MA359 Measure Theory

Transcription

1 A359 easure Theory Thomas Reddington Usman Qureshi April 8, 204 Contents Real Line 3. Cantor set General easures 2 2. Product spaces Outer measures easurable Functions and Integration Devil s staircase Integration for f X [0, ] Integration for general f X R {±} Different modes of convergence Product measures Signed easures 49 5 Crash Course on L p Spaces 54

2 Introduction These lecture notes are a projection of the A359 easure Theory course 203/204, delivered by Dr José Rodrigo at the University of Warwick. These notes should be virtually complete, but the tedious treasure hunt of errors will always be an open game. And, obviously, completeness and accuracy cannot be guaranteed. If you spot an error, or want the source code to fiddle with in your way, send an to me@tomred.org. We hope these are helpful, and good luck! Tom and Usman Useful links. The up-to-date version of these notes should be found here: 2. Failing that: 3. Students taking this course should also take a look at Lewis Woodgate s Skydrive notes: and Alex Wendland s Dropbox notes: 2

3 We want to measure every subset of R. i.e. we want a map: m P(R) [0, ] parts of R where m(interval (a, b)) = b a (same with intervals such as [a, b)). e.g. m((, 2)) = 2. Wish list form: A = n= m((a, b)) = b a m(a) = m(a + h) ( A R, h R) A n m(a) = n= m(a n ) Claim. There isn t such an m. Goal: Construct a subset R, such that it is impossible to assign a measure and satisfy the proposition in the wish list. Real Line Agree on the measure of intervals: Definition. Let A R: m (A) = inf { I = (a, b) = [a, b) = (a, b] = [a, b] m(i) = usual length of I = b a k= I k I k are open intervals and A I k } k= where I k is the length of the interval I k. m is the outer measure. Construction:. Cover A by lots of open intervals. 2. Summing the lengths creates a set in [0, ]. 3. Compute the infimum (the existence of which is trivial, as the set in question is bounded below). Proposition (Properties of m ).. 0 m (A) A R. 2. m (Q) = 0 (surprising because Q is dense). 3. m is defined for P(R) (it is defined for every subset of R). 4. m (A) m (B) whenever A B. 3

4 5. m (I) = I for any interval I. 6. m (A + h) = m (A) h R, A R. Proof.. For any interval I, I 0, and as m (A) is a greatest lower bound, m (A) 0 A. 2. Take {x n } to be an enumeration of Q. Define: I n = (x n ε 2, x n+ n + ε 2 ) n+ for any fixed ε > 0. otice I n = ε 2 n and Q ( n= I n ). Since: we have: m (Q) = inf { n= m (Q) J n J n open and Q J n } n= n= I n = n= ε 2 n = ε So 0 m (Q) ε for any ε 0. By sending ε 0 m (Q) = 0 3. k= I k is defined as it is a limit of an increasing sequence, so our infimum will always be defined. 4. Every open cover of B by intervals also covers A the collection of elements over which we compute m (A) m (B). 5. We will show two inequalities: m (I) I m (I) I ow for I m (I): Take a cover of I, {I n } n=. Since I = [a, b] there exists a finite subcover of I (as it s compact and closed). Upon relabelling the sets, say: I, I 2,..., I We have our finite subcover of I: I I 3 I ( ) ( ) ( ) [ ] a a 2 a 3 a 4 ( ) I 2 a + As these are all open, there is overlap inbetween the open intervals. Choose a point in each of the overlapping intervals. i.e. choose a j from each I j I j+. So (a, a 2 ) I, (a 2, a 3 ) I 2 and (a j, a j+ ) I j. ow: b a a + a = a + a + a a + a... + a 2 a = a j+ a j As a j+ a j I j. Then: j= I = b a a j+ a j I j I j j= j= j= (for all open covers) 4

5 Which implies: I inf I j I j open... = m (I) j= ow for I m (I): Say I = [a, b]. Define I = (a ε 2, b + ε 2 ), I j = j 2. [a, b] j= I j m (I) I b a + ε = I + ε ε > 0 as ε 0, m (I) I. 6. Reason is I + h = I. The only property that we do not have is: m (A B) = m (A) + m (B) (it s false) Another observation: A A n n= m (A) m ( n= A n ) n= m (A n ) Proof. ε > 0 there exists a countable collection of open intervals {I n,k } k= such that: k= If n= m (A n ) =, there is nothing to prove. Else, sum ( ) w.r.t. n: Want to show: This is true ε > 0 so send ε 0.. Cantor set n,k= m ( A n ) n= I n,k m (A n ) + ε 2 n (*) I n,k n= m (A n ) + ε I n,k ( m (A n )) + ε (**) n,k n= 0 C C C 2 5

6 C = n= C n but there exists a bijection between C and R. m (C) m (C n ) 2 n ( n 3 ) 0 (as n ) So m (C) = 0. Thus measure and cardinality do not mix well... Definition 2. We say that A R is measurable iff: m (E) = m (E A) + m (E A c ) ( E R) Remark. It is enough to show m (E) m (E A) + m (E A c ) E R. This is because: E (E A) (E A c ) m (E) m ((E A) (E A c )) m (E A) + m (E A c ) by the above proposition. Example (Examples of measurable sets). Q is measurable, as m (Q) = 0. Any set A R with m (A) = 0. Proof. (E A) A m (E A) m (A) = 0 (E A c ) E m (E A c ) m (E) m (E) m (E A) + m (E A c ) R Q is by the lemma below. Lemma. A measurable A c measurable. Proof. m (E) = m (E A c ) + m (E (A c ) c ) Proposition 2. Intervals are measurable. Proof. Want to show: m (E) = m (E I) + m (E I c ) ( E R) First, take an open cover of E by intervals, say {E k } k=. I E k is an interval k. I c E k is at most two intervals k. (From {E k } it isn t possible to construct (open) covers of I E and I E c ) (I E k ) A k for A k an open interval. (I c E k ) (B k C k ) for B k, C k open intervals. Choose such that: A k + B k + C k < I k + ε 2 k ow, {I k } cover E: ( I k + ε n= 2 ) k A k + B k + C k n= 6

7 Also: k= (A I k ) k= A k & (A I k ) = A ( I k ) k= k= & A E ( I k ) k= (Similarly for B k + C k ) So: k= By taking the infimum over all possible covers: Finally, let ε 0. A k + B k + C k m (E A) + m (E A c ) m (E A) + m (E A c ) ( I k ) + ε k= m (E A) + m (E A c ) m (E) + ε Proposition 3. A, B measurable A B and A B measurable. Proof. We know m (F ) = m (F A) + m (F A c ) F. Take F = E (A B) for some E. We want: ow: For intersection: m (E) = m (E (A B)) + m (E (A B) c ) m (E (A B)) = m (E (A B) A) + m (E (A B) A c ) m (E) = m (E A) + m (E A c ) = m (E A) + m (E B A c ) (as A is measurable) = m (E A) + m (E A c B) + m (E A c B c ) (as B is measurable) = m (E (A B)) + m (E (A B) c ) A c and B c measurable A c B c measurable (A c B c ) c measurable A B measurable. Proposition 4. Let A,..., A measurable and pairwise disjoint. Then: m (E A i ) = i= m (E A i ) i= ( E) ote, if E = R, then: m ( A i ) = i= m (A i ) i= 7

8 Proof. By induction. = is trivial. Assume true for,...,. Then: m (E + n= A i ) = m ((E + n= A i ) A + ) + m ((E + A i ) A c +) n= = m (E A + ) + m (E A i ) n= = m (E A + ) + + = i= (E A i ) n= m (E A i ) (by induction) Proposition 5. Let {A i } i= be measurable. Then i= A i is measurable. oreover, if A i are pairwise disjoint then: m ( A i ) = i= m (A i ) Proof. Let B = i= A i & B n = n i= A i which is measurable by a previous proposition. Want to show: i= m (E) = m (E B) + m (E B c ) Assume for the moment that A i are pairwise disjoint. We know that m (E) = m (E B n ) + m (E B c n): Thus: B n B B c B c n (E B c n) (E B c ) ow, LHS RHS and LHS is independent of n, so: ow, consider: m (E B c n) m (E B c ) m (E) m (E B n ) + m (E B c ) m (E B n ) +m (E B c ) = m (E n i= A i ) n m (E A i ) + m (E B c ) i= LHS lim n RHS m (E) m (E A i ) + m (E B c ) i= m (E B) = m (E A i ) i= = m ( (E A i )) i= m (E A i ) i= 8

9 So m (E) m (E B) + m (E B c ). But: Thus, take E = i= A i, then: m (E) m (E B) + m (E B c ) m (E) = m (E A i ) + m (E ( c A i ) ) ( E) i= i= m ( A i ) = i= m (A i ) + m ( ) Finally, need to show the extra hypothesis of pairwise disjoint. Define: i= W = A W 2 = A 2 A = A 2 A c W 3 = A 3 (A A 2 ) W n = A n ( n A i ) i= (measurable) (measurable) Thus: W i are measurable and pairwise disjoint. i= A i = W i i= Observation: {B i } i= measurable i= B i measurable. Proposition 6. List of properties of measurable sets: Complements, countable unions and intersections of measurable sets are measurable. Intervals are measurable. (Countable additivity) A i measurable and pairwise disjoint m ( i= A i ) = i= m (A i ). (Countinuity) A i measurable and A A 2... A n A n+... and B i measurable and B B 2... B n B n+... and m (A ) <. Then i= A i & i= B i are measurable. oreover: m ( A i ) = lim i= i m (A i ) & m ( (Translation invariance) A measurable A + h measurable. Open and closed sets are measurable. i= m (A) = m (A + h) B i ) = lim i m (B i ) (Approximation property) A measurable, then ε > 0 B closed, C open, B A C s.t. m (C B) < ε. oreover, if m (A) < then B can be taken compact. Proof of continuity. First, we don t need m (A ) <, we need m (A n ) < for some n, as for m (A ) =, m (A 2 ) <. We have: i= A i = A i i=2 9

10 as A A 2. Let s do B i s first. B B 2 B 3... Create a disjoint collection whose union is i= B i. Define: otice: so: lim n m (B n ) = lim m ( n n i= C = B C 2 = B 2 B C 3 = B 3 B 2 C n = B n B n i= n C i = B i, i= C i ) = lim m (C i ) = n i= i= m (C i ) = m ( i= C i ) = m ( B i ) i= A A 2... measurable and m (A ) <. Construct increasing set, define D n = A A c n measurable. D D 2... So we know m ( n= D n ) = lim n m (D n ), and: A = A n D n m (A ) = m (A n ) + m (D n ), ( n)( ) and: So: and, by ( ): A = A n D n n= n= as m (A ) < and m ( ) 0. So, by ( ): m (A ) = m ( A n ) + m ( D n ). n= n= m (A ) = m ( A n ) + lim m (D n ) n= n m (A ) = lim n m (A n ) + lim n m (D n ) ( ) m ( A n ) = lim (A n ) n= n Proof that open & closed sets are measurable. Claim. Every open set in R can be written like: U = I n (for open intervals I n ) n= Thus open sets in R are measurable. So closed are too as complements are measurable. Proof of approximation property. A is measurable, want to show that ε > 0 B, C with B A C s.t. m (C B) < ε. First, assume A J, J a closed & bounded interval. Since m (A) <, there exists a cover {I j } j= by open intervals: A such that: j= I j j= I j m (A) + ε 2 0

11 and: Define: m ( I j ) j= I j m (A) + ε 2 j= C = I j j= C is open and m (C) m (A) ε 2. = m (C A) as A C To find B, consider J A (which is measurable). We can find an open set O s.t. (J A) O and: m (O) m (J A) < ε 2 ( ) Define B = J O = J O c (closed from finite intersections of closed sets). As B is measurable: So: m (C) = m (B C) +m (C B c ) =B = m (B) + m (C B) m (C B) = m (C) m (B) So all that is left is to show m (A) m (B) < ε 2 : = m (C) m (A) +m (A) m (B) ε 2 m (J) m (O B) m (O) + m (B) < m (J A) + m (B) + ε 2 But: m (J) = m (A) + m (J A) m (A) + m (J A) < m (J A) + m (B) + ε 2 m (A) < m (B) + ε 2 (as m (J A) < ) m (A) m (B) < ε 2 Finally, we remove our assumption. Define: {A n } = A [n, n + ] For each A n fine B n A n C n with B n closed, C n open and m (C n B n ) B n A n C n ε 2 n. Then: So: And let C = C n, B = B n open. Exercise to show that B n is closed. Also: B n A C n m (C B) m ( n= (C n B n )) n= ε 2 n = mε (for some number m)

12 2 General easures Let X be any non-empty set. Definition 3. An algebra of sets on X is a non-empty collection A that satisfies: Y A Y c A Y,..., Y n A n Y i A i= Definition 4. A σ-algebra of sets on X is a non-empty collection of sets A that satisfies: A is closed under complements A is closed under countable unions Observations: Collection of measurable sets from Chapter is a σ-algebra. Example 2. Let X be any infinite set. Consider: Exercise: check A is a σ-algebra. Observations: Every σ-algebra is an algebra If A is an algebra, A, X A A = {E X such that E countable or E c countable} The word union can be changed with intersection (Exercise) Proposition 7. An algebra that is closed under countable disjoint unions is a σ-algebra. Proof. Given {A i } i=, A i A we want to show: i= A i A Construct out of {A i } i= a collection of pairwise disjoint collection sets B i such that i= B i = i= A i : B = A It is clear they are disjoint by construction B 2 = A 2 A B n = A n n A i i= n i= i= A i = n B i i= A i = B i A i= (by assumption) Observation: any arbitrary intersection of σ-algebras is a σ-algebra Definition 5. Let X be a non-empty set, a σ-algebra of X. A function µ [0, ] is a measure if it satisfies: µ( ) = 0 2

13 Countable additivity i.e {E i } i=, E i pairwise disjoint. Then: Definition 6. Any µ [0, ] such that µ( ) = 0 if {E i } i=, E i pairwise disjoint, then: µ ( E i ) = i= µ ( E i ) = i= µ(e i ) i= µ(e i ) is called a finitely additive measure (not necessarily a measure). otation: A pair (X, ), where X is a non-empty set, is a σ-algebra, is called a measurable space. A triplet (X,, µ) is a measure space. Given (X,, µ), if µ(x) < then µ is a finite measure. If µ(x) = but there is a collection of sets {E i } i= such that E i, i= E i = X and µ(e i ) <, then µ is σ-finite. Example 3. Let X = R, be the collection of measurable sets from Chapter, µ = m. Then µ(x) = m (R) = but if E n = [ n, n], then n= E n = X and µ(e n ) = 2n <, so m is σ-finite. Remark 2. If whenever µ(e) =, E, then F E, F such that µ(f ) < then µ is called a semi-finite measure. Example 4. Consider a non-empty X, = P(X), let f X [0, ]. Define: i= µ(e) = f(x) x E Clearly, µ( ) = 0. Also, countable additivity holds as sums are positive and can be rearranged. f then µ(e) counts elements. Let: if x = x 0 f(x) = 0 otherwise But, take A = [, ], B = [ 2, 2], X = R, x 0 = 0. Then: µ(a B) µ(a) + µ(b) Example 5. Let X be an infinite set, = P(X). Define: µ(e) = 0 µ(e) = (if E is finite) (if E is infinite) Claim. This is not a measure but is a finitely additive measure. 3

14 Example 6. Let X be an infinite set, and consider: = {E X E is countable or E c is countable} Define: Exercise: prove this is a measure. 0 if E countable µ(e) = otherwise Theorem. Let (X,, µ) be a measure space. The following are true:. onotonicity: if E F, E, F. Then µ(e) µ(f ). 2. Subadditivity: {E j } then: µ ( E j ) j= µ(e j ) j= 3. Continuinity from below: {E j }, E E 2 E 3... Then: µ ( E j ) = lim µ(e j ) j= j 4. Continuity from above: {F j }, F F 2 F 3... Then: Proof.. µ(f ) = µ (E (F /E)) = µ(e) + µ(f /E) 0 µ ( F j ) = lim µ(f j ) j= j µ(e) µ(f ) 2. Construct A i such that A i are pairwise disjoint and i= A i = i= E i : A = E A 2 = E 2 E = E 2 E c A n = E n n i= The {A i } i= is clearly pairwise disjoint, A i i, and: 3. E 0 =. Define E i = E n ( n c E i ) i= µ ( E i ) = µ ( A i ) = i= i= A = E = E E 0 A 2 = E 2 E A i = E i E i µ(a i ) i= µ(e i ) i= 4

15 We can do this because the E i are nested. A i, A i pairwise disjoint. So: µ ( E i ) = µ ( A i ) = i= i= µ(a i ) = µ(e i E i ) i= i= = (µ(e i ) µ(e i )) i= n = lim (µ(e i ) µ(e i )) n i= = lim n µ(e n ) µ(e 0 ) = lim n µ(e n ) (telescopic sum) [ F ] B 2 [ F 2 B 2 ] B 3 [ F 3 ] B 3 B i [ ] F i B i 4. F F 2..., define: B i = F c i F. So F = B i F i i, and we have: F = ( n F i ) ( n B i ) i= i= As we have a disjoint union for F, we can say that µ(f ) = µ ( F i ) + µ ( B i ) (*) i= i= µ(f ) = µ(f i ) + µ(b i ) (**) With some playing around and by use of the diagram, we can see that B B 2 B 3..., so µ(f ) = µ ( F i ) + lim µ(b i ) (by applying 3. on (*) to µ ( i= B i )) i= i µ(f ) = lim µ(f i ) + lim µ(b i ) i i lim µ(f i ) = µ ( F i ) i i= Definition 7. Let (X,, µ) be a measure space. If E satisfies µ(e) = 0, then we say that E is null. Definition 8. If a statement about points in X is true for all x X except for a set of measure 0, then we say that the statement is true almost everywhere (a.e.). Observation: µ(e) = 0 and F E, then µ(f ) = 0 provided F. Definition 9. A measure whose domain (i.e ) contains every subset of every null set is called complete. Example 7. Consider (R,, µ) where are measurable sets and µ = m. Then every point in R is irrational is true almost everywhere. 5

16 Theorem 2 (Solve lack of completeness of some measures). Let (X,, µ) be a measure space. Let: = {E µ(e) = 0} = {E F E, F where } Then is a σ-algebra, and there exists a unique extension µ of µ to. Proof. We want to show firstly that is closed under. Countable unions 2. Complements. Take A i. We want i= A i. We can write A i as A i = E i F i with E i, F i i, i. i= A i = ( i= E i ) ( F i ) i= To show i= A i, we need ( i= is a σ-algebra. Define := E i ) and ( i= i= i, so ( i= We need µ() = 0, but µ() µ( i ) = 0 so i= F i ) with We know ( E i ) because i= F i ) i =, because i. i= 2. We need A A c. Let A = E F, E, F,. w.l.o.g., E = (otherwise take F E, E instead of F and ). eed (E F ) c to be in. We need to write E c F c as Ẽ F with Ẽ, F Ñ. Using E =, we can derive the identity: so: E F = (E ) ( c F ) (E F ) c = (E ) c ( c F ) c We need (E ) c and ( c F ) c Q, Q. As E, (E ) c. Also, ( c F ) c = F c, so by taking Q = we have that ( c F ) c Q, Q Having shown that is a σ-algebra, we want to now show there exists a unique µ. Given A, assume A = E F. This decomposition is not unique unfortunately. Define µ(a) = µ(e). It is trivial that µ is an extension. ow w.t.s. µ is well defined and unique. Assume A = E F = E 2 F 2 where E, E 2, F, F 2 2. We need: µ(e ) = µ(e 2 ) By sandwiching A in-between two sets that are in we have E (E 2 F 2 ) E 2 2 µ(e ) µ(e 2 2 ) µ(e 2 ) + µ( 2 ) =0 So µ(e ) µ(e 2 ). By replacing 2 we have µ(e 2 ) µ(e ) 6

17 We ve defined σ-algebra and measure abstractly. How do you construct σ-algebra? Observation: an arbitrary intersection of σ-algebra (in X) is a σ-algebra. Use of observation: Suppose E is a collection of sets in X. We can define (E) as the smallest σ-algebra that contains E, where smallest is the intersection of all σ-algebras. ((E) exists because P(X) is a σ-algebra containing E) Definition 0. (Borel σ-algebra) The smallest σ-algebra of X that contains the open sets is called the Borel σ-algebra, denoted B X. What we constructed in Chapter is bigger (the Lebesgue σ-algebra). Proposition 8. Let X = R using usual topology. B R = (open sets in R) Then. B R = (intervals (a, b)) 2. B R = ([a, b]) 3. B R = ([a, b)) = ((a, b]) 4. B R = ((a, )) = ((, a)) 5. B R = ([a, )) = ((, a]) Proof. Exercise. The motivation is that soon we will define measurable functions as f (E) is measurable for all E which is measurable (analogously to continuous functions on open sets in topological spaces). Having the proposition will simplify things. We will need f (G) for G in any of the smaller families. 2. Product spaces How do you think of R 2? R 2 or R R? Let X α be non empty sets, α Λ (in principle an uncountable index set). Consider: X = X α α Λ Define π α X X α, where π α is the projection onto the α co-ordinate. The product σ-algebra on X is the σ- algebra generated by {π α (E α ) E α α }. This is denoted by: (eed to check with lecturer) α α Λ Proposition 9. The product σ-algebra we ve just defined is also the σ-algebra generated by: provided Λ is countable. E α (E α α ) α Λ Definition. A Banach space is separable if there exists a countable subset that is dense Proposition 0. (About Borel sets) Let X,..., X n be metric spaces. X = n i= X i equipped with the product metric. Then n B Xj j= B X where if the X j are all separable sets then we have equality. 7

18 2.2 Outer measures We need to find a way to construct measures, so we have a definition. Definition 2. An outer measure in a (non-empty) set X is a function µ P(X) [0, ] that satisfies:. µ ( ) = 0 2. If A B then µ (A) µ (B) 3. µ ( A i ) i= µ (A i ) i= Proposition. Let E P(X), X non empty, such that X E, E. Let p E [0, ] be any function such that p( ) = 0. Then define for A P(X): This is an outer measure. Proof. We need: µ (A) = inf p(e j ) E j E, A E j j= j=. µ ( ) = 0. Because p( ) = 0, from the definition it follows that µ ( ) = µ (A) µ (B) whenever A B. This is trivial. Every cover of B is a cover of A. When you look at µ (A) you are taking the infimum over a bigger set, so the result follows. 3. µ ( A i ) µ (A i ) i= i= Consider A k. By properties of infimum there exists {E k,j } j=, where E k,j E such that: and taking unions in k gives: j= p(e k,j ) µ (A k ) + ε 2 k ( ε > 0) A k E k,j j= A k k= k= j= E k,j (i.e. a countable cover of the set by elements in E). This implies: µ ( A k ) k= By letting ε 0 we get the result. p(e k,j ) k= j= k= (µ (A k ) + ε 2 ) = k k= µ (A k ) + ε Definition 3. Let X be non-empty, µ an outer measure. We say that A X is measurable iff: µ (E) = µ (E A) + µ (E A c ) ( E P(X)) Theorem 3. (Caratheodory) Let µ be an outer measure on a non-empty set X. Then the collection of measurable sets, denoted by is a σ-algebra, and moreover the restriction of µ to is a complete measure. 8

19 Proof. We want to show that:. is a σ-algebra i.e.: a) is closed under complements. b) is closed under countable unions. 2. µ is a measure, i.e.: a) µ ( ) = 0 b) µ ( i= A i ) = i= µ (A i ), A i pairwise disjoint 3. µ is complete.. a) This is trivial from the definition (write A c instead of A). b) We first show that is an algebra, then is closed under a countable disjoint union, which implies from a previous proposition that is a σ-algebra. Claim. is an algebra. Consider A, B. We want A B (by induction we can then prove it for finite n (exercise)). We know that: We want: µ (E) = µ (E A) + µ (E A c ) ( E) µ (F ) = µ (F B) + µ (F B c ) ( F ) µ (E) = µ (E (A B)) + µ (E (A B) c ) So by using the definition of A and B being measurable we have We want to show: so if: µ (E) = µ (E A) + µ(e A c ) = µ (E A B) + µ (E A B c ) + µ (E A c B) + µ (E A c B c ) =µ (E (A B) c ) µ (E) µ (E (A B)) + µ (E (A B) c ) µ (E A B) + µ (E A B c ) + µ (E A c B) µ (E (A B)) then we re done. By set theory, A B ((A B) (A B c ) (A c B)). So E (A B) ((E A B) (E A B c ) (E A c E)), hence by properties of outer measure we ve shown A B is measurable. Assume A B =, A, B. Take E = A B. As A is measurable: µ (E) = µ (E A) + µ (E A c ) µ (A B) = µ (A) + µ (B) By induction we can show that µ ( A i ) = µ (A i ), with A i pairwise disjoint. Hence we have i= i= shown that is an algebra. To show it is a σ-algebra, it is enough to show {A i } i=, A i pairwise disjoint, A i A i i= Let B = A i, B n = n A i i= i= 9

20 We know B n. We want B, i.e. µ (E) = µ (E B) + µ (E B c ). Since A i : µ (F ) = µ (F A n ) + µ (F A c n) ( F ) Take F = E B n Details for the proof on the left B n A n = A n µ (E B n ) = µ (E B n A n ) + µ (E B n A c n) = µ (E A n ) + µ (E B n ) = µ (E A n ) + µ (E A n ) + µ (E B n 2 ) (inductively) n = µ (E A i ) i= B n µ (E) = µ (E B n ) + µ (E Bn) c µ (E B n ) + µ (E B c ) n = (E A i ) + µ (E B c ) i= ( n A n) A n = A n i= B n A c n = B n n i=a i A c n = ( n A i A n) A c n i= B n B = n i= B c B c n (E B c ) (E B c n) (A i A c n) =A i µ outer measure so µ (E B c ) µ (E B c n) 2. a) Trivial. So is complete. LHS indep. of n b) Rewrite * for E = B: RHS dep. of n LHS lim n RHS µ (E) µ (E A i ) + µ (E B c ) i= µ ( A i ) = µ (B) = i= µ (E B) + µ (E B c ) ( ) (Reason: E B (E A i )) µ (B A i ) + µ (B B c ) = µ (B A i ) = µ (A i ) i= 3. We require that if A, µ (A) = 0 then for every W A we have W. Take W A. We want: E W E A A, so: i= µ (E) µ (E W ) + µ (E W c ) ( E) µ (E W ) µ (E A) µ (A) = 0 =0 So we need µ (E) µ (E W c ). This is trivial as E W c E. i= echanism for constructing σ-algebra. We have a way (Caratheodory) to construct measures that come with a σ-algebra. eed a family ξ and a family ρ The outcome is that a measure and a σ-measure of measurable sets. In general (ξ) is not the σ-algebra of measurable sets! 20

21 In R, ξ = open intervals (ξ) = Borel set. What comes out of Caratheodory is strictly bigger. That σ-algebra is called the Lebesgue σ-algebra, denoted by L L is the completion of B Theorem 4 (Describing all possible measures in R, B R ). If F R R is increasing, right continuous, then there exists a unique Borel measure, µ F that satisfies µ F ([a, b]) = F (b) F (a) (If there exists another such function G then G = F + constant). Conversely, if µ is a measure in (R, B R ) that is finite on all bounded sets then the function: µ((0, x]) if x > 0 F (x) = 0 if x = 0 µ([ x, 0)) if x < 0 is increasing, right continuous, and µ F = µ. 2

22 3 easurable Functions and Integration Let f X Y, (X, ), (Y, ) be measurable spaces. Definition 4. Given (X, ), (Y, ) measure spaces, X, Y, we say that f X Y is measurable iff f (E) E. Observation: the composition of measurable functions is measurable. (X, ) f (Y, ) g (Z, O) (,, O σ-algebras in X, Y, Z respectively) Exercise: Decide whether or not this is true for f X Y, f( E n ) = f(e n ) f ( E n ) = f (E n ) f( E n ) = f(e n ) f ( E n ) = f (E n ) f(e c ) = (f(e)) c f (E c ) = (f (E)) c Proposition 2. Let (X, ), (Y, ) be measurable spaces, f X Y. Assume that is generated by a collection ξ, i.e. the smallest σ-algebra (ξ) = containing ξ is. Then: Proof. f is measurable (i.e. f (F ) F ) f (E) E ξ : f is measurable by definition: f (F ) F. : Look at the collection of sets, G, for which f (G). Let {G f (G) } = Ω. First, ξ Ω. ow, Ω is a σ-algebra by *. = (ξ) Ω. Proposition 3 (Corollary). f X R. (X, ), (R, B R ). Then the following are equivalent: a) f is measurable from (X, ) into (R, B R ) b) f ((a, )) a c) f ([a, )) a d) f ((, c)) c e) f ((, c]) c Corollary. Let X, Y be metric spaces, then every continuous function is measurable from (X, β X ) to (Y, B Y ). Proof. f continuous f (U) open U open in Y f (U) open U f (U) is in B x U open f is measurable. For most of this course, Y = R or Y = C. Definition 5. Given (X, ) measurable space with X and f X R (or C), we say f is (X, )- measurable if f is measurable from (X, ) to (R, B R ) (or (C, B C )). Definition 6. Let f R n R (or C). We say f is Borel measurable if f is measurable from (R n, B R n) into (R, B R ) (or (R, B C )). Definition 7. We say f is Lebesgue measurable if f is measurable from (R n, L) into (R, B R ) (or (C, B C )), where L in R n is the completion of B R n. 22

23 Remark 3. Composition of Borel measurable functions is Borel measurable but the composition of Lebesgue measurable functions is not necessarily Lebesgue measurable! Example 8. Let R g R f R. Consider (f g) (E). Clear that E Borel, f and g Borel measurable, then f (E) Borel g (f (E)) Borel. But if f, g Lebesgue measurable then E Borel f (E) is Lebesgue g (f (E)) not necessarily Lebesgue. If f Borel, g Lebesgue, then f g is Lebesgue. Proposition 4. Let (X, ), f X R, g X R. If f, g are measurable, then f, f + g, min (f, g), max (f, g), f g are measurable. So is cf for c R. Proof. By previous proposition, only need to check f for a family that generates B R. f + g: f (E) ( E B R ) g (E) ( E B R ) (f + g) ((a, )) = ({f > r} {g > a r}) r Q (otation: {f > r} = f ((r, )) = {x f(x) > r}) Therefore, f + g measurable as it is a countable union of measurable sets. cf: If c = 0: If c > 0: f : ( f ) ((a, )) = {f > a} {f < a} min (f, g): {min (f, g) > a} = {f > a} {g > a} max (f, g): {max (f, g) > a} = {f > a} {g > a} f g: Show f 2 is measurable (exercise). (cf) X if a < 0 ((a, )) = otherwise (cf) ((a, )) = {cf > a} = {f > a c } (f + g) 2 = f 2 + g 2 + 2f g f g = (f + g)2 f 2 g 2 2 Proposition 5. Let f j X R {, }, f j measurable. Then sup j f j and inf j f j are measurable. So are lim inf j f j, lim sup j f j, and lim j f j. Proof. {inf j f j < a} = {f j < a} j= {sup j f j > a} = {f j > a} j= lim inf j lim sup j f j = sup inf f j k j k f j = inf k sup f j j k lim f j = lim inf j j f j = lim sup f j j (where it exists) 23

24 Alternate proof. Let f = lim j f j {f > a} = m= = n= {f n > a + m } Remark 4. We need not define a measure to talk about measurable functions (like continuous functions in topology). 3. Devil s staircase f f This procedure provides {f n } n=. Let f = lim n f n. f n are continuous and uniformly continuous (exercise). Properties: f(0) = 0, f() =. f exists on every open interval we remove (and equals 0). f() f(0) 0 f (x) dx. The derivative is 0 on a set of measure. f(cantor set C) has measure. f(complement of C) { c 2 n c {0,,..., 2 n }} Q, so has measure 0. Except for the points { c 2 n } the function has an inverse. Let x [0, ] be written in base 3 as: x = Cantor set points are where ε n {0, 2} n. Then: n= ε n 3 n (ε {0,, 2}) f(x) = 2 n= εn if x C 2 n unique constant such that f is continuous otherwise As for the inverse of f (call it g), write: y = n= ε n 2 n (ε n = 0, ) 24

25 ow: g(y) = n= 2ε n 3 n (for y { c 2 n }) g maps [0, ] { c } into the Cantor set. g(e) C so it is measurable. g (g(e)) = E, so the inverse 2 n Inside is a measurable set, say E. of a measurable set g(e) is not measurable (g, which arose as the inverse of a continuous function, is not Borel measurable). 3.2 Integration for f X [0, ] Definition 8. Let (X, ) be a measurable space. χ E is a characteristic function ( indicator function) if: if x E χ E (x) = 0 if x E c Definition 9. A function f is simple if it takes finitely many values. i.e. if and E,..., E X such that f(x) = j= α j χ Ej (x). Definition 20. f X R (or C) is simple measurable if, {E,..., E } such that f = j= α j χ Ej. Definition 2. For a simple function f we say it is written in standard form if E j = f (α j ). 4 ( ) 3 2 [ ] [ ] f(x) = χ [0,] + 4χ (,2) + χ [2,3] = χ [0,3] + 3χ (,2) = χ [0,] [2,3] + 4χ (,2) standard form Theorem 5. Let (X, ) be a measurable space, f (X, ) (R, B R ) for R = R {±} a) If f X [0, ] measurable, then there is a sequence {φ i } i= of simple measurable functions such that: 0 φ φ 2... f and such that lim i φ i (x) = f(x) x X. oreover, φ i f uniformly on any subset where f is bounded. b) If f X R (or C) is measurable, then there is a sequence {φ j } j= of measurable simple functions such that: 0 φ φ 2... f otation: and such that lim j φ j (x) = f(x) x X. oreover, φ j f uniformly on any subset where f is bounded. φ j f means φ j converges to f uniformly. a.e. means almost everywhere. 25

26 Proof. a) For n =, 2,..., define: E k n = f ([ k 2 n, k + 2 n )) (k = 0,,..., 22n ) F n = f ([2 n, )) Then E k n, F n X. Define: Claim. 2 2n k φ n = k=0 2 χ n E + n k 2n χ Fn φ j φ j+ f, j =, 2,... φ j f, φ f on f bounded. 2 n k+ 2 n k 2 n E k n F n E k n F n b) f X R (or C). Define: f + = max{f, 0} f = max{ f, 0} ow, f + + f = f and f + f = f, f + and f are measurable, f +, f X [0, ]. To each one, apply part a) to get {φ + n}, {φ n}. Define φ n = φ + n φ n. For f X C, f = R(f)+iI(f) with R(f), I(f) X R and apply the real case to each. Proposition 6. Let (X,, ν), ν complete, f X R (or C) be a measurable function (B R or B C ). If g = f a.e. then g is measurable. Proposition 7. Let (X,, ν), ν complete, f n f a.e. with f n measurable. Then f is also measurable. 26

27 Define: L + = {f f X [0, ], f is measurable} Let (X,, ν) and f L +, then: f dν = a k ν(a k ) For f = k= a k χ Ak in standard representation. Also, for A measurable: k= A f dν = f χ A dν simple Proposition 8. The integral of a simple function f = a n l An where A n are disjoint is fdµ = a n µa n. Proof. See otes from Jose. Proposition 9. If f = b j l Bj, then fdµ = b j l Bj Proof. Let the measurable sets {C ii [m] } be the unique coarsest partition of B j such that for any j [], we can write each nonempty B j as a disjoint union of C i. For each i, let I(i) be the unique indexing set such that j I(i) iff C i B j. Thus by construction, C i = B j i j I(i) And also Then f = b j l Bj = j= i= fdµ = i= b j j I(i) µ(c i) = j= j I(i) b j i j I(i) b j l C i µ(c i ) = b j µ(b j ) j= Proposition 20. Let f L +, f = j= b j χ Bj, not necessarily in standard form, then: f dν = b j ν(b j ) j= Proof. Let f = k= a k χ Ak in standard form. Assume one a k = 0 and one of b j = 0, so that: {B j } may not be disjoint, but we know {A k } are. f dν = a k ν(a k ) k= A k = X = B j k= j= B j = B j X = B j ( A k ) = (B j A k ) k= k= 27

28 Therefore: j= b j ν(b j ) = j= b j ν ( B j A k ) k= = b j ν(b j A k ) j= k= = a k ν(b j A k ) j= k= a k k= j= = Assume for now that {B j } are pairwise disjoint, then: The proof pauses here. ν(b j A k ) f dµ = a k µ(a k ) = a k µ(a k B j ) k= k= j= = b j µ(a k B j ) j= k= = b j µ(b j ) Proposition 2. Let φ, ψ be measurable, simple, nonnegative, then:. cφ dµ = c φ dµ (c > 0) 2. φ + ψ dµ = φ dµ + ψ dµ 3. If φ(x) ψ(x) x X then φ dµ ψdµ. 4. Fix φ, then the map A A φ dµ is a measure A. Call it ν. Proof.. Trivial as cφ is simple. So φ dµ = k= a k µ(a k ) in standard form. Then: j= cφ dµ = ca k µ(a k ) = c a k µ(a k ) = c φ dµ k= k= 2. Assume: φ = k= a k χ Ak in standard form ψ = j= b j χ Bj 28

29 and assume k= A k = X = j= B j : ow: φ dµ + ψ dµ = a k µ(a k ) + b j µ(b j ) k= = k= = k= j= a k µ (A k j= a k µ ( j= B j ) + j= (A k B j )) + j= b j (B j A k ) k= b j µ ( (B j A k )) k= = a k µ(a k B j ) + b j µ(b j A k ) k= j= j= k= = (a k + b j )µ(a k B j ) j= k= φ = a k χ Ak = a k χ Ak j= Bj k= k= = a k χ j= (A k B j) = k= ψ = b j χ Bj = b j χ Bj k= A = b k j χ k= (A k B j) =... j= j= φ + ψ = (a k + b j )χ Ak B j j= k= So φ dµ + ψ dµ = φ + ψ dµ. j= a k χ Ak B j k= j= Back to our original proof for a bit: Proof. Let φ = T i= c i χ Ei be measurable simple. Let: φ = φ + φ φ T where φ i = c i χ Ei is in standard form. Then: φ dµ = φ + φ φ T dµ = φ dµ φ T dµ T = φ i dµ i= T = c i χ Ei dµ i= T = c i µ(e i ) i= Back to the proof of the proposition: Proof. 29

30 3. φ(x) = k= a k χ Ak and ψ(x) = j= b j χ Bj in standard form, k= A k = X = j= B j. From previous argument: φ(x) = a k χ Ak B j k= j= ψ(x) = b j χ Ak B j k= j= {A k B j } are pairwise disjoint. Consider x A k B j for some i, j. φ(x) = a k, ψ(x) = b j a k b j So: φ dµ = a k µ(a k B j ) b j µ(a k B j ) = ψ dµ j= k= k= j= 4. eed: ν( ) = 0 ν( i= A i ) = i= ν(a i ) Where A = i= A i. Write φ = j= b j χ Bj standard form with j= B j = X. A φ dµ = φχ A dµ = = j= j= b j χ Bj χ A dµ b j χ Bj A dµ = b j µ(b j A) j= = j= b j µ (B j i= A i ) = j= = b j µ ( lim (B j A i )) j= n n i= = b j lim µ ( n (B j A i )) n j= i= = b j lim µ(b j A i ) j= n n i= = b j µ(b j A i ) i= j= = i= = i= j= j= b j χ Bj A i dµ b j χ Bj χ Ai dµ = φ dµ = ν(a i ) Ai i= i= b j µ ( (B j A i )) i= 30

31 Define L + = {f: f is measurable and non-negative}. Then we define f dµ = sup{ φ dµ φ simple measurable, 0 φ f} But we don t know that f + g dµ = f dµ + g dµ f dµ g dµ (if f g, f, g L + ) cf dµ = c f dµ (for c 0) Theorem 6 (onotone convergence). Let (X,, µ) be a measure space, {f n } n= L +. If f n f n+ n, then: lim n f n dµ = lim f n dµ n Proof. First, lim n f n exists as (f n ) is a monotone sequence in R + and measurable by an earlier theorem. Take limits to get: Fix φ as a simple measurable function: f m lim n f n f m dµ lim n f n dµ ( m ) lim m f m dµ 0 φ f = lim n f n lim f n dµ n Fix α (0, ). Define E n = {x X f n (x) αφ(x)}. Claim that n= E n = X (because f n (x) f(x) and αφ(x) < f(x) for f(x) 0). Consider: lim n φ dµ = φ dµ E n X lim ν(e n) = ν ( E n ) = ν(x) n n= by continuity (With vertical equals signs to add) We know: En αφ dµ En f n dµ X f n dµ X f n+ dµ lim n X f n dµ Thus: So: lim m αφ dµ lim E m n f n dµ lim n f n dµ αφ dµ lim n f n dµ sup { αφ dµ 0 φ f} φ = αf dµ = α f dµ lim n f n dµ sup {α f dµ} α = f dµ Example 9 (Counterexamples). Let: 3

32 f n = χ [n,n+], f n (x) 0 x. g n = nχ (0, n ), g n (x) 0 x. f n = and g n = n, but f = 0 and g = 0. The monotone convergence theorem (CT) implies we do not need to take the supremum over all 0 φ f in the definition of f dµ. It s enough to take one family {φ n } n= of simple measurable functions such that 0 φ n φ n+ and φ n (x) f(x) x. CT lim n φ n dµ = lim φ n dµ = n f dµ Recall that we proved a theorem that shows the existence of at least one such family {φ n } n=. Theorem 7. (X,, µ), {f n } n= L + then: n= f n dµ = n= f n dµ (this proves that f + g dµ = f dµ + g dµ) Proof. Take: {φ n } n= measurable simple, 0 φ n φ n+ n, φ n f. {ψ n } n= measurable simple, 0 ψ n ψ n+ n, ψ n g. We know, by the CT: lim n φ n dµ = f dµ lim n ψ n dµ = g dµ Also, φ n + ψ n f + g, so: We also know that: So now just need to check: lim n (φ n + ψ n ) dµ = (f + g) dµ φ n + ψ n dµ = φ n dµ + ψ n dµ lim ( φ n dµ + ψ n dµ) = lim n n φ n dµ + lim n ψ n dµ As everything is non-negative, it is true by Analysis I. So, by induction: n= f n dµ = n= f n dµ Let g = n= f n dµ. We have: and: f n 0 g g + g f n n= 32

33 By CT: lim g dµ = lim g dµ f n dµ = lim n= n= n= = lim = f n dµ f n dµ n= f n dµ Theorem 8. (X,, µ), f L +. Then: f dµ = 0 f = 0 a.e. Proof. otice that the statement is trivial, for measurable simple functions φ = n= a n χ En, a k 0, as we know: and: as: k= φ dµ = n= a n µ(e n ) a n µ(e n ) = 0 φ = 0 a.e. k= a n µ(e n ) = 0 a n µ(e n ) = 0 either a n = 0 or µ(e n ) = 0 ( n {,..., }) ( n {,..., }) So {x φ(x) 0} is a finite union of sets of measure 0. ow we look at general f L +. Suppose: f dµ = sup { φ dµ 0 φ f, φ simple measurable} f = 0 a.e. φ f measurable simple, we have φ = 0 a.e. for those φ, φ dµ = 0 (exercise) ext, want to show f dµ = 0 f = 0 a.e. Look at: Then: µ(e n ) = 0 µ({x f(x) 0}) n= µ(e n ) = 0 f dµ = sup {0} = 0 {x f(x) 0} = {f n } n= E n f dµ f dµ En En n dµ = n χ E n dµ = n µ(e n) =0 33

34 Corollary 2. {f n } n= L +, f n f n+ a.e. Then: lim f n dµ = lim n n f n dµ (By lim n f n we mean the lim n f n (x) where it exists and 0 otherwise) Proof. f n f in E with µ(e c ) = 0. E f n dµ = f n dµ E f dµ = f dµ f dµ = f dµ = lim f n dµ = lim E E n n f n dµ = lim E n f n dµ = lim n f n dµ Warning: in general, we do not have: but... lim n f n dµ = lim f n dµ n Lemma 2 (Fatou s Lemma). Let {f n } n= L +, (X,, µ), then: lim inf f n dµ lim inf n n f n dµ Proof. Recall lim inf n = sup k inf n k f n. inf n k f n f j for every j k, which implies: inf f n dµ f j dµ ( j k) n k (Let g k = inf n k f n, then g k monotone increasing) CT inf f n dµ inf f j dµ n k j k lim g k dµ lim inf f j dµ k k j k lim inf f n dµ lim inf f j dµ k n k k j k lim inf f n dµ lim inf n n f n dµ Corollary 3. (X,, µ), {f n } n= L +, assume lim n f n = f a.e. Then: Proof. Exercise. f dµ lim inf n f n dµ 34

35 3.3 Integration for general f X R {±} Definition 22. Let f X R {±} measurable, f = f + f with f +, f 0. Then assuming at most one of f + dµ, f dµ is infinite, we define: f dµ = f + dµ f dµ Definition 23. We say f is integrable iff f + dµ and f dµ are finite. This is equivalent to f dµ <. Proposition 22. The space of integrable functions is a vector space. The integral is a linear functional in that vector space. Proof. If f, g are integrable then af + bg integrable a, b R. As af + bg a f + b g, we have: af + bg dµ a f dµ + b g dµ = a f dµ + b g dµ < A functional is a map from a space of functions to R (or C). So define: We need:. I(af) = ai(f) 2. I(f + g) = I(f) + I(g). There are three cases to this: Case. a = 0, then trivial as both sides are null. I(f) = f dµ I(af) = af dµ (a R) = (af) + dµ (af) dµ Case 2. a > 0, then (af) + = a(f) + and (af) = a(f), which implies: Case 3. a < 0, exercise. 2. Let h = f + g. Then: I(af) = af + dµ af dµ f = f + f g = g + g = a f + dµ a f dµ = a f dµ h = h + h = f + f + g + g h + + f + g = f + g + + h h + + f + g dµ = f + + g + + h dµ h + dµ + f dµ + g dµ = f + dµ + g + dµ + h dµ (as everything is positive) h + dµ h dµ = f + dµ f dµ + g + dµ g dµ 35

36 Remark 5. For f X C, define: f dµ = R(f) dµ + i I(f) dµ Then, as long as everything is finite, everything translates to the complex case. otation: (X,, µ), we define L (X,, µ) (a.k.a. L (µ) or L (X)) to be the space of integrable functions ( f dµ < ). Proposition 23. (X,, µ), f X R (or C): Proof. f dµ f dµ (R) f dµ = f + dµ f dµ f + dµ + f dµ = f + + f dµ = f + + f dµ triangle inequality (C) First, if f dµ = 0 then nothing to prove, so assume f dµ 0 and define: α = f dµ f dµ Observe: and f dµ R +, so: f dµ = ( f dµ) ( f dµ) f dµ = α f dµ f dµ = α f dµ = R(α f dµ) = R( αf dµ) = R(αf) dµ (definition of complex integral) = R(αf) dµ (as it s in R + ) R(αf) dµ (by above) αf dµ = α f dµ = α f dµ = f dµ f dµ f dµ = f dµ Proposition 24. Let f L : 36

37 a) {x f(x) 0} is σ-finite and {x f(x) {±}} has measure 0. b) Let f, g L, then: E f dµ = E g dµ E f g dµ = 0 f = g a.e. Definition 24. A set is σ-finite if we can write it as a countable union of sets that have finite measure. Proof of proposition. a) We only do the real case: w.l.o.g. assume f is non-negative (otherwise do same thing for f + and f ): f X [0, ] {f 0} = {f > 0} = {f > n= n } eed to check that µ({f > }) <. We use Chebychev s inequality: n The second part is an exercise. b) Assume f = g a.e. Therefore: > f dµ fχ {f> n } dµ n χ {f> n } dµ = n µ ({f > n }) f g = 0 a.e. f g = 0 a.e. f g dµ = 0 non-negative by earlier proof of nonnegative case. ow, assume f g dµ = 0. Then: E f dµ E g dµ = E f g dµ E f g dµ f g dµ = 0 E f dµ = E g dµ ( E) ow, assume E f dµ = E g dµ E and f, g X R. ote: So enough to show {f g > 0} has measure 0. Let: For this E we have E f g dµ = 0. {f g} = {f g > 0} {g f > 0} E = {f g > 0} = (f g) ((0, ]) so E is measurable For contradiction, assume µ(e) > 0: E = {f g > 0} = {f g > n= n } µ ({f g > }) µ(e) > 0 n Which implies n such that µ({f g > }) > 0. Therefore: n So f = g a.e. f g dµ f g dµ E {f g> n } {f g> n } n dµ = n µ ({f g > n }) > 0 37

38 We have L, the space of measurable functions such that f <. Try to define a norm: Would hope for: f 0, f = 0 f = 0 the bit that fails λf = λ f for λ R f + g f + g f = f dµ How to fix the first property? We define an equivalence relation: f g f = g a.e. Define: ow, L is a metric space. We define: L = L / (ote: [0] = [χ Q ]) In L, we can define: [f] + [g] = [f + g] λ[f] = [λf] [f] = f dµ well defined by a previous theorem otation: we generally ignore the square brackets for practical purposes. Theorem 9 (Dominated Convergence Theorem). Let {f n } n=, f n L such that: a) f n f a.e. b) g L such that 0 f n g n. Then: Proof. f n g g f n 0 and f n + g 0 Then apply Fatou s Lemma, which is: lim n f n dµ = lim f n dµ n for h n 0. lim inf h n dµ lim inf n n h n dµ lim n (g f n ) dµ lim inf n (g f n ) dµ (as the limit exists) and lim n (g + f n ) dµ lim inf n (g + f n ) dµ and g lim f n dµ lim inf ( n g dµ f n dµ) n g + lim f n dµ lim inf ( n g dµ + f n dµ) n g dµ f dµ g dµ lim sup f n dµ (lim inf n ( a n ) = lim sup n (a n )) n g dµ + f dµ g dµ + lim inf n f n dµ lim sup f n dµ f dµ lim inf n n f n dµ f dµ = lim n f n dµ 38

39 Theorem 0. {f j } j= L, assume n= f n dµ <. Then j= f j converges a.e. oreover: and: Proof. By the monotone convergence theorem: since: k= j= f n dµ = n= > f k dµ = k= f k dµ < f j L (i.e. j= f j dµ < ) proposition f n dµ n= k= f k dµ f k < a.e. ext, define h k = k n= f n. We want that lim k h k dµ = lim k h k dµ. otice: k h k = f n So, by the dominated convergence theorem: n= k n= f n k= n= f n < L lim h k dµ = lim k k h k dµ Theorem. L is complete. Proof. eed to show that if {f n } n= is Cauchy then f L such that f n f Assume w.l.o.g. that f 0. So: f n f = f n f n + f n... + f 2 f n f n = j= f j+ f j Assume (by taking a subsequence) that f n+ f n ε 2 n. Denote g j = f j+ f j. Look at: By previous theorem: j= j= g j dµ f j+ f j j= g j < a.e. Define: f = j= f j+ f j and f L by theorem. 2. (Want f f n 0 as n ) j= j= ε 2 j = ε f j+ f j < a.e. f f n = f j+ f j j= n f j+ f j = f j+ f j f j+ f j j= j=n j= j=n ε 2 j ε 2 n 39

40 Proposition 25 (Simple functions are dense in L ). Let f L (X,, µ), then ε > 0 φ simple measurable such that: f φ = f φ dµ < ε If X = R, µ the Lebesgue measure, then φ can be taken as φ = j= a j χ Ej where E j are open intervals. oreover, g continuous such that f g dµ < ε. Proof. If f X [0, ] and φ n simple measurable such that: 0 φ φ 2... φ n... f (Want to show that f φ n dµ 0) Let h n = f φ n. Then: By DCT: lim n h n dµ = In general case we know f X R (or C). We know: h n f + φ n 2 f lim h n dµ = n 0 dµ = 0 0 φ φ 2... φ n... f Let h n = f φ n, then h n f + φ n 2 f and so lim n h n dµ = 0 by DCT. ow, a sketch of the moreover statement. X = R, µ = Lebesgue. ε φ n measurable such that f φ n dµ < ε 00. Suppose φ n = ε j= a j χ Aj. Since A j is measurable we know σ j open such that µ(σ j A j ) and σ 0 00 j A j. Then φ n = a j χ σj. φ n φ n is very small. In R every open set is the union of open intervals. So: σ j = I j,k k (for I j,k open interval) 0 Since µ(σ j ) < µ(i j, k). Define σ j = k= I j,h we can make µ(σ j σ j ) very small. Define k σ j = a j χ aj (need to check f σ n dµ < ε ). f σ n + σ n σ n + σ n σ n dµ (then use triangle inequality). g continuous such that χ [a,b] g dµ < ε. Theorem 2. Let (X,, µ), f X [a, b] R (or C), x X, t [a, b]. Assume f(x, t) is integrable w.r.t. x t. Define F (t) = X f(x, t) dµ. a) Assume g L (X,, µ), g 0 such that f(x, t) g(x) and lim t t0 f(x, t) = f(x, t 0 ). Then: lim F (t) = F (t 0 ) t t 0 b) Suppose f t (x, t) exists and h L such that f (x, t) h(x). Then: t Proof. F f (t) = (x, t) dµ X t a) Define f n (x) = f(x, t n ) for some {t n } n= such that t n t 0. Then we have a sequence {f n } n= on X such that f n L. oreover, f n (x) g(x) so by DCT: X lim n f n dµ = lim n X f n dµ Which implies: F (t 0 ) = f(x, t 0 ) dµ = lim n f(x, t n ) dµ = lim n F (t n ) 40

41 b) Define f t (x, t 0) = lim n h n (x) where: and t n t 0. h n (x) h(x), so apply DCT. h n (x) = f(x, t n) f(x, t 0 ) t n t 0 Theorem 3. Let f [a, b] R. a) If f is Riemann integrable then it is Lebesgue integrable (and the two values are the same). b) f is Riemann integrable iff {x f(x) is discontinuous at x} has Lebesgue measure Different modes of convergence Let {f n } n=, f n X R (or C). Uniform convergence: ε > 0 ε s.t. f n (x) f m (x) < ε m, n x 2. Pointwise convergence: x ε > 0 ε (x) s.t. f n (x) f m (x) < ε n, m 3. a.e. convergence: (Pointwise convergence except on a set of measure 0) 4. L convergence: ε > 0 s.t. f n f m dµ < ε n, m 5. Convergence in measure: ε > 0 µ({x X f n (x) f m (x) > ε}) 0 as n, m Proposition 26. L measure. Proof. We have: f n f dµ 0 (Want that ε > 0, µ({x f n f > ε}) 0 as n ) 0 f n f dµ f n f dµ ε dµ = εµ({x f n f > ε}) {x fn f >ε} {x fn f >ε} So µ({x f n f > ε}) 0 as n. Proposition 27. Uniform measure Proof. ε > 0 s.t. f n (x) f(x) < ε for n x (want µ({x f n (x) f(x) > ε}) 0). For n, {x f n (x) f(x) > ε} =. Example 0 (Showing pointwise / measure). Consider f n (x) = χ [n,n+]. f n (x) = 0 n > x +, so f n 0 pointwise but µ({x f n (x) 0 > }) = µ([n, n + ]) = n. 2 4

42 Example (Showing uniform / L ). Consider f n (x) = n χ [0,n]. sup x R n χ [0,n] 0 = 0, therefore n f n 0 uniformly. n χ [0,n] 0 dµ = n n = Example 2 (Showing L / a.e.). Construct a sequence as f = χ (0, 2 ), f 2 = χ ( 2, ), continuing in this fashion until the positive end of the interval is greater than, such as in f 3 = χ ( 2 + 3, ) = χ ( 5 6, 3 2 ), so we define f 4 = χ (0, 5 ), f 5 = χ ( 5, ), and so on. We have: f n 0 dµ n + 0 So {f n } n= converges to 0 in L, but it doesnt convergence to 0 a.e. Let x (0, ). For every n, f (x) = for some > n, so f n (x) / 0 x (0, ). Theorem 4. Let {f n } n= L. Assume f n is Cauchy in measure. Then f such that f n converges to f in measure. Furthermore, there is a subsequence {f nj } n= such that f nj f a.e. oreover, if f n converges in measure to g then g = f a.e. Proof. We know ε > 0 µ({x f n (x) f m (x) > ε}) 0 as m, n. i.e. ε > 0 δ > 0 s.t. µ({x f n (x) f m (x) > ε}) < δ m, n >. Choose ε =, δ =. Then 2 j 2 j j s.t. µ({x f n (x) f m (x) > }) < 2 j 2 j for all m, n j. Define g j = f j (want to show g j (x) converges to something). So: Define: F k points where we shouldn t hope for convergence. µ ({x g j (x) g j+ (x) > 2 }) < j 2 j E j F k = E j j=k µ(f k ) µ(e j ) j=k j=k 2 j = 2 2 k What happens outside F k? Take x / F k, look at g j (x) g i (x). w.l.o.g. assume j i: Remember x / F k so take i, j k, then: g j (x) g i (x) = g j (x) g j (x) + g j+ (x)... + g i+ (x) g i (x) j l=i g l+ (x) g l (x) j g j (x) g i (x) l=i 2 l l=i 2 l = 2 2 i 2 2 k This means {g j (x)} j= is a Cauchy sequence in R. Define f(x) to be the limit of g n(x) as n for x / F k. We have now defined f(x) for x Fk c for every k. So we have f(x) for x k= (F k c ). We now need only define f for x ( k= (F k c))c, but ( k= (F k c))c = k= F k. otice that: ( n) µ ( F k ) µ(f k ) 2 k= 2 k µ ( F k ) = 0 k= So define f to be anything you like in k= F k. Thus we have shown that there is a subsequence {f nj } j= such that f nj f a.e. ext, want to show f n f in measure. We know g j (x) g i (x) 2 2 i for x / F k, j i k. Taking limits as j (they exist by above): lim g j(x) g i (x) 2 (for x / F j 2 i k, i k) f(x) g i (x) 2 2 i (for x / F n, i k) 42

43 We know µ({x g j (x) f(x) 2 2 j }) µ(f k ). Given ε > 0 choose j such that 2 2 j < ε: Letting j = k gives: µ({x g j (x) f(x) > ε}) µ ({x g j (x) f(x) > 2 2 j }) µ(f k) µ({x g j (x) f(x) > ε}) µ(f j ) 2 0 (as j ) 2j So far we ve only shown g i converges to f in measure, not that f n converges to f in measure. So: {x f n (x) f(x) > ε} = {x f n (x) g i (x) + g i (x) f(x) > ε} ({x f n (x) g i (x) > ε 2 } {x g i(x) f(x) > ε 2 }) µ({x f(x) n f(x) > ε}) µ ({x f n (x) f i (x) > ε 2 }) + µ ({x f i (x) f(x) > ε 2 }) Finally, uniqueness, let there be two, f and g, then: But: Theorem hypothesis that f n Cauchy in measure 0 as i as shown µ({x f n (x) f(x) > ε}) 0 (as n ) µ({x g n (x) g(x) > ε}) 0 (as n ) {x f g > ε} = {x f f n + f n g > ε} ({x f f n > ε 2 } {x f n g > ε 2 }) µ({x f g > ε}) µ ({x f f n > ε 2 }) + µ ({x f n g > ε 2 }) LHS lim n RHS = 0 µ({x f g > ε}) = 0 So: {x f g} = µ({x f g}) n= n= {x f g > 2 n } µ ({x f g > 2 }) = n 0 = 0 f = g a.e. n= Corollary 4. If {f n } n= L is Cauchy in L, then there is a subsequence f nj f a.e. Theorem 5 (Egorov). Let f X R, µ(x) <, f n X R and f n f a.e. Then ε > 0 there is a set E with µ(e) < ε such that f n f on E c. Proof. w.l.o.g. f n f x (add to E the set where you don t converge). Define: otice E n (k) E n+ (k) with k fixed: E n (k) = {x f m (x) f(x) > m=n k } lim E n(k) = E n (k) = n n= 43

44 and: So: µ(e (k)) µ(x) < µ ( E n (k)) n= =µ( )=0 = lim n µ(e n (k)) So k, given ε > 0 n(k) (i.e. n depending on k) such that µ(e n(k) (k)) < ε 2 k. Define: Then: (eed to show f n f on E c ) So: But: E = E n(k) (k) k= µ(e) µ(e n(k) (k)) < (E n(k) (k)) c = k= k= E c = (E n(k) (k)) c k= ε 2 k = ε x E c x (E n(k) (k)) c ( k ) = m=n(k) m=n(k) ({x f m (x) f(x) > k } c) {x f m (x) f(x) k } So: x E c f m (x) f(x) k ( k, for m large enough) Which implies uniform convergence. 3.5 Product measures Reminder of the product σ-algebra: take (X,, µ), (Y,, ν). We defined the product σ-algebra in terms of: We denoted this as. generated by: { α (E α ) E α α } When you have a finite (or countable) product, that is the same as the one { E α E α α } Definition 25. A rectangle is any set of the form A B for A, B. Observations: (A B) (E F ) is a rectangle as (A B) (E F ) = (A E) (B F ). (A B) c = (X B c ) (A c B) 44