MAT 571 REAL ANALYSIS II LECTURE NOTES. Contents. 2. Product measures Iterated integrals Complete products Differentiation 17

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1 MAT 57 REAL ANALSIS II LECTURE NOTES PROFESSOR: JOHN QUIGG SEMESTER: SPRING 205 Contents. Convergence in measure 2. Product measures 3 3. Iterated integrals 4 4. Complete products 9 5. Signed measures 6. Differentiation 7 7. Extension to complex values 27. Convergence in measure We ve already seen a few notions of convergence of a sequence {f n } of functions, e.g., pointwise, uniform, a.e., L. Here s another that is occasionally useful, particularly in probability theory: Definition.. Let {f n } be a sequence of measurable functions on a measure space (, M, µ). We say f n f in measure if for all ε > 0 we have µ ( { f n f ε} ) 0. Proposition.2. If f n f in L then f n f in measure. Proof. This follows immediately from Markov s Inequality. which says that if f is integrable and c > 0 then f cµ({ f c}), although in the analysis literature this is occasionally referred to as Chebyshev s Inequality, with possible variations in the spelling, such as Tchebychev

2 2 PROFESSOR: JOHN QUIGG SEMESTER: SPRING 205 Remark.3. We ve seen that if f n f in L then some subsequence converges to f a.e. Similar arguments can be used to show that if f n f in measure, then some subsequence converges to f a.e. This can be used to prove that if f : R R is Lebesgue measurable then there is a sequence of continuous functions converging to f a.e., generalizing from the integrable case. Here s one more notion of convergence that is sometimes useful: Definition.4. Let {f n } be a sequence of measurable functions on a measure space (, M, µ). We say f n f almost uniformly if for all ε > 0 there exists A M such that µ(a c ) < ε and f n f uniformly on A. The following is an exercise: Proposition.5. Almost uniform convergence implies convergence both a.e. and in measure. Theorem.6 (Egoroff s Theorem). Let {f n } be a sequence of measurable functions on a measure space (, M, µ). If µ is finite and f n f a.e., then f n f almost uniformly. Proof. Let ε > 0. Replacing f n by f n f, without loss of generality f n 0 and f n 0 a.e. For each k, m N put { A m k = f n }. m n=k Then A m A m 2, and µ( k Am k ) = 0, so because µ is finite lim k µ(a m k ) = 0. Choose k m such that µ(a m k m ) < 2 m ε, and put B = m A m k m. Then µ(b) m µ(a m k m ) < ε. For all x B c and any m N, we have x / A k k m, so It follows that f n 0 uniformly on B c. f n (x) < m for all n k m. Remark.7. Egoroff s Theorem can be used, among other things, to help prove Lusin s Theorem: if f : [a, b] R is measurable, then for all ε > 0 there exists E [a, b] such that λ([a, b]\e) < ε and f E is continuous. Note that this does not say that f itself is continuous at the points of E, rather that the restriction is continuous as a function from E to R. In fact, f := χ Q [0,] is measurable on [0, ], but it discontinuous everywhere. Nevertheless, letting E = [0, ] \ Q, we have λ([0, ] \ E) = 0 and the restriction f E is continuous,

3 MAT 57 REAL ANALSIS II LECTURE NOTES 3 2. Product measures Throughout, (, M, µ) and (, N, ν) will be σ-finite measure spaces. Lemma 2.. If C and D are semialgebras on and, respectively, then is a semialgebra on {A B : A C, B D} Definition 2.2. A measurable rectangle in is any set of the form A B with A M and B N. Lemma 2. guarantees that the set of all measurable rectangles is a semialgebra on. Definition 2.3. For A M, B N define ρ(a B) = µ(a)ν(b). Lemma 2.4. ρ is a premeasure on the semialgebra of all measurable rectangles. Proof. Clearly ρ 0 and ρ( ) = 0. For the finite additivity, let A B be a measurable rectangle that is a finite disjoint union of measurable rectangles A j B j. Temporarily fix y B. For each x A there exists a unique j such that (x, y) A j B j. Thus A = A j, so µ(a) = {j:y B j } µ(a j ) = {j:y B j } j It follows that for arbitrary y we have µ(a)χ B (y) = j µ(a j )χ Bj (y). µ(a j )χ Bj (y). Integrating both sides over, we get µ(a)ν(b) = µ(a j )χ Bj (y) dν(y) j = µ(a j ) j χ Bj (y) dν(y) (linearity of integration) = j µ(a j )ν(b j ). Finally, for the countable subadditivity, let A B be a measurable rectangle that is contained in a countable union of measurable rectangles A n B n. Fix y B. For each x A there exists n such that (x, y) A n B n. Thus A y B n A n, so µ(a) µ(a n ) = µ(a n )χ Bn (y). y B n n

4 4 PROFESSOR: JOHN QUIGG SEMESTER: SPRING 205 It follows that for arbitrary y, µ(a)χ B (y) n Integrating over, we get µ(a)ν(b) µ(a n )χ Bn(y) dν(y) n = µ(a n ) n = µ(a n )ν(b n ). n χ Bn(y) dν(y) µ(a n )χ Bn (y). (Monotone Convergence Theorem) Definition 2.5. The product σ-algebra M N is the σ-algebra on generated by the set of measurable rectangles. By the σ-finite version of Carathéodory s Extension Theorem, we can immediately conclude: Theorem 2.6. There is a unique measure µ ν, called the product measure, on M N such that µ ν(a B) = µ(a)ν(b) for all A M, B N. Moreover, µ ν is σ-finite. 3. Iterated integrals Before we prove that product integrals can be evaluated one variable at a time, we need a technical result that might seem very general, but which we ll only need for this one purpose. Definition 3.. A family A of subsets of a set is a monotone class if it is closed under countable increasing unions and countable decreasing intersections. It is easy to see that the intersection of any family of monotone classes is a monotone class, so that If S is any family of subsets of then there is a smallest monotone class containing S, which we call the monotone class generated by S. Theorem 3.2 (Monotone Class Lemma). If A is an algebra on, then the monotone class generated by A coincides with the σ-algebra on generated by A. Proof. Let M denote the monotone class generated by A, It suffices to show that M is a σ-algebra. For A M define C(A) = {B M : A \ B, B \ A, and A B are all in M}. By symmetry, B C(A) if and only if A C(B). Properties of set operations show that C(A) is a monotone class. If A A then A C(A) because A is an algebra. Thus M C(A) because C(A) is a monotone class. Then by symmetry for all A M we have A C(A), hence M C(A). Thus M is closed under differences and finite intersections. Then M is an algebra since A C. Any countable union of sets in M can be expressed as a

5 MAT 57 REAL ANALSIS II LECTURE NOTES 5 countable increasing union of finite unions of sets in M, so is in M since M is an algebra and a monotone class. Therefore M is a σ-algebra, as desired. For the next result it is convenient to introduce a notation that is commonly used to turn a function of two variables into a single-variable function. Notation. If f : Z and y define f(x, ) : Z by f(x, )(y) = f(x, y), and similarly for f(, y). There are more-or-less self explanatory variations on this notation, e.g., if f : C and µ is a measure on we can consider the function f(x, ) dµ(x), whose domain is all y for which f(, y) is µ-integrable, and whose value at such a y is ( ) f(x, ) dµ(x) (y) = f(x, y) dµ(x). Theorem 3.3 (Tonelli s Theorem). Let (, M, µ) and (, N, ν) be σ-finite measure spaces, and let f : [0, ] be M N -measurable. Then: () f(x, ) is N -measurable for all x ; (2) f(, y) is M-measurable for all y ; (3) f(, y) dν(y) is M-measurable; (4) f(x, ) dµ(x) is N -measurable; (5) f d(µ ν) = f(x, y) dν(y) dµ(x); (6) f d(µ ν) = f(x, y) dµ(x) dν(y). Proof. By symmetry, it suffices to prove (2), (4), and (6). Case. µ and ν are finite. Let F = {nonnegative M N -measurable f : (2), (4), (6)}; C = {A M N : χ A F}. First we show that F is closed under nonnegative linear combinations: let f,..., f n F and c,..., c n 0, and put f = n c if i. Then n f(, y) = c i f i (, y) is M-measurable, and by linearity f(x, ) d(µ ν) = n c i f i (x, ) d(µ ν) is N -measurable

6 6 PROFESSOR: JOHN QUIGG SEMESTER: SPRING 205 and Thus f F. n f d(µ ν) = c i n = c i n = c i = f i d(µ ν) f i (x, y) dµ(x) dν(y) f i (x, y) dµ(x) dν(y) f(x, y) dµ(x) dν(y). It follows that C is closed under finite disjoint unions, since χ n n A = χ i Ai. Next we show that F is closed under increasing sequential limits: let f n f, with f n F for all n. Then f n (, y) f(, y), which is thus M-measurable, and by the Monotone Convergence Theorem f n (x, ) dµ(x) f(x, ) dµ(x), which is thus N -measurable and Thus f F. f d(µ ν) = lim f n d(µ ν) n = lim f n (x, y) dµ(x) dν(y) n = lim f n (x, y) dµ(x) dν(y) n = f(x, y) dµ(x) dν(y). It follows that C is closed under countable increasing unions, since if A n A then χ An χ A. Before continuing, we introduce the following temporary notation: if A and y let A y = {x : (x, y) A}. Note that χ A (x, y) = χ A y(x),

7 MAT 57 REAL ANALSIS II LECTURE NOTES 7 and so if A y M then It follows that if f = χ A then χ A (x, y) dµ(x) = µ(a y ). (2) A y M for all y; (4) µ(a y ) is measurable in y; (6) µ ν(a) = µ(ay ) dν(y). Next we show that C is closed under countable decreasing intersections: let A n A, with A n C for all n. Then A y n A y, which is thus in M. By continuity from above and µ < we have µ(a y n) µ(a y ), which is thus measurable in y. By continuity from above and µ ν < we have µ ν(a) = lim n µ µ(a n ) by continuity from above, since µ ν < = lim µ(a y n n) dν(y) = µ(a y ) dν(y) By the Dominated Convergence Theorem, which applies since for all n, y we have Thus A C. µ(a y n) µ() <. We ve shown that C is a monotone class. We will now show that it contains every measurable rectangle B C. We have { (B C) y B if y C = if y / C, which is in M in each case. Then which is measurable in y. Thus µ ν(b C) = µ(b)ν(c) = Therefore B C C. µ ( (B C) y) = µ(b)χ C (y), µ(b)χ C (y) dν(y) = µ ( (B C) y) dν(y). Since C is closed under finite disjoint unions it contains the algebra generated by the measurable rectangles, because the set of measurable rectangles is a semialgebra. Therefore, since

8 8 PROFESSOR: JOHN QUIGG SEMESTER: SPRING 205 C is a monotone class it contains the σ-algebra generated by the measurable rectangles, by the Monotone Class Lemma, and hence C = M N. Then, since F is closed under nonnegative linear combinations it contains all nonnegative simple functions, and then since it is closed under increasing sequential limits it contains every nonnegative measurable function, and we re done. Theorem 3.4 (Fubini s Theorem). Let (, M, µ) and (, N, ν) be σ-finite measure spaces, and let f be µ ν-integrable. Then: () f(x, ) is ν-integrable for µ-a.e. x ; (2) f(, y) is µ-integrable for ν-a.e. y ; (3) the a.e.-defined function f(, y) dν(y) is µ-integrable; (4) the a.e.-defined function f(x, ) dµ(x) is ν-integrable; (5) f d(µ ν) = f(x, y) dν(y) dµ(x); (6) f d(µ ν) = f(x, y) dµ(x) dν(y). Proof. Again by symmetry it suffices to prove (2), (4), and (6). First suppose f 0. By Tonelli s Theorem we know that (6) holds, f(, y) is M-measurable for every y, and the function f(x, ) dµ(x) is N -measurable. Then (6) implies that f(x, ) dµ(x) is ν- integrable, giving (4), and f(x, y) dµ(x) < a.e. y, giving (2). In the general case, by the above we see that for a.e. y the functions f + (, y) and f (, y) are µ-integrable, and hence so is f(, y) = f + (, y) f (, y), the a.e.-defined functions f + (x, ) dµ(x) and f (x, ) dµ(x) are ν-integrable, and therefore are finite-valued a.e., and hence the a.e.-defined function f(x, ) dµ(x) = f + (x, ) dµ(x) f (x, ) dµ(x) is integrable, and f d(µ ν) = = = = = f + d(µ ν) f d(µ ν) f + (x, y) dµ(x) dν(y) f (x, y) dµ(x) dν(y) ( ) f + (x, y) dµ(x) f (x, y) dµ(x) dν(y) ( f + (x, y) f (x, y) ) dµ(x) dν(y) f(x, y) dµ(x) dν(y)

9 MAT 57 REAL ANALSIS II LECTURE NOTES 9 4. Complete products Theorem 4.. Let (, M, µ) and (, N, ν) be complete σ-finite measure spaces, and let (, M N, µ ν) be the completion of the product measure space (, M N, µ, ν). () Complete Tonelli: If f is nonnegative and M N -measurable, then: (a) f(x, ) is N -measurable for a.e. x ; (b) f(x, ) is M-measurable for a.e. y ; (c) the a.e.-defined function f(, y) dν(y) is M-measurable; (d) the a.e.-defined function f(x, ) dµ(x) is N -measurable; (e) f d µ ν = f(x, y) dν(y) dµ(x); (f) f d µ ν = f(x, y) dµ(x) dν(y). (2) Complete Fubini: If f is µ ν-integrable, then (a) f(x, ) is ν-integrable for a.e. x ; (b) f(, y) is µ-integrable for a.e. y ; (c) the a.e.-defined function f(, y) dν(y) is µ-integrable; (d) the a.e.-defined function f(x, ) dµ(x) is ν-integrable; (e) f d µ ν = f(x, y) dν(y) dµ(x); (f) f d µ ν = f(x, y) dµ(x) dν(y). Proof. We only prove (); (2) is left as an exercise. This time, for variety and convenience, we show (a), (c), and (e), which suffices by symmetry. Write f = g + h, where g is M N - measurable and h = 0 µ ν-a.e. Due to the nature of sets in M N, there exists D M N such that µ ν(d) = 0 and h = 0 on D c. Then by Tonelli s Theorem we have 0 = µ ν(d) = ν(d x ) dµ(x), so ν(d x ) = 0 for µ-a.e. x. Since h = 0 on D c, for all x we have h(x, ) = 0 on D c x. Thus, for µ-a.e. x we have h(x, ) = 0 ν-a.e., so f(x, ) = g(x, ) ν-a.e., so f(x, ) is N -measurable and f(x, ) dν = g(x, ) dν. Hence f(, y) dν(y) = g(, y) dν(y)

10 0 PROFESSOR: JOHN QUIGG SEMESTER: SPRING 205 as µ-a.e.-defined functions, and consequently f(, y) dν(y) is M-measurable. Thus we can compute: f dµ ν = g dµ ν = g dµ ν = g(x, y) dν(y) dµ(x) = f(x, y) dν(y) dµ(x). Definition 4.2. n-dimensional Lebesgue measure is the completion of the product measure λ n = n {}}{ λ λ, where λ is Lebesgue measure on R. We just write (R n, L, λ) for the complete measure space, which won t cause any confusion. Example 4.3. In this case the completion is really necessary: there exist A, B R such that A B L and λ(a B) = 0 but A B is not in the product σ-algebra L L. Corollary 4.4. If f is Lebesgue measurable on R n and is either nonnegative or integrable, and if (i,..., i n ) is any rearrangement of (,..., n), then f dλ = f(x,..., x n ) dx i dx in. R n R R Proposition 4.5. (R n, λ) is regular in the sense that for any Lebesgue measurable subset A R n, λ(a) = inf{λ(u) : A U, U open} = sup{λ(k) : K A, K compact}. Proof. We only prove the first equality, for then the second equality follows similarly to the case of the real line. First we reduce to bounded sets: we have A = A k where each A k bounded and measurable, and, given ε > 0, if for each k we have an open set U k A k with λ(u k ) λ(a k ) + 2 k ε, then U := U k is open, A U, and ( λ(u) λ(u k ) λ(ak ) + 2 k ε ) = λ(a k ) + ε = λ(a) + ε. Thus without loss of generality A is bounded. Next we reduce to measurable rectangles: given ε > 0, we can find measurable rectangles B, B 2,... such that A B k and λ(b k) < λ(a)+ε. If for each k we have an open set U k B k with λ(u k ) < λ(b k )+2 k ε, then U := U k is open, A U, and λ(u) λ(u k ) < λ(b k ) + ε < λ(a) + 2ε.

11 MAT 57 REAL ANALSIS II LECTURE NOTES Thus without loss of generality A is a bounded measurable rectangle, say A = n A i where each A i is a bounded measurable subset of R. Now we use the -dimensional result: given ε > 0, for each i we can choose an open set U i R such that A i U i and λ(u i ) < λ(a i ) + ε. Then U := n U i is open in R n, A U, and n n ( λ(u) = λ(u i ) < λ(ai ) + ε ). Since the latter product tends to λ(a) = n λ(a i) as ε 0, we are done. 5. Signed measures Throughout, (, M) will be a measurable space. Definition 5.. A signed measure on (, M) is a function ν : M R such that: () ν takes at most one of the values ± ; (2) ν( ) = 0; (3) for every pairwise disjoint sequence {A n } in M, ν( n A n) = n ν(a n). Remark 5.2. () In (3) above, since the union doesn t depend upon the ordering of the A n s, the sum on the right-hand side must also be rearrangement-invariant. In particular, if it converges then it must converge absolutely. (2) A measure in the previous sense may be regarded as a signed measure ν that happens to take only nonnegative values. In this case we typically refer to ν simply as a measure, or sometimes as a positive measure if there is a danger of confusion. Example 5.3. If µ, µ 2 are measures on M, not both infinite, then µ µ 2 is a signed measure. Example 5.4. Let (, M, µ) be a measure space, and let f be a measurable function on such that f + or f is integrable. Then we get a signed measure ν on by ν(a) = f dµ. We write dν = f dµ to indicate how ν is related to f and µ. The Radon-Nikodym Theorem below essentially characterizes the signed measures arising in this way. Lemma 5.5. Every signed measure is continuous both from below and from above, where in the latter case the first set in the sequence is required to have finite measure. Definition 5.6. If ν is a signed measure on M, then A M is positive (respectively, negative, or null) for ν if for all B M with B A we have ν(b) 0 (respectively, 0, or = 0). Example 5.7. If dν = f dµ, then A M is ν-positive, negative, or null if and only if f 0, 0, or = 0 µ-a.e., respectively. Lemma 5.8. For a signed measure, then every measurable subset of a positive set is positive, and so is every countable union of positive sets. A

12 2 PROFESSOR: JOHN QUIGG SEMESTER: SPRING 205 Theorem 5.9 (Hahn Decomposition). For a signed measure on, there are a positive set P and a negative set N such that = P N and P N =, and if P, N is another such pair then P P and N N are null. Proof. Without loss of generality µ omits the value. Let M = sup{ν(a) : A is ν-positive}. Choose positive sets A n such that ν(a n ) M. Replace A n by n A i, and put P = n A n. Then P is ν-positive, and ν(p ) = M since A A 2 and ν(a ) <. In particular, M <. It suffices to show that N := P c is ν-negative. Suppose not. Then there is A N such that ν(a) > 0. Note that A is not ν-positive, otherwise P A would be ν-positive and ν(p A) = ν(p ) + ν(a) > M, a contradiction. Thus there exists B A with ν(b) < 0, and then A \ B A and ν(a \ B) > ν(a). Let n = min{n N : there is A N such that ν(a ) > /n} n 2 = min{n N : there is A 2 A such that ν(a 2 ) > µ(a ) + /n}, and continue inductively. Put A = k A k. Then µ(a) = lim µ(a k ) since µ(a ) <. We have > µ(a), n k so n k. Now choose B A with ν(b) > ν(a), and find m N such that ν(b) > ν(a) + /m, and then find k such that n k > m. But then B A k and contradicting the choice of n k. ν(b) > ν(a k ) + m, Thus we have existence. For the uniqueness, suppose we have another such decomposition = P N. Then P N is both ν-positive and ν-negative, hence is ν-null, and similarly P N is ν-null. Thus is ν-null. P P = P \ P P \ P = (P N ) (P N) = N N Definition 5.0. In the above theorem, = P N is a Hahn decomposition for ν. Definition 5.. Two signed measures µ, ν on (, M) are mutually singular, written µ ν, if there are null sets A for ν and B for µ such that = A B and A B =. Corollary 5.2 (Jordan Decomposition). For every signed measure ν there exist unique positive measures ν +, ν such that ν = ν + ν and ν + ν.

13 MAT 57 REAL ANALSIS II LECTURE NOTES 3 Proof. Choose a Hahn decomposition = P N for ν, and define ν +, ν by ν + (A) = ν(a P ) and ν (A) = ν(a N). Then ν +, ν have the right properties, proving existence. For the uniqueness, suppose ν = ν + ν is another such decomposition. Since ν + ν, we can find complementary sets P, N such that ν + (N) = ν (P ) = 0. Then 0 ν + (N) = ν(p ) + ν (N) = ν (N) + ν (N) 0, and so ν + (N) = 0. Similarly ν (P ) = 0. Then for any measurable set A we have ν + (A) = ν + (A P ) + ν + (A N) = ν + (A P ) = ν(a P ) + ν (A P ) = ν(a P ) = ν + (A), and so ν + = ν +, and hence also ν = ν. Definition 5.3. In the above corollary, ν+ and ν are the positive and negative variations of ν, ν = ν + ν is the Jordan decomposition of ν, and ν := ν + + ν is the total variation of ν. Lemma 5.4. Let ν be a signed measure on M. () A M is ν-null if and only if ν (A) = 0. (2) For any measure µ on M, the following are equivalent: (a) ν µ; (b) ν µ; (c) ν + µ and ν µ. (3) ν is the smallest positive measure such that ν(a) ν (A) for all A M. Definition 5.5. Let ν be a signed measure on (, M). () L (ν) := L (ν + ) L (ν ). (2) For f L (ν), f dν := f dν + f dν. Lemma 5.6. If ν is a signed measure, then L (ν) = L ( ν ), and f dν f d ν for all f L (ν). Definition 5.7. ν is finite or σ-finite if ν has the corresponding property. Definition 5.8. If (, M, µ) is a measure space, then a signed measure ν on M is absolutely continuous with respect to µ, written ν µ, if µ(a) = 0 implies ν(a) = 0 for all A M.

14 4 PROFESSOR: JOHN QUIGG SEMESTER: SPRING 205 Theorem 5.9. If ν is a finite signed measure and µ a positive measure on (, M), then ν µ if and only if for all ε > 0 there exists δ > 0 such that for all A M, µ(a) < δ implies ν(a) < ε. Proof. The ε δ condition obviously implies absolute continuity, so assume ν µ. Then also ν µ, and ν(a) ν (A) for all A M, so without loss of generality ν 0. Suppose the ε δ condition fails. Then there exists ε > 0 such that for all n N there exists A n M such that µ(a n ) < 2 n and ν(a n ) ε. Put B k = k A n (k N) and C = k B k. Then B B 2 and µ(b ) n µ(a n) <, so µ(c) = lim µ(b k ) = 0 because µ(b k ) k 2 n = 2 k. But since ν is finite which is a contradiction. ν(c) = lim ν(b k ) lim ν(a k ) ε, Corollary If µ is a positive measure on (, M) and f L (µ), then for all ε > 0 there exists δ > 0 such that for all A M, µ(a) < δ implies f dµ < ε. The following two theorems will be proven together, but they deserve to be stated separately. Theorem 5.2 (Radon-Nikodym Theorem). Let ν be a signed measure and µ a positive measure on (, M). If ν, µ are σ-finite and ν µ, then there exists a measurable function f : R such that dν = f dµ, and any two such functions are equal µ-a.e. Theorem 5.22 (Lebesgue Decomposition Theorem). Let ν be a signed measure and µ a positive measure on (, M), both σ-finite. Then there exists unique σ-finite signed measures ν, ν 2 such that ν = ν + ν 2, ν µ, and ν 2 µ. Proof of Theorems 5.2 and Case. ν and µ are both positive and finite. Let F be the set of measurable functions f : [0, ] such that f dµ dν, i.e., f dµ ν(a) for A all A M, and put M = sup f F f dµ. Note that F since 0 F. If f, g F then max{f, g} F, since for any A M we can let and then A A f = {x A : f(x) g(x)} A g = A \ A f, max{f, g} dµ = f dµ + A f g dµ ν(a f ) + ν(a g ) = ν(a). A g Choose a sequence {f n } in F such that A f n dµ M.

15 MAT 57 REAL ANALSIS II LECTURE NOTES 5 Replacing f n by max n f i, we can assume that {f n } is increasing. Let f = lim f n. Then by the Monotone Convergence Theorem we have f dµ dν and f dµ = M. Define dν = f dµ and ν 2 = ν ν. Note that since f is µ-integrable we can assume without loss of generality that f(x) < for all x. Also note that any other such f with dν = f dµ must equal f µ-a.e. To finish with the existence in Case it suffices to show that ν 2 µ. For each k N let = P k N k be a Hahn decomposition for ν 2 k µ, and let P = k P k and N = k N k = P c. Then for every k N, N is a negative set for ν 2 k µ, i.e., ν 2 (N) µ(n)/k, and so ν 2 (N) = 0. Thus it suffices to show µ(p ) = 0. Suppose not. Then µ(p k ) > 0 for some k. Note that P k is a positive set for ν 2 k µ. Put h = f + χ P k k. Since h dµ = M + µ(p k) > M, k we have h / F. However, for all A M we have h dµ = ν (A) + µ(a P k) A k ν (A) + ν 2 (A P k ) since A P k P k giving a contradiction. ν (A) + ν 2 (A) = ν(a), We now have existence in Case, and for the uniqueness suppose we have another such decomposition ν = ν + ν 2. Then ν ν = ν 2 ν 2 is both absolutely continuous and mutually singular with respect to µ, and so must be 0 (exercise). Thus µ i = µ i for i =, 2. Case 2. ν and µ are both positive and σ-finite. Then is a countable disjoint union n n with ν( n ), µ( n ) < for all n. Define finite measures ν n, µ n by ν n (A) = ν(a n ) and µ n (A) = µ(a n ). By Case, for each n we have unique measures ν n, ν2 n and a measurable function f n : [0, ) such that ν n = ν n + ν2 n, dν n = f n dµ n, and ν2 n µ n. Since µ n (n) c = 0, without loss of generality f n 0 on n. c For each n choose a measurable subset B n n such that ν n (B n ) = 0 and µ n ( n \ B n ) = 0. Put ν = n ν n ν 2 = n f = n ν n 2 f n.

16 6 PROFESSOR: JOHN QUIGG SEMESTER: SPRING 205 Then ν i are σ-finite positive measures for i =, 2 and f : [0, ) is measurable. For all A M we have f dµ = f n dµ n = ν n (A n ) = ν (A), A n A n n so dν = f dµ and ν µ. On the other hand, ν 2 (B) = n ν 2 (B n ) = n ν n 2 (B n ) = 0 and µ( \ B) = n µ( n \ B) = n µ n ( n \ B n ) = 0, and so ν 2 µ. This gives existence in Case 2, and uniqueness follows from uniqueness on each n. Case 3. ν is a σ-finite signed measure and µ is σ-finite and positive. Let ν = ν + ν be the Jordan decomposition. Then by Case 2 we have unique σ-finite measures ρ i, τ i (i =, 2) and measurable f, g : [0, ) such that Put ν + = ρ ρ 2 d ρ = f dµ ρ 2 µ ν = τ τ 2 d τ = g dµ τ 2 µ ν = ρ τ ν 2 = ρ 2 τ 2 h = f g; these are well-defined because, since ν + ν, we can assume that we have disjoint sets A, B such that ρ i and f are concentrated on A and τ i and g are concentrated on B. We then have ν = ν + ν 2 d ν = h dµ ν 2 µ. This gives existence in Case 3, and uniqueness follows from uniqueness on A and B. Definition In the Radon-Nikodym theorem, any such function f is a Radon-Nikodym derivative of ν with respect to µ, written so that f = dν dµ, dν = dν dµ dµ. In the Lebesgue Decomposition Theorem, ν and ν 2 are the absolutely continuous part and the singular part of ν with respect to µ, respectively, and ν = ν + ν 2 is the Lebesgue decomposition of ν with respect to µ.

17 MAT 57 REAL ANALSIS II LECTURE NOTES 7 Proposition Let ν be a signed measure and µ a positive measure. If ν, µ are σ-finite and ν µ, then f dν = f dν dµ dµ for all f L (ν). Proof. By the Jordan Decomposition Theorem and linearity, without loss of generality ν 0. We know it s true for characteristic functions, then for nonnegative simple functions by linearity, then for nonnegative measurable functions by the Monotone Convergence Theorem, and finally for integrable functions by linearity again. Corollary Let ν be a signed measure and µ, ρ positive measures. If ν, µ, ρ are σ-finite, ν µ, and µ ρ, then ν ρ and dν dρ = dν dµ dµ dρ. Proof. Case. ν 0. For any measurable set A we have d ν ν(a) = A d µ dµ d ν d µ = dρ by Proposition 5.24, d µ d ρ A where if d ν/d µ is not µ-integrable we express it as a limit of an increasing sequence of nonnegative µ-integrable functions and apply the Monotone Convergence Theorem. Case 2. ν is signed. Then either ν + or ν is finite, so either d ν + /d µ or d ν /d µ is µ- integrable, so with a little care in the calculation the result follows by linearity. 6. Differentiation We are ready to study Lebesgue s theory of differentiation. To pave the way, we begin with the following elementary covering lemma. Throughout, λ denotes Lebesgue measure. Lemma 6.. Let S be a family of open intervals and c < λ( S). Then there exists a finite pairwise disjoint subfamily F S such that λ(i) > c 3. I F Proof. Choose a compact subset K S such that λ(k) > c, and then choose a finite subfamily D S such that K D. Choose I D with maximal length. Then choose I 2 D that is disjoint from I and has maximal length, then I 3 D disjoint from 2 I j with maximal length, and so on until we have I,..., I k such that no I D is disjoint from k I j. Then for every I D \ {I,..., I k } there exists j such that I I j, and if j is the

18 8 PROFESSOR: JOHN QUIGG SEMESTER: SPRING 205 smallest such then λ(i) λ(i j ), so I Ĩj, where Ĩj denotes the open interval concentric with I j and 3 times the length. Thus K k Ĩj, so k k c < λ(k) λ(ĩj) = 3 λ(i j ). We call f : R R locally integrable if it is Lebesgue measurable and f dλ < for K every compact K R. Throughout this discussion, f will be locally integrable. For r > 0 define A r f : R R by A r f(x) = x+r f(y) dy = f. 2r λ(b r (x)) x r Definition 6.2. The Hardy-Littlewood maximal function of f is Hf : R R given by Hf(x) = sup A r f(x). r>0 B r(x) Lemma 6.3. For all a R, {Hf > a} is open. In particular, Hf is measurable. Proof. We first show that for each r > 0 the function A r f is continuous. Note that A r f(x) = fχ (x r,x+r). 2r Let x R and x n x. Without loss of generality x n x < for all n. Then for all y R and n N we have f(y) χ (xn r,x n+r)(y) f(y) χ (x r,x+r+) (y), and the function fχ (x r,x+r+) is integrable since f is locally integrable. For every y x we have lim x f(y) χ (xn r,x n+r)(y) f(y)χ (x r,r+r) (y), because the characteristic functions converge. Thus by the Dominated Convergence Theorem proving continuity at x. lim A rf(x n ) = A r f(x), n For every a > 0 and x R we have Hf(x) > a if and only if there exists r > 0 such that A r f (x) > a, so {Hf > a} = a>0{a r f > a}, R which is open, being a union of open sets. Theorem 6.4 (Hardy-Littlewood Maximal Theorem). For all a > 0, λ({hf > a}) 3 f. a

19 MAT 57 REAL ANALSIS II LECTURE NOTES 9 Proof. Let E a = {Hf > a} and c < λ(e a ). For all x E a choose r x > 0 such that A rx f(x) > a. The intervals I x := (x r x, x + r x ) cover E a, so there exists a finite subset F E a such that the intervals {I x } x F are disjoint and λ(i x ) > c 3. But then c < 3 x F Letting c λ(e a ) gives the result. Theorem 6.5. For a.e. x R, x F λ(i x ) < 3 a I x f 3 a lim A rf(x) = f(x). r 0 f. Proof. It suffices to show that for every n N we have A r f(x) f(x) for a.e. x < n, and then replacing f by fχ ( n,n+) we can assume that f is integrable. Given ε > 0, choose g C c (R) such that f g < ε. Note that A r g(x) g(x) for all x R (exercise). We have Thus, if for a > 0 we let lim sup A r f(x) f(x) r 0 = lim sup Ar f(x) A r g(x) + A r g(x) g(x) + g(x) f(x) r 0 lim sup A r (f g)(x) g(x) f(x) r 0 H(f g)(x) + f g (x). E a = {lim sup A r f f > a}, r 0 we have { E a H(f g) > a } { f g > a }, 2 2 so by Markov s Inequality and the Maximal Theorem we have λ(e a ) 6 a f g + 2 a f g < 8ε a. Letting ε 0 we get λ(e a ) = 0 for all a > 0. Since A r f(x) f(x) for all x / n E /n, the result follows. Corollary 6.6. For a.e. x R, lim r 0 x+r f(y) f(x) dy = 0 2r x r

20 20 PROFESSOR: JOHN QUIGG SEMESTER: SPRING 205 Proof. For all a Q we have lim r 0 x+r f(y) a dy = f(x) a 2r x r for all x / E a, where λ(e a ) = 0. Put E = a Q E a. Then λ(e) = 0. Let x / E and ε > 0. Since f is locally integrable, for a.e. x we have f(x) <, and then we can choose a Q with f(x) a < ε. Then lim sup r 0 2r x+r x r ( 2r lim sup r 0 = f(x) a < ε. Letting ε 0 gives the result. f(y) f(x) dy x+r x r Definition 6.7. The Lebesgue set of f is { x R : lim r 0 2r f(y) a dy + 2r x+r x r x+r x r } f(y) f(x) dy = 0. ) f(x) a dy Thus, the above theorem says that the complement of the Lebesgue set has measure 0. Definition 6.8. A family {E r } r>0 of sets shrinks nicely to x if there exists a > 0 such that for all r > 0 we have E r (x r, x + r) and λ(e r ) > ar. Corollary 6.9 (Lebesgue s Differentiation Theorem). For all x in the Lebesgue set of f, lim f(y) f(x) dy = 0 whenever E r shrinks nicely to x. r 0 λ(e r ) E r Proof. Let x be in the Lebesgue set of f, and let {E r } r>0 shrink nicely to x, with a > 0 as in the definition. Then f(y) f(x) dy f(y) f(x) dy λ(e r ) E r ar E r 2 x+r f(y) f(x) dy a 2r r 0 0. Definition 6.0. A signed Borel measure ν on R is called regular if its total variation ν is regular. Corollary 6.. Let ν be a regular signed Borel measure on R, and let dν = dρ + f dλ be the Lebesgue decomposition of ν with respect to Lebesgue measure λ, with ρ λ. Then for λ-a.e. x R, if {E r } r>0 shrinks nicely to x then x r ν(e r ) lim r 0 λ(e r ) = f(x).

21 MAT 57 REAL ANALSIS II LECTURE NOTES 2 Proof. Note that our hypotheses guarantee that ν is σ-finite, so the Lebesgue Decomposition Theorem applies. By the Lebesgue Differentiation Theorem it suffices to show that ρ(e r ) lim r 0 λ(e r ) = 0. Moreover, if a > 0 is as in the definition of shrinking nicely, then for all r > 0 ρ(e r ) λ(e r ) ρ (E r) ρ (x r, x + r) 2 ρ (x r, x + r) λ(e r ) λ(e r ) aλ(x r, x + r), so without loss of generality E r = (x r, x + r). By the Jordan Decomposition Theorem, without loss of generality ρ 0. Choose a Borel set B such that ρ(b) = λ(b c ) = 0. For each n N put { ρ(e r ) E n = x B : lim sup r 0 λ(e r ) > }. n It suffices to show that λ(e n ) = 0 for all n. Given ε > 0, choose an open U B such that ρ(u) < ε/n. For each x E n choose an open interval I x U centered at x such that ρ(i x ) λ(i x ) > n. Put V = x E n I x, and let c < λ(v ). Then there exists a finite subset F E n such that {I x } x F are disjoint and c < 3 x F λ(i x ) < 3n x F ρ(i x ) 3nρ(U) < 3ε. Letting c λ(v ) gives λ(v ) 3ε, and hence λ(e n ) 3ε, because E n V. Then letting ε 0 gives λ(e n ) = 0, as desired. A special case of Corollary 6. deserves to be single out: Corollary 6.2 (Fundamental Theorem of Calculus ). Let f L [a, b], and define F : [a, b] R by F (x) = x a f. Then F = f a.e. Using Lebesgue-Stieltjes measures, we can turn Corollary 6. into a statement concerning differentiation of increasing functions: Corollary 6.3. If f : [a, b] R is increasing, then f is differentiable a.e., and (6.) b a f f(b) f(a). Proof. Extend f to an increasing function on R by { f(a) if x < a f(x) = f(b) if x > b. Let g(x) = f(x+). Then g is increasing and right continuous. Let µ = µ g be the associated Lebesgue-Stieltjes measure. Then µ is a regular Borel measure. Let ν be the absolutely

22 22 PROFESSOR: JOHN QUIGG SEMESTER: SPRING 205 continuous part of µ with respect to λ, and let h = dν/dλ be the Radon-Nikodym derivative. Then by Corollary 6., for a.e. x [a, b] we have g(x + r) g(x) lim r 0 r = lim r 0 µ(x, x + r] λ(x, x + r] = h(x), since µ(x, x + r] = g(x + r) g(x) and the intervals (x, x + r] shrink nicely to x. Similarly, g(y) g(x) lim y x y x = h(x) a.e. x, since the intervals (y, x] for y < x shrink nicely to x. Thus g = h a.e. Note that f = g wherever f is continuous, and hence a.e. Let k = g f. Then k 0, and k(x) > 0 if and only if x is in the countable set D of discontinuities of f. We will show that k = 0 a.e., from which it will follow that f = g = h a.e. Note that, since f is increasing, ( ) f(x+) f(x) k(x) = x D x D f(b) f(a) <, so that the measure σ := x D k(x)δ x is finite, where δ x is the point mass at x. Thus by Corollary 6. k(x + r) k(x) k(x + r) + k(x) r r σ( x 2 r, x + 2 r ) r = 4 σ( x 2 r, x + 2 r ) λ ( x 2 r, x + 2 r ) since σ λ. r 0 0 a.e. x, Therefore b f = a (a,b] h dλ = ν(a, b] µ(a, b] = g(b) g(a) = f(b) g(a) f(b) f(a).

23 MAT 57 REAL ANALSIS II LECTURE NOTES 23 Bounded variation. To motivate what follows, consider two increasing right continuous functions F, F 2 : R R, and let µ i = µ Fi be the associated regular Borel measures on R. Suppose the F i are bounded. Then the µ i are finite, so µ := µ µ 2 is a finite signed Borel measure on R. Put F = F F 2. Then for every bounded right-half closed interval (a, b], µ(a, b] = µ (a, b] µ 2 (a, b] = ( F (b) F (a) ) ( F 2 (b) F 2 (a) ) = F (b) F (a). Thus, somehow the signed measure µ is characterized by the function F. We want to characterize the functions F that arise in this way. Note that F has the following two properties: () F is the difference of two increasing functions. (2) Whenever x 0 < x < < x n we have n F (x i ) F (x i ) n ( F (x i ) F (x i ) ) + = n µ (x i, x i ] + µ (R) + µ 2 (R) <. n ( F2 (x i ) F 2 (x i ) ) n µ 2 (x i, x i ] It turns out that each of these two properties, together with right continuity, characterize the functions F arising in this way. We call property (2) bounded variation, and we will now study such functions. To simplify the development, we restrict to (measures and) functions on a compact interval [a, b]; the results carry over, with some fussiness, to R. Let a = x 0 < x < < x n = b be a partition of [a, b]. Define n ( p = f(xi ) f(x i ) ) + n = n ( f(xi ) f(x i ) ) t = p + n = n f(x i ) f(x i ). Note that f(b) f(a) = p n. Also note that p, n, t cannot decrease if the partition is refined by adding more points. Now define P = sup p N = sup n T = sup t, where the sup is over all partitions of [a, b]. We write P b a, N b a, T b a if we want to emphasize the interval, and P f, N f, T f to emphasize the function. Definition 6.4. f is of bounded variation on [a, b] if T <.

24 24 PROFESSOR: JOHN QUIGG SEMESTER: SPRING 205 Thus, f is of bounded variation on [a, b] precisely when { n sup f(xi ) f(x i ) } : a = x0 < < x n = b is a partition of [a, b] <. Lemma 6.5. If f is of bounded variation on [a, b], then Proof. For any partition we have Taking the sup, Also Taking the sup, T = P + N f(b) f(a) = P N. p = n + f(b) f(a). P = N + f(b) f(a). t = p + n = 2p ( f(b) f(a) ). T = 2P ( f(b) f(a) ) = P + N. Theorem 6.6. A function is of bounded variation if and only if it is a difference of two monotone functions. Proof. First let f be of bounded variation. Define g, h : [a, b] R by Then g, h are increasing, and by Lemma 6.5 g(x) = P x a h(x) = N x a. f(x) = f(a) + g(x) h(x), a difference of the monotone functions f(a) + g and h. Conversely, every increasing function is of bounded variation, and the set of functions of bounded variation on [a, b] is a vector space, so a difference of two increasing functions is of bounded variation. Corollary 6.7. Every function of bounded variation is differentiable a.e., and its derivative is integrable. Proof. We know that every increasing function satisfies the conclusions, so the corollary follows from Theorem 6.6. Example 6.8. Every Lipschitz function has bounded variation. Example 6.9. The function f : [0, ] R defined by { x sin(/x) if 0 < x f(x) = 0 if x = 0 is not of bounded variation.

25 MAT 57 REAL ANALSIS II LECTURE NOTES 25 We can now complete the characterization mentioned at the beginning of this section: Theorem If µ is a finite signed Borel measure on [a, b], then the function F : [a, b] R defined by F (x) = µ(a, x] is right continuous and of bounded variation. Conversely, if F is right continuous and has bounded variation on [a, b], then there is a unique finite signed Borel measure µ = µ F on [a, b] such that µ(a, x] = F (x) F (a) for all x [a, b]. Proof. We established the first part in our motivating discussion. Conversely, assume that F is right continuous and of bounded variation on [a, b]. Then F = F F 2 for increasing functions F, F 2, and since F is right continuous we have F (x) = F (x+) = F (x+) F 2 (x+), so we can assume that F, F 2 are right continuous. Let µ i = µ Fi be the associated Lebesgue- Stieltjes measures, and put µ = µ µ 2. Then µ is a finite signed Borel measure in [a, b], and for x [a, b] we have µ(a, x] = µ (a, x] µ 2 (a, x] = ( F (x) F (a) ) ( F 2 (x) F 2 (a) ) = F (x) F (a). In the above theorem, we call µ the Lebesgue-Stieltjes signed measure associated to F. Note that F is continuous (rather than merely right-continuous) if and only if µ({x}) = 0 for every x, equivalently the value µ(i) for an interval I is unchanged if endpoints of I are added or removed. It is tempting to imagine that the properties of µ F should reflect those of F. The following result gives some confirmation of this. Recall that if F has bounded variation on [a, b] then F L [a, b]. Proposition 6.2. Let F be right continuous and have bounded variation on [a, b], and let µ = µ F be the associated Lebesgue-Stieltjes signed measure. Then: () µ λ if and only if F = 0 λ-a.e. (2) µ λ if and only if F (x) = F (a) + x a F for all x [a, b]. Proof. We have seen before that F is the Radon-Nikodym derivative of the absolutely continuous part of µ with respect to λ, and the result follows. Absolute continuity. To motivate what follows, let F be right continuous and of bounded variation on [a, b], and let µ = µ F be the associated finite Lebesgue-Stieltjes signed measure on [a, b]. Suppose µ λ. We know that this is equivalent to F (x) = F (a) + x F for all a x [a, b], but now we will find another property of F that reflects the absolute continuity of µ. We have µ λ also, so for all ε > 0 there exists δ > 0 such that for every Borel set A, λ(a) < δ implies µ (A) < ε. In particular, if A is a finite disjoint union of right half closed intervals (a i, b i ] in [a, b], then i (b i a i ) < δ implies µ(ai, b i ] µ (a i, b i ] < ε. i i

26 26 PROFESSOR: JOHN QUIGG SEMESTER: SPRING 205 We conclude that F has the following property: for all ε > 0 there exists δ > 0 such that n F (b i ) F (a i ) < ε whenever {(a i, b i )} n are disjoint intervals in [a, b] with n (b i a i ) < δ. We call this property absolute continuity, and we will show that it characterizes right continuous functions of bounded variation with absolutely continuous Lebesgue-Stieltjes signed measures. Remark Every absolutely continuous function on [a, b] is uniformly continuous, since we can take n = in the definition of absolute continuity. In the other direction, every Lipschitz function, and hence every differentiable function with bounded derivative, is absolutely continuous. Theorem If F is right continuous and of bounded variation on [a, b], and if µ F λ, then F is absolutely continuous on [a, b]. Conversely, if F is absolutely continuous on [a, b], then it has bounded variation, and µ F λ. Proof. We established the first part in our motivating discussion. Conversely, let F be absolutely continuous on [a, b]. We must first show that F has bounded variation. Choose δ corresponding to ε = in the definition of absolute continuity. Let I = [a, b]. Choose nonoverlapping intervals I,..., I N such that I = N I j and l(i j ) < δ for all j =,..., N. Let P = {x 0,..., x n } be a partition of I. Refine P by adding the endpoints of the I j s, which can only increase the t-sum for P. Then P is a union of partitions P,..., P N of I,..., I N, and we have N t-sum for P = t-sum for P j < j= N (by choice of δ) j= = N. Taking the sup over P, we get T N, so F has bounded variation. Now we must show that µ = µ F is absolutely continuous with respect to Lebesgue measure λ. Let A [a, b] be a Borel set with λ(a) = 0. We want to show µ(a) = 0, and without loss of generality A (a, b) since F is continuous. Let ε > 0, and choose δ > 0 as in the definition of absolute continuity. Since µ(a) = inf{µ(u) : U A is an open subset of (a, b)}, we can find open sets U U 2 A in (a, b) such that µ(u k ) µ(a), and without loss of generality λ(u k ) < δ for all k. For each k, U k is a disjoint union of a finite or infinite

27 MAT 57 REAL ANALSIS II LECTURE NOTES 27 sequence of open intervals {(a ki, b ki )} i, and whenever {(a ki, b ki )} n i= is a finite initial segment of this sequence we have n i= (b ki a ki ) < δ, so n µ(aki, b ki ) n = F (b ki ) F (a ki ) < ε. i= Taking the sup over n gives ( µ(u k ) = µ (a ki, b ki )) µ(a ki, b ki ) ε. i i Letting k we get µ(a) ε, and then letting ε 0 gives µ(a) = 0. Combining Theorems 6.2 and 6.23, we get: i= Corollary 6.24 (Fundamental Theorem of Calculus 2). F : [a, b] R is absolutely continuous if and only if it is a.e. differentiable and F (x) = F (a) + x a F for all x [a, b]. Example The Cantor function F is continuous and increasing, with F = 0 a.e. F is not absolutely continuous, since F () F (0) = 0 0 F. 7. Extension to complex values It s important to be aware that much of the analysis we developed can be extended to allow complex values. Here we outline what s involved. First of all, we can allow complex normed spaces, where C is the field of scalars. There is not much new here, fundamentally every complex normed space can be turned into a real normed space by just restricting the scalar multiplication to real numbers. The associated metric space does not change. In finite dimensions this doubles the dimension C n is essentially the same metric space as R 2n. The test for completeness involving absolute convergence of series carries over to complex normed spaces. Most of the properties of bounded and of continuous real-valued functions carry over to the complex-valued case. In particular, the complex versions of l () and C b () are Banach spaces. Next, if (, M) is a measurable space, we define a function f : C to be measurable if the real and imaginary parts Ref and Imf are measurable. It s routine to check that this is equivalent to the condition that f (B) M for every Borel subset B C. The usual arithmetic properties of measurable functions carry over, and every measurable function is the limit of a sequence of simple ones. Note that we can regard real-valued measurable functions as a special case; however, once we allow complex-valued functions there is no longer a possibility of allowing extendedreal-valued functions in measure and integration theory there is no extended complex plane.

28 28 PROFESSOR: JOHN QUIGG SEMESTER: SPRING 205 We integrate complex-valued measurable functions f : C by f dµ := Ref dµ + i Imf dµ, except now we must require both Ref, Imf to be integrable. A convenient test for integrability, namely f is measurable and f is integrable, carries over from the real case. Essentially all the basic theory of integration carries over to the complex case, most notably the Dominated Convergence Theorem, completeness of L (, M, µ), and Fubini s Theorem. Complex measures are countably additive functions ν : M C, except now we must require that both the real and imaginary parts Re ν, Im ν to be finite signed measures. In fact, every complex measure is finite, again because there is no extended complex plane. If µ is a (positive) measure, we say ν µ if Re ν, Im ν µ, and similarly for ν µ. The Lebesgue Decomposition and Radon-Nikodym Theorems carry over. There is one subtlety: the total variation measure ν is much harder to define, although it is still characterized as the smallest positive measure satisfying ν (E) ν(e) for all E M, and also is characterized by { n } ν (E) = sup ν(e i ) : E,..., E n is a measurable partition of E. For a complex measure ν, the total variation ν is always a finite measure. The Lebesgue Differentiation Theorem carries over, as do both versions of the Fundamental Theorem of Calculus. Also the theory of bounded variation and absolute continuity for functions on an interval carry over to complex-valued functions.