC(0, 1) : F(0) = F(1) = 0, F 0 L (0, 1) . 1,2 { ∈ ∈ } 1,2 Hint. For the inclusion W0 (0, 1) V , you will have to use (and prove) that if F W0 (0, 1) and ⊂ 1 ∈ F 0 = f , there exists a sequence fn of (say) functions with compact support in [0, 1] such that the C 2 integral function of each Fn is in and fn f L . D → ∈ 6.6 The weighted maximal theorem. A weight w is a positive locally integrable function d on R . In this exercise we consider the absolutely continuous, with respect to Lebesgue measure µ, measure dw = wdµ. Let M denote the Hardy-Littlewood maximal operator, namely

Mf (x) = sup f I 1I (x). d I cube of R 〈| |〉 1 d Let f L (R ). Prove that ∈ Z d sup λw x R : Mf (x) > λ C f (x)Mw(x) dµ(x) λ>0 d ∈ ≤ R for some constant C depending only on the dimension d. Hint. Prove the same statement for the dyadic version M ﬁrst and then use the consequence of three grid lemma. D CHAPTER 7

Product measures and convolution

7.1 Product σ-algebras and product measures. In this section, we consider two measure spaces (X j, j, µj), j = 1, 2 and denote by X = X1 X2. We aim to construct a measure space (X , F, µ) with the property × F A = A1 A2 Aj j, j = 1, 2, µ(A) = µ1(A1)µ2(A2). × ∈ F ∀ ∈ F During this procedure, we will appeal several times to the following property of Dynkin systems which was proved in Exercise 1.5. For convenience we recall that a Dynkin system ∆ (X ) is a nontrivial collection of sets which is closed under complement and countable disjoint⊂ P union. 7.1.1 DYNKIN π-λ LEMMA. Let (X ) have the property that G ⊂ P A, B = A B . ∈ G ⇒ ∩ ∈ G Let ∆ be any Dynkin system containing . Then ∆ contains Σ( ). G G The smallest σ-algebra satisfying our basic requirement is

= Σ( ), = A = A1 A2 : Aj X j, j = 1, 2 F R R { × ∈ } We will refer to as the collection of rectangles of X . Taking a cross-section of sets in yields sets in j.R F F 7.1.2 LEMMA. Let F . Then, for all x1 X1, x2 X2 ∈ F ∈ ∈ F y X : x , y F , F y X : y , x F . x1 = 2 2 ( 1 2) 2 x2 = 1 2 ( 1 2) 2 { ∈ ∈ } ∈ F { ∈ ∈ } ∈ F PROOF. By symmetry, it sufﬁces to consider sections in the ﬁrst coordinate. Let ∆ be the collection of sets of such that F for all x X . Since x1 2 1 1 F ∈ F ∈ A x A A A 2 1 1 ( 1 2)x1 = ∈ × x1 A1 ; 6∈ we have that . Since F c F c, is closed under complement. Suppose F is ∆ ( )x1 = ( x1 ) ∆ the disjoint unionR ⊂ of sets F . The sets F are pairwise disjoint and their union is j ( j)x1 2 F . Thus F belongs to . It follows that ∆ ∈is Fa Dynkin system containing . By the π-λ x1 R lemma it must be that ∆F= , so that the claim of the lemma is proved. F We have a simple extension of the previous principle to measurable functions. 7.1.3 LEMMA. Let f : X [0, ] be a ( , ([0, ]))-measurable function. Then for all x1 X1, x2 X2 the functions→ ∞ F B ∞ ∈ ∈ y f y , x f y , y f x , y f y 1 ( 1 2) = x2 ( 1) 2 ( 1 2) = x1 ( 1) 7→ 7→ are respectively 1, 2 measurable. F F 85 86 7. PRODUCT MEASURES AND CONVOLUTION

PROOF. Fix x1. If f 1F for some F , then fx 1F and 2 measurability follows = 1 = x1 from Lemma 7.1.2. By linearity, f is ∈measurable F whenever fFis a positive linear com- x1 2 bination of indicator functions of sets inF (aka simple functions). Let now f be a generic positive -measurable function. Take f kFto be a sequence of positive simple functions in- creasingF to f , then f k increases to f . Hence f x is measurable. ( )x1 x1 1 2 F We now put the measures µj into the picture. We do so by verifying an abstract version 3 of the Cavalieri-Simpson procedure for computing volumes of solids in R .

7.1.4 PROPOSITION. Let µj, j = 1, 2 be σ-ﬁnite measures. Then, for all F the functions Z Z ∈ F x1 1F (x1, x2) dµ2(x2), x2 1F (x1, x2)dµ1(x1) 7→ 7→ are respectively 1, 2 measurable and Z ZF F Z Z (7.1) 1F (x1, x2) dµ2(x2) dµ1(x1) = 1F (x1, x2) dµ1(x1) dµ2(x2)

Furthermore µ(F) := the common value of (7.1), F defines a positive σ-finite measure on , called product measure, satisfying∈ F F µ(A1 A2) = µ1(A1)µ2(A2) A1 A2 . × ∀ × ∈ R PROOF. We briefly notice that the equality in (7.1) can be rewritten in the form Z Z F d x F d x µ2( x1 ) µ1( 1) = µ1( x2 ) µ2( 2)

We first verify the claim concerning the measure µ, assuming the equality above. Let F be the disjoint union of F . Fix x X . Then F is the disjoint union of F , whence j 1 1 x1 ( j)x1 ∈ X F ∞ F µ2( x1 ) = µ2(( j)x1 ) j=1 and the equality Z X Z X F F d x F d x F µ( ) = µ2( x1 ) µ1( 1) = µ2(( j)x1 ) µ( 1) = µ( j) follows from the monotone convergence theorem for series. We now move to the proof of the first claim. We set up some notation and invoke σ- k finiteness of µj, j = 1, 2 to find two increasing sequences Yj j of sets exhausting X j k k k k ∈ F and such that µj(Yj ) < . The sets Y =: Y1 Y2 belong to and increase to X . ∞ × R ⊂ F k k Fix k and let ∆k be the collection of sets F such that (7.1) holds with F := F Y ∈ F ∩ in place of F. Let us begin by observing that R ∆k, since, for instance, if F = A1 A2, k k k ⊂ × F = A1 Y1 A2 Y2 and ∩ × ∩ 0 x A Y k µ F k 1 1 1 2(( )x1 ) = k 6∈ ∩ k µ(A2 Y2 ) x1 A1 Y1 ∩ ∈ ∩ 7. PRODUCT MEASURES AND CONVOLUTION 87

k k k and thus both sides of (7.1) are equal to µ1(A1 Y1 )µ2(A2 Y2 ). In particular, Y ∆k. c c k k k ∩ ∩ F F ∈ This, and the fact that 1(F ) = 1Y 1F can be used to observe that, if ∆k, ∆k as well. Furthermore, if Fn ∆k are pairwise− disjoint sets whose union is F, ∈ ∈ Z Z ∈ 1F k (x1, x2) dµ2(x2) dµ1(x1)

Z X Z X Z Z 1 k x , x dµ x dµ x 1 k x , x dµ x dµ x = (Fn) ( 1 2) 2( 2) 1( 1) = (Fn) ( 1 2) 2( 2) 1( 1) n n X Z Z 1 k x , x dµ x dµ x = (Fn) ( 1 2) 1( 1) 2( 2) n with two applications of the monotone convergence theorem for series and the fact that k Fn ∆k in the last step. Two more monotone convergence theorems lead to (7.1) for F . Hence∈ ∆k is a Dynkin system containing and by the Dynkin lemma coincides with . Now, for any F , using that F k increaseR to F F Z Z ∈ F 1F (x1, x2) dµ2(x2) dµ1(x1)

Z X Z X Z Z = 1F k (x1, x2) dµ2(x2) dµ1(x1) = 1F k (x1, x2) dµ2(x2) dµ1(x1) k k X Z Z Z Z = 1F k (x1, x2) dµ1(x1) dµ2(x2) = 1F (x1, x2) dµ1(x1) dµ1(x2) k and we have proved (7.1) for F. We are ready to prove the main result about product integration.

7.1.5 FUBINI’STHEOREM. Let (X j, j, µj), j = 1, 2 be σ-ﬁnite measure spaces and (X , , µ) be their product measure space. ThenF F 1. If f : X [0, ] is measurable, the functions → ∞Z Z F1(x1) := f (x1, x2) dµ2(x2), F2(x2) := f (x1, x2) dµ1(x1)

are respectively 1, 2 measurable and Z F F Z Z (7.2) f (x1, x2) dµ(x1, x2) = F1(x1) dµ1(x1) = F2(x2) dµ2(x2) X X1 X2

2. if f : X C is such that → Z Z f (x1, x2) dµ2(x2) dµ1(x1) < X1 X2 | | ∞

1 then f L (X , , µ). ∈ F 88 7. PRODUCT MEASURES AND CONVOLUTION

1 1 3. If f L (X , , µ), then, for j = 1, 2 Fj L (X j, j, µj), ∈ F ∈ F f L1 X , , a.e. x , f L1 X , , a.e. x , x1 ( 2 2 µ2)[µ1] 1 x2 ( 1 1 µ2)[µ2] 2 ∈ F − ∈ F − and (7.2) holds.

PROOF. Note that 2. is an easy consequence of 1. applied to f . We then deduce 3. from | | 1. and for this it sufﬁces to assume f real. First of all, we apply 1. to f ±. Denoting by F1± the analogous of F1 for f ±, we obtain, for instance Z Z

F x x f f 1 1±( 1) dµ( 1) = ±dµ L (µ) < ≤ k k ∞ 1 whence F L µ and f 1 < , µ -a.e. x . But then the same holds for F and 1± ( 1) x±1 L (µ2) [ 1] 1 1 f . Then (7.2)∈ follows byk takingk the∞ above displays. x1 ± We come to the proof of 1. which, in the case f = 1F , F , is simply Proposition 7.1.4. By taking positive linear combinations equality 7.2 extends∈ to F simple positive functions. Let now f positive and measurable. Take sn simple positive and increasing to f . Associate F (Sn)1 to sn the same way as F1 is associated to f . It is easy to see that x2 sn(x1, x2) 7→ increases to x2 f (x1, x2), whence by the monotone convergence theorem (Sn)1 increases to F1. Whence by7→ a further application of monotone convergence Z Z Z Z F1 dµ1 lim Sn 1 dµ1 lim sn dµ f dµ = n ( ) = n = →∞ →∞ and (7.2) is proved.

In the upcoming sections, we may use the notations 1 2 for and µ1 µ2 for µ, according to context. F × F F ×

7.2 Completion and product of Lebesgue measures. The product measure space (X1 × X2, 1 2, µ1 µ2) may fail to be complete even if (X j, j, µj), j = 1, 2 are complete σ-finiteF × measure F × spaces. For instance, suppose that thereF exists A1 1 with N1 A1 ∈ F ⊂ nonmeasurable set, i.e. N1 F1 and = A2 2 with µ2(A2) = 0. Then µ1 µ2(A1 A2) = 0 6∈ ; 6 ∈ F × × but B = N1 A2 A1 A2 does not belong to 1 2: if it were otherwise, all of its horizontal sections would× ⊂ belong× to 1, but this is notF the×F case. ?F We may define (X , , µ) to be the abstract completion of the measure space (X , 1 F ? F × 2, µ). Now, if f is a -measurable function, there exists an -measurable function g suchF that g = f [µ]-a.e.F It follows that the conclusions of LemmaF 7.1.3 and of the Fubini Theorem 7.1.5 continue to hold, with the exception that f is now -measurable only x1 2 F [µ1]-a.e. x1, and F1 is only defined [µ1]-a.e. x1. d We now apply the product measure constructions to Lebesgue measure on R . In the d upcoming proposition, for d 1, denote by , , µd respectively the Borel and Lebesgue d σ-algebras on R and the d-dimensional≥ LebesgueB L measure. We leave the proof as part of an easy exercise.

7.2.1 PROPOSITION. The following hold: d d d d 1. 1 2 = 1+ 2 ; d d ? d d 2. (B 1× B 2 ) B= 1+ 2 ; 3. µB ×µ B µ L. d1 d2 = d1+d2 × 7. PRODUCT MEASURES AND CONVOLUTION 89

7.3 Distribution function. Marcinkiewicz interpolation theorem. Fix a σ-finite measure space (X , , µ). The distribution function of a measurable f : X C is defined by F → mµ,f : [0, ) [0, ], mµ,f (t) := µ x X : f (x) > t ∞ → ∞ { ∈ | | } We omit µ from the subscript whenever no confusion arises. The following lemma is fundamental: we have already encountered a discrete version of it in one of the additional problems. Note that we simply write dt for the Lebesgue measure on [0, ). ∞ 7.3.1 LEMMA. Let φ : [0, ) [0, ) be a nondecreasing function with ∞ → ∞ φ 0 0, lim φ t ( ) = t ( ) = →∞ ∞ and assume that φ AC([0, T]) for all T > 0. Then ∈ Z Z ∞ φ f dµ = mµ,f (t) φ0(t) dt. ◦ | | 0 p In particular, with φ(t) = t , for 0 < p < | | ∞ 1 Z p p 1 f p pt m t t (7.3) L (µ) = − µ,f ( ) d . k k PROOF. The set E = (x, t) X [0, ) : f (x) > t is measurable in . To see{ this, notice∈ × that∞ if f |is simple| and} positive, E is a union of rectangles. Moreover,F × L if sn are simple and increase to f , the corresponding sets E increase to the set defined in the display. Notice that Z mf (t) = 1E(x, t) dµ(x) and, using Fubini’s theorem, since Ex = (0, f (x) ) | | Z Z Z Z Z f (x) ∞ ∞ | | mµ,f (t) φ0(t) dt = 1E(x, t)φ0(t) dt dµ(x) = φ0(t) dt dµ(x) 0 X 0 X 0 Z = φ( f (x) ) dµ(x) | | by the FTC and φ(0) = 0. This observation prompt us to the definition of the weak-Lp quasinorms 1 p f p, t m t p L (µ) = sup µ,f ( ), 0 < < . k k ∞ t>0 ∞ we will write p, when the measure µ is self-evident. When p = , the notation ∞ f , standsk simply · k for f . A manifestation of Chebychev’s inequality∞ is that k k∞ ∞ k k∞ f p, f p L (µ) L (µ). k k ∞ ≤ k k Although the triangle inequality does not hold, it is easy to verify that 1 p f + g p, 2 f p, + g p, . k k ∞ ≤ k k ∞ k k ∞ 90 7. PRODUCT MEASURES AND CONVOLUTION

p, However, when 1 < p , L ∞(µ) can be turned into a Banach space by prescribing an equivalent Banach norm.≤ ∞ This construction is sketched in the exercises. The importance of weak-Lp lies in that we are allowed to reverse Chebychev’s inequality if we allow ourselves to employ the weak-Lp norms in an "-neighborhood of p.

7.3.2 LEMMA. Let 0 < p0 < p1 , 0 < θ < 1 and deﬁne p0 < p = p(θ) < p1 by ≤ ∞ (7.4) 1 1 θ θ . p = p−0 + p1 Then

1 p0 p1 p p p p (1 θ) p θ (7.5) f f − f . p p p0 + p1 p p0, p1, k k ≤ − − k k ∞ k k ∞ pj PROOF. We prove the case p , which is harder. Call f A and set 1 < pj , = j ∞ ∞ 1 k k p1 p0 A1 − α = . A0 We have by assumption that pj mf (α) min Ajα− ≤ { } and the j = 0 estimate is more efﬁcient for 0 α < α. Using that p p0 ≤p p1 − θ, − 1 θ p1 p0 = p1 p0 = ( ) − − − − we integrate and get Z α Z ∞ 1 p p p0 1 p p1 1 p− f p A0 α − − dα + α − − dα k k ≤ 0 α θ 1 θ 1 A 1 A A 1 A 0 − = p p0 0 A0 + p1 p 1 A1 − − which is the claimed estimate. Weak Lp spaces provide a suitable target space for operators which fail to map Lp functions into Lp functions. For instance, Theorem 6.4.1 concerning the maximal operator

(7.6) Mf (x) = sup f I 1I (x) d I cube of R 〈| |〉 can be restated as follows: d 1 d (7.7) Mf 1, 18 f 1 f L (R ). k k ∞ ≤ k k ∀ ∈ We now provide a version of Lemma 7.3.2 for sublinear operators.

7.3.3 MARCINKIEWICZINTERPOLATIONTHEOREM. Let (X , , µ) be a σ-ﬁnite measure space and T be an operator mapping -measurable functions intoF -measurable functions, satisfying the sublinearity conditions F F

T[αf ](x) = α T[f ](x), T[f0 + f1](x) T[f0](x) + T[f1](x) , a.e. x X | | | | | | ≤ | | | | ∈ for all α C and measurable f , f0, f1. Let 0 < p0 < p1 , 0 < θ < 1 and deﬁne ∈ pj ≤ ∞ p0 < p = p(θ) < p1 as in (7.4). Assume that for all f L (µ) there holds ∈ (7.8) T f pj , A f pj j 1, 2. [ ] L ∞(µ) j L (µ) = k k ≤ k k 7. PRODUCT MEASURES AND CONVOLUTION 91

Then 1 p p p θ 1 θ (7.9) T f p 2 A A f p . [ ] L (µ) p p0 + p1 p 1 0− L (µ) k k ≤ − − k k Before the proof, we record the following corollary, which is obtained by combining (7.7) with the obvious observation that Mf f . k k∞ ≤ k k∞ 7.3.4 THE Lp-MAXIMALTHEOREM. Let 1 < p . Referring to (7.6), there holds

≤ ∞ 1 p p d Mf p f p L (µ) 2 18 L (µ), k k ≤ p 1 k k − with obvious meaning of the constant when p = . ∞ 1 d 7.3.5 REMARK. We have showed in Theorem 6.2.2 that, if f L (R ) and X ∈ En(f )(x) = f Q1Q(x) Q n〈 〉 ∈D 1 d p d then En(f ) f pointwise a.e. and in L (R ). Now, if f L (R ), 1 < p < , En(f ) f pointwise a.e.→ as well. This holds with f replaced by f 1I∈for any fixed dyadic∞ cube I, since→ 1 d d then f 1I L (R ). But then En(f ) f a.e. on every dyadic cube, which yields a.e. in R . ∈ → p p Since En(f ) Md f for all n, and the previous theorem yields Md f L if f L , we can | p | ≤ ∈ ∈p apply the L dominated convergence theorem to conclude that En(f ) f in L as well. → PROOFOFTHE MARCINKIEWICZINTERPOLATIONTHEOREM. By replacing T with T/A0, and A1 with A1/A0 we can assume A0 = 1 (verify this). Let δ > 0 a parameter to be chosen later and α > 0 be fixed for now. Define

>δα δα >δα f := f 1 f >δα, f ≤ := f f | | − Notice that, by the sublinearity condition, and by the assumptions (7.8) we have that

p0 >δα p0 p1 p1 δα p1 m m >δα m δα f A f T f (2α) T(f )(α) + T(f )(α) α− p + α− 1 ≤ p ≤ ≤ ≤ k k 0 k k 1 (7.10) Z Z p0 p0 p1 p1 p0 = α− f dµ + A1 α− f dµ f >δα | | f δα | | | | | |≤ p 1 Multiplying by α − and integrating the above inequality on α (0, ) Z ∈ ∞ 1 ∞ T f p αp 1m 2α dα p p = − T f ( ) p2 k k 0 Z Z Z Z ∞ ∞ p p0 1 p0 p1 p p1 1 p0 α − − f dµ dα + A1 α − − f dµ dα ≤ 0 f >δα | | 0 f δα | | | | | |≤ f (x) Z Z | δ | Z Z ∞ p0 p p0 1 p1 p0 p p1 1 = f α − − dα dµ + A1 f α − − dα dµ f (x) X | | 0 X | | | δ | 1 1 p p0 p p1 p1 p = f p δ − + A1 δ − k k p p0 p1 p − − 92 7. PRODUCT MEASURES AND CONVOLUTION where we have used Fubini in the third step. It turns out that choosing

p1 p0 p p p p1 p −p 0 1 p0 p p1 p1 p 0 1 θ p δ = A1 − = δ − = A1 δ − = A1 − = A1 ⇒ is optimal and, after rearranging and taking p-th power, yields the constant claimed in (7.9) when A0 = 1. The proof is complete. 7.4 Convolutions. Minkowski’s and Young’s integral inequalities. We begin this paragraph with a seemingly unrelated result on iterating Lp norms.

7.4.1 MINKOWSKI’SINTEGRALINEQUALITY. Let (X j, j, µj), j = 1, 2 be σ-ﬁnite measure spaces. Then for all 1 2-measurable functions f , and 1F p < , we have F × F Z Z ≤ ∞

f , x dµ x f , x p dµ x ( 2) 2( 2) ( 2) L (µ1) 2( 2) p · L (µ1) ≤ k · k

Before the proof, we motivate the inequality’s name with the following observation. If gn is a sequence of measurable functions on a σ-ﬁnite measure space (Y, , ν), we can deﬁne the function G f (n, y) := fn(y), (n, y) N Y ∈ × which is (N) -measurable. An application of Minkowski’s inequality then yields the generalizedP triangle× G inequality

X X fn fn Lp p (ν) L (ν) n N ≤ n N k k ∈ ∈ PROOF. We may assume that f is positive, via the following argument. If f is complex, and the right hand side is finite, then the left hand side for f in place of f is finite, whence the left hand side for f is well defined and majorized by the| | left hand side| | for f , which is controlled by the right hand side for f , which coincides with the right hand side| | for f .

With the notation of Theorem 7.1.5,| | the left hand side is equal to F p . Define 1 L (µ1) p 1 p 1 k k G F − . A simple computation yields G Lp µ F1 −p . Then, using Fubini’s theorem = 1 0 ( 1) = L (µ1) at the second step k k k k Z ZZ p F1 Lp µ = GF1 dµ1 = G(x1)f (x1, x2) dµ1(x1)dµ2(x2) k k ( 1) Z Z p 1 G p f , x p dµ x F f , x p dµ x L (µ1) ( 2) L (µ1) 2( 2) 1 L−p µ ( 2) L (µ1) 2( 2) ≤ k k 0 k · k ≤ k k ( 1) k · k and rearranging finishes the proof. 7. PRODUCT MEASURES AND CONVOLUTION 93 Chapter 7 Exercises 7.1. Prove that for all x > 0 Z ∞ x t 1 e− dt = . 0 x Use the above to prove that n Z sin x π lim dx = . n x 2 →∞ 0 7.2 The good-λ inequality. Let (X , , µ) be a σ-finite measure space and f , g positive, measurable functions. Suppose that forF all " > 0 there exists γ > 0 such that for all λ > 0 µ x X : f (x) > 2λ, g(x) γλ "µ x X : f (x) > λ . { ∈ ≤ } ≤ { ∈ } We mean that the right hand side is finite for all λ > 0. Then for all 0 < p < there exists

a constant Cp > 0 such that ∞ f p Cp g p. k k ≤ k k d 7.3 An improvement on the maximal theorem. Let ([0, 1) ) be the standard dyadic grid d 1 d in [0, 1) and for f L ([0, 1) ) deﬁne D ∈ d Md f (x) = sup f I 1I (x), x [0, 1) I ([0,1)d )〈| |〉 ∈ ∈D the dyadic maximal operator restricted to the unit cube. Assume

f 1 d L ([0,1) ) = 1. k k Prove that there exists a positive constant C such that Z Md f (x) 1 C f (x) log(e + f (x) ) dµ(x). k k ≤ [0,1)d | | | | (EXTRA CREDIT) Prove that the opposite inequality Z f (x) log(e + f (x) ) dµ(x) K Md f (x) 1 [0,1)d | | | | ≤ k k holds for a suitable choice of K. Hint. For the ﬁrst inequality, you should improve the dyadic maximal theorem to obtain the more precise estimate ¨ Z « d 1 µ x [0, 1) : M f (x) > 2λ min 1, f dµ { ∈ } ≤ λ f >λ | | | | Since f 1 = 1, the second estimate is more advantageous when λ > 1. k k For the second inequality, assume f 0. For k 0, let Ik = I j,k : j N be the maximal dyadic k ≥ ≥ { ∈ } intervals contained in Md f > 2 . Show that { } 2k f 2k+d < I j,k 〈 〉 ≤ so that ! X X M f x c 1 f 1 d ( ) + I j,k I j,k ≥ k j 〈 〉 94 7. PRODUCT MEASURES AND CONVOLUTION

d Integrating this on [0, 1) , carrying out an easy computation and using that log(e + Md f (x) ) | | ≥ log(e + f (x) ) should sufﬁce. | |