Complex Measures and the Radon-Nykodym Theorem

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 x

F(x)= f (t) dt = f 1[0,x) dµ. Z0 Z 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 ! 8 2 Conversely, if F is continuously differentiable on (0, 1), then

F(x) F(0)= F 01[0,x) dµ. Z 1 Both statements fail badly when the assumption of f continuous is replaced by f L (0, 1). 2 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 In,j En+1 = In+1,2j 1 In+1,2j. Deﬁne the Riemann integrable function fn : [0, 1] C \ [ ! 2n x 3 n+1 f 1 , F x f t dt. n = 2 In,j n( )= n( ) j=1 0 Å ã X Z 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 a continuous function F on [0, 1] is the following. Notice that gn = fn+1 fn is zero unless x En, and there holds that 2 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 theÄ middle third of In,j. Inä particular,Ä since the above hasä 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 2 · 57 58 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 ! complement of the Cantor set E = En. It follows that F(x + h) F(x) F 0(x)=S lim = 0 h 0 h ! for all x E (and therefore, F’=0 almost everywhere). Thus, the second fundamental theorem of62 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 -algebra on a set X . If E , and F 2F 1 E = Ej, Ej j, j = k = Ej Ek = j=1 2F8 6 ) \ ; [ then Ej : j N is called a measurable partition of the set E.Acomplex measure is a map { 2 } ⌫ : C with the property that, whenever E and Ej : j N is a measurable partitionF 7! of E, 2F { 2 }

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

1 1 5.2.1 EXAMPLE. Let (X , , µ) be a measure space and f L (X , , µ). Let F 2 F

⌫f (E)= f dµ, A . ZE 2F Exercise 6 from Chapter 3 yields exactly that ⌫f is a complex measure on . We observe F for further use that the same statement yields ⌫ f is a ﬁnite positive measure and ⌫f (E) | | | |  ⌫ f (E) for all E . | | 2F The second part of the above example suggests that, given a complex measure ⌫ on , one should look for a positive ﬁnite measure on such that ⌫(E) (E) for all A F . Such a measure would need to satisfy F | |  2F

1 1 (E)= (Ej) ⌫(Ej) j=1 j=1 | | X X whenever E and Ej : j N is a measurable partition of E. Thus, we deﬁne the map 2F { 2 } 1 (5.2) ⌫ : [0, ], ⌫ (E) := sup ⌫(Ej) E meas. part. of E | | F! 1 | | j j=1 | | { } X which is the least possible such measure and is termed the total variation of ⌫. It should be pretty obvious that if ⌫ is a complex measure taking nonnegative values, then ⌫ = ⌫. | | 5. COMPLEX MEASURES AND THE RADON-NYKODYM THEOREM 59

5.2.2 LEMMA. Let ⌫ be a complex measure on . Then the total variation ⌫ of (5.2) is a finite positive measure on . F | | F PROOF. Let us first prove that ⌫ 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` { \ 2 } { \ Ej : ` N is a measurable partition of Ej. Thus, with the first equality justified by the convergence2 } of the sums, there holds

1 1 1 1 1 1 1 1 ⌫(F`) = ⌫(F` Ej) ⌫(F` Ej) = ⌫(F` Ej) ⌫ (Ej). ` 1 | | ` 1 j 1 \  ` 1 j 1 | \ | j 1 ` 1 | \ |  j 1 | | = = = = = = = = X X X X X X X X Taking the supremum over F` in the left hand side we obtain { } 1 (5.3) ⌫ (E) ⌫ (Ej). | |  j=1 | | X For the opposite inequality, let Aj,` : ` N be a measurable partition of each Ej such that { 2 } 1 j ⌫ (Ej) "2 + ⌫(Aj,`) . | |  `=1 | | X Clearly Aj,` : j, ` N is a measurable partition of E. Thus { 2 } 1 1 1 (5.4) ⌫(Ej) " + ⌫(Aj,`) " + ⌫ (E)+" j=1 | |  j=1 `=1 | |  | | X X X and " > 0 is arbitrary. Combining (5.3) with (5.4) yields countable additivity of ⌫ . Let us move to ⌫ 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 1 z z . j ⇡ j j S j=1 | | 2 X X Suppose ﬁrst that there exists a measurable set E with ⌫ (E)= . Then there must | | 1 exist a partition Ej : j T of E and a ﬁnite N such that { 2 } N

⌫(Ej) > ⇡(1 + ⌫(E) ). j=1 | | | | X Using the sublemma, choose a subcollection of indices S 1, . . . , N such that ⇢ { }

(1 + ⌫(E) ) ⌫(Ej) = ⌫(A), A = Ej. | |  j S | j S X2 2 c [ Let B = E A . It must be by ﬁnite additivity of ⌫ that \ ⌫(B) = ⌫(E) ⌫(A) ⌫(E) ⌫(A) 1 | | | | | | | | 60 5. COMPLEX MEASURES AND THE RADON-NYKODYM THEOREM

So we have found two disjoint measurable sets A, B such that ⌫(A) , ⌫(B) 1 and A B = E, whence, without loss of generality ⌫ (A)= . | | | | [ We now show that if ⌫ (X )=| | we reach1 a contradiction. By repeating the above procedure, replacing E with| | A each1 time, we construct a sequence Bj of pairwise disjoint sets with ⌫(Bj) 1. But this contradicts the convergence of the series | | 1 1 ⌫ Bj = ⌫(Bj), Çj=1 å j=1 [ X 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 , ) : ⌫ : C complex measures F { F! } is a linear space and ⌫ X = ⌫(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 2 F then ⌫f = ⌫ f and ⌫f = f 1. | | | | k k k k Let ⌫ be a complex measure on taking real values. We can decompose F ⌫ ⌫ ⌫ ⌫+ ⌫ , ⌫ : , = + ± = | |±2 such that the measures ⌫± are ﬁnite positive measures on . This decomposition is referred to as the Jordan decomposition of ⌫. F

5.3 Absolute continuity. Let (X , , µ) be a measure space and ⌫ be an arbitrary (complex or positive) measure on . We sayF that ⌫ is absolutely continuous with respect to µ, and write ⌫ µ, if F Î (5.5) E , µ(E)=0 = ⌫(E)=0. 2F ) It is immediate to see that if ⌫ is a complex measure then ⌫ ⌫ . A measure ⌫ (positive or complex) on is said to be concentrated on the set A ifÎ | | F 2F ⌫(A B)=⌫(B) B . \ 8 2F If ⌫1, ⌫2 are two measures on such that ⌫j is concentrated on Aj and A1, A2 partition X , then we say that ⌫1, ⌫2 are mutuallyF singular and write ⌫1 ⌫2. 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, ⌫, ⌫1, ⌫2 be either positive or complex measure on the same -algebra . Then if ⌫ is concentratedF on A, then ⌫ is concentrated on A as well;

• if ⌫1 ⌫2 then ⌫1 ⌫2 ; | | • ? | | ? | | if ⌫1 µ, ⌫2 µ then ⌫1 + ⌫2 µ; • ? ? ? if ⌫1 µ, ⌫2 µ then ⌫1 + ⌫2 µ; • if ⌫ Î µ andÎ⌫ µ then ⌫ Î⌫ ; 1 Î 2 1 2 • ⌫ µ, ⌫ µ if and? only if ⌫ =?0. • Î ? 5. COMPLEX MEASURES AND THE RADON-NYKODYM THEOREM 61

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

5.3.2 LEMMA. Let (X , , µ) be a measure space and µ be a -ﬁnite measure. Then there exists a ﬁnite positive measureF ⌫ on such that F ⌫ µ, µ ⌫ Î Î and ⌫ takes the form

⌫(A)= w dµ, A ZA 8 2F 1 for some w L (X , , µ) with 0 < w < 1. 2 F PROOF. By -ﬁniteness, X X with µ X < . The function = n N n ( n) 2 1 n S 1 2 w x 1 ( )= Xn 1 µ X n n=1 + ( ) X 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 THEOREM FOR COMPLEX MEASURES. Let (X , , µ) be a measure space and assume that µ is a -ﬁnite measure. Let ⌫ : C be a complexF measure. Then there exists a unique decomposition F!

⌫ = ⌫a + ⌫s, ⌫a µ, ⌫s µ Î ? and if ⌫ is a positive ﬁnite measure, so are ⌫a, ⌫s. Furthermore, there exists a unique f 1 L (X , , µ) such that 2 F ⌫a(A)= f dµ, ZA and f is termed the Radon-Nykodym derivative of ⌫ with respect to µ.

5.4.2 LEBESGUE-RADON-NYKODYM THEOREM FOR POSITIVE MEASURES. Let (X , , µ) be a measure space and assume that µ is a -ﬁnite measure. Let ⌫ : C be a -ﬁniteF positive measure. Then there exists a unique decomposition F!

⌫ = ⌫a + ⌫s, ⌫a µ, ⌫s µ Î ? Furthermore, there exists a unique positive measurable f such that

⌫a(A)= f dµ, ZA and f is termed the Radon-Nykodym derivative of ⌫ with respect to µ. 62 5. COMPLEX MEASURES AND THE RADON-NYKODYM THEOREM

Before the proof, we observe that the second theorem may fail in case ⌫ is not assumed to be -ﬁnite. Indeed, let µ be the Lebesgue measure on (0, 1) and ⌫ be the counting measure on the Lebesgue -algebra. Assuming that ⌫ = ⌫a + ⌫s is a Lebesgue decomposition of ⌫ with respect to µ leads to a contradiction. Suppose that (0, 1)=A B with A B = and [ \ ; ⌫s(E)=⌫s(E B)= . It follows that for all E ﬁnite \ ; ⌫s(E B) ⌫(E)=⌫a(E)+⌫s(E B), \  \ but ⌫a(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 ⌫ ( )=⌫( X n) is a ﬁnite measure. Applying Theorem 5.4.1 for each n we obtain the decomposition· · \

n n n n ⌫(A X n)=⌫ (A)=⌫a(A)+⌫s (A), ⌫a(A)= fn dµ. \ ZA for some nonnegative f L1 µ . We can replace f with f 1 without perturbing the last n ( ) n n Xn equality. Then 2

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

1 f = fn n=1 X Then by the monotone convergence theorem

n ⌫a(A) := ⌫a(A)= fn dµ = f dµ n n ZA Xn ZA X X \ and it is obvious that ⌫a µ. This completes the proof. ⇤ Î PROOF OF THEOREM 5.4.1. We first reduce to the case where ⌫ is a positive finite measure. If ⌫ is complex, write ⌫ = ⌫1 + i⌫2 with ⌫j real and apply the statement to each, taking advantage of Proposition 5.3.1. If ⌫ is real, apply the statement to ⌫± and take advantage of Proposition 5.3.1. The core case is when ⌫ 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 d = d⌫ + wdµ. The functional

2 f L () I(f ) := f d⌫ 2 7! Z 5. COMPLEX MEASURES AND THE RADON-NYKODYM THEOREM 63 is (clearly) linear and by Hölder 1 2 2 I(f ) f d⌫ f d (X ) f d . | |  Z | |  Z | |  ✓Z | | ◆ 2 ∆ By the Riesz representation, there exists g L () such that 2 (5.6) f d⌫ = fgd Z Z 2 2 for all f L (). Let now E be a measurable set with (E) > 0, then 1E L1() L () and first2 equality in (5.6) above reads 2 ⇢ ⌫ E 1 1 (5.7) 0 ( ) 1 d⌫ g d 1 E = E E = E  ( ) ( ) Z ( ) ZE  Then it must be 0 g 1 a.e. . We can redefine g on a null set of (and thus of µ so that 0 g 1 everywhere.  Define   A = x X :0 g(x) < 1 , B = x X : g(x)=1 , ⌫a( )=⌫( A), ⌫s( )=⌫( B) { 2  } { 2 } · · \ · · \ which yields a decomposition ⌫ = ⌫a + ⌫s. Now (5.6) is the exact same as

2 (5.8) f (1 g) d⌫ = fgwdµ f L () Z Z 8 2 Now ⌫s 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 ) L1() and test now (5.8) for f = fn1E, it yields 2 n (1 g ) d⌫ = fn gwdµ = fn gw1A dµ ZE ZE ZE 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

⌫a(E)=⌫(E A)= f ,dµ \ ZE It is then obvious that ⌫ µ and since ⌫ X < , f L1 µ . This completes the a Î a( ) ( ) proof. 1 2 ⇤ 5.5 Corollaries of the Radon-Nykodym theorem.

5.5.1 POLAR DECOMPOSITION LEMMA. Let ⌫ be a complex measure on (X ). Then there exists a measurable function ↵ : X C such that F⇢P ! ↵(x) = 1 x X ,dµ = ↵d µ . | | 8 2 | | We leave the proof as an exercise.

1 5.5.2 COROLLARY. Let (X , , µ) be a measure space and f L . Then ⌫f = ⌫ f . F 2 | | | | 64 5. COMPLEX MEASURES AND THE RADON-NYKODYM THEOREM

PROOF. By the polar decomposition lemma, we find a unimodular measurable function ↵ such that ↵d ⌫f = d⌫f = f dµ = d ⌫f = ↵f dµ | | ) | | but then ↵f 0 a.e, whence ↵f = f a.e, whence ⌫f = ⌫ f . ⇤ | | | | | | 5.5.3 HAHN DECOMPOSITION THEOREM. Let ⌫ 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 ↵d⌫ = d ⌫ be the polar decomposition of ⌫. Since ⌫ is real valued, also ↵ is. Noticing | | 1 ↵ d⌫± = ± d ⌫ 2 | | 1 ↵ and that ± 1A , where A x : ↵ 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 -finite measure. Let 1 p < . For all linear boundedF functionals ⇤ on L (µ) there exists p 0  1 a unique g L (µ) such that ⇤ = ⇤g . 2 PROOF. We prove the case where µ(X ) < and leave the -finite case as an exercise. p Fix a linear bounded functional ⇤ on L (µ). We1 can normalize ⇤ = 1 and also µ(X )=1. Define the set function k k A ⌫(A)=⇤(1A) 1 2 F 7! p First of all ⌫(A) = ⇤(1A) 1A p = µ(A) 1. So ⌫ takes values in the unit disk. If | | | | k k  A, B are disjoint measurable sets, it is evident that ⌫(A B)=⌫(A)+⌫(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 1 ⌫ E lim ⇤ 1A lim ⌫ En ⌫ En . ( )=n ( n )=n ( )= ( ) !1 !1 j=1 j=1 X X Summarizing, ⌫ 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 2 E , if s = 1E 2F ⇤(s)=⌫(E)= g dµ = sgdµ ZE Z This equality carries to simple functions s by linearity, and by density we obtain

p (5.9) ⇤(f )= fgdµ f L (µ). Z 8 2 p Since ⇤(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 theorem2 in the casek k when µ(X ) < ⇤ 1