Class Notes for Math 921/922: Real Analysis, Instructor Mikil Foss

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1 University of Nebraska - Lincoln DigitalCommons@University of Nebraska - Lincoln Math Department: Class Notes and Learning Materials Mathematics, Department of 200 Class Notes for Math 92/922: Real Analysis, Instructor Mikil Foss Laura Lynch University of Nebraska-Lincoln, llynch@ccga.edu Follow this and additional works at: Part of the Science and Mathematics Education Commons Lynch, Laura, "Class Notes for Math 92/922: Real Analysis, Instructor Mikil Foss" (200). Math Department: Class Notes and Learning Materials This Article is brought to you for free and open access by the Mathematics, Department of at DigitalCommons@University of Nebraska - Lincoln. It has been accepted for inclusion in Math Department: Class Notes and Learning Materials by an authorized administrator of DigitalCommons@University of Nebraska - Lincoln.

2 Class Notes for Math 92/922: Real Analysis, Instructor Mikil Foss Topics include: Semicontinuity, equicontinuity, absolute continuity, metric spaces, compact spaces, Ascoli s theorem, Stone Weierstrass theorem, Borel and Lebesque measures, measurable functions, Lebesque integration, convergence theorems, L p spaces, general measure and integration theory, Radon- Nikodyn theorem, Fubini theorem, Lebesque-Stieltjes integration, Semicontinuity, equicontinuity, absolute continuity, metric spaces, compact spaces, Ascoli s theorem, Stone Weierstrass theorem, Borel and Lebesque measures, measurable functions, Lebesque integration, convergence theorems, Lp spaces, general measure and integration theory, Radon-Nikodyn theorem, Fubini theorem, Lebesque-Stieltjes integration. Prepared by Laura Lynch, University of Nebraska-Lincoln August 200

3 Course Information Office Hours: :30- MWF Optional Meeting Time: Thursdays at 5pm Assessment Chapter Drawbacks to Riemann Integration Homework 6 Midterm Final 34pts 33pts 33pts (Dec 3 : 3pm). Not all bounded functions are Riemann Integrable. 2. All Riemann Integrable functions are bounded. 3. To use the following theorem, we must have f R[a, b]. Theorem (Dominated Convergence Theorem for Riemann Integrals, Arzela). Let {f n } n= R[a, b] and f R[a, b] be given. Suppose there exists g R[a, b] such that f n (x) < g(x) for all x [a, b] and x [a, b], then lim n b a f n (x)dx = b a f(x)dx. if x = p q in lowest terms with q n on [0, ], Example. Define f n (x) = 0 otherwise. lim f n(x) = f(x) for all n Then f n (x) χ Q [0,] =: f. Notice here f n (x) < 2 for all x [0, ] but f is not Riemann Integrable. Thus we can not use the theorem. 4. The space R[a, b] is not complete with respect to many useful metrics. Good Properties of Riemann Integration. R[a, b] is a vector space 2. The functional f b f(x)dx is linear on R[a, b]. a 3. b f(x)dx 0 when f(x) 0 for all x [a, b]. a 4. Theorem holds.. Measurable and Topological Spaces Definition (p 3,9). Let be a nonempty set.. A collection T of subsets of is called a topology of if it possesses the following properties: (a) T and T. (b) If {U j } n j= T, then n U j T. (c) If {U α } α A T, then α A U α T. 2. If T is a topology on, then (, T ) is called a topological space. If T is understood, we may just call itself a topological space. The members of T are called open sets in. The complements of open sets in are called closed sets.

4 3. If and Y are topological spaces and f : Y, then f is called continuous if f (V) is open in for all V that are open in Y. Examples.. If = R, then {, R} is a topology on R. 2. If = R, then the power set {P(R)} is a topology on R. 3. {, a>0 {( a, a)}, R} is a topology on R. 4. {, a<b R {(a, b)}, a R {(, a), (a, )}, R} is a topology on R = R {± }. Definition (pg2,25,43).. A collection M of subsets of is called a σ algebra if (a) M and M. (b) If E M, then E C M. (c) If {E j } j= M, then j= E j M. 2. If M is a σ algebra on, then (, M) is called a measurable space. If M is understood, then we may just call itself a measurable space. The members of M are called measurable sets. 3. If (, M) and (Y, N ) are measurable spaces, then f : Y is called (M, N ) measurable or measurable if f (E) M whenever E N. Examples.. (R, {, R}) is a σ algebra 2. (R, {, {}, R \ {}, R}) is a σ algebra 3. (R, {E R E or E C is countable}) is a σ algebra Lemma. If {E j } j= P(), then {F j} j= P() defined by F j = E j \ j k= E k is a sequence of mutually disjoint sets and j= E j = j= F j. Note. As a result of Lemma, we can actually modify part (c) of our definition of a σ algebra to say (c) If {E j } M is a sequence of mutually disjoint sets, then E j M. Remarks.. Property (a) of our definition for σ algebra could be replaced with M or M. 2. If {E j } j= M, then {EC j } j= M and j= E j = [ j= EC j ]C M. 3. If E, F M, then E \ F = E F C M. Theorem 2. If E is a collection of subsets of, then there exists a unique smallest σ algebra M(E) that contains the members of E. Note: By smallest, we mean any other σ algebra will contain all the sets in M(E). Proof. Let Ω be the family of all σ algebras containing E. Note Ω as P() Ω. Define M(E) = M Ω M. We want to show M(E) is a σ algebra.., M(E) since, M for all M Ω. 2. Let E M(E). Then E M for all M Ω which implies E C M for all M Ω which implies E C M(E). 3. If {E j } j= M(E), then {E j} j= M for all M Ω which implies E j M for all M Ω which implies E j M(E).

5 Remark. M(E) is called the σ algebra generated by E. Definition. Let (, T ) be a topological space. The σ algebra generated by T is called the Borel σ algebra on and is denoted B. The members of a Borel σ algebra are called Borel sets. Proposition (p 22). B R is generated by each of the following:. E = {(a, b) : a < b} 2. E 2 = {[a, b] : a < b} 3. E 3 = {(a, b] : a < b} or E 4 = {[a, b) : a < b} 4. E 5 = {(a, ) : a R} or E 6 = {(, a) : a R} 5. E 7 = {[a, ) : a R} or E 8 = {(, a] : a R} Proof. In text. Remark. The Borel σ algebra on R is B R = {E R : E R B R }. It can be generated by E = {(a, ] : a R}. Proposition 2. If (, M) is a measurable space and f : R, then TFAE. f is measurable 2. f ((a, )) M for all a R 3. f ([a, )) M for all a R 4. f ((, a)) M for all a R 5. f ((, a]) M for all a R Proposition 3 (p 43). Let (, M) and (Y, N ) be measurable spaces. If N is generated by E P(Y), then f : Y is (M, N ) measurable if and only if f (E) M for all E E. Proof. ( ) Since E N, if f is measurable then f (E) M for all E E by definition. ( ) Define O = {E Y : f (E) M}. Want to show O is a σ algebra. Then since E O and N is generated by E, we will be able to conclude N O. Recall (p4 of text) f (E C ) = [f (E)] C and f ( E j ) = f (E j ). Claim: O is a σ algebra. Proof:. Since f ( ) = and f (Y) = f ( C ) = (f ( )) C = ( ) C = M, we have, Y O. 2. Suppose F O. Then f (F ) M by definition of O and f (F C ) = [f (F )] C M since M is a σ algebra. 3. Suppose {F j } j= O. Then {f (F j )} j= M and f ( j= F j) = j= f (F j ) M as M is a σ algebra. Thus j= F j O. Hence O is a σ algebra on Y, which contains E. Then N O implies f is measurable. Definition. Let be a nonempty set and let {Y α, N α } be a family of measurable spaces. If f α : Y α is a map for all α A (some index set), then the σ algebra on generated by {f α } α A is the unique smallest σ algebra on that makes each f α measurable. It is generated by {f (E) : α A and E N α }. Proposition 4. If (, M) is a measurable space, then f : R is measurable if and only if f ((a, ]) M. Theorem 3. Let (, M) be a measurable space, let Y, Z be topological spaces. Let φ : Y Z be a continuous function. If f : Y is (M, B Y ) measurable, then φ f is (M, B Z ) measurable.

6 Proof. By definition, we need to check (φ f) (E) M for all E B Z. By Proposition 3, we need only to check (φ f) (E) M for all E open in Z. Let E be open in Z. Then φ (E) is open in Y since φ is continuous. Since f is (M, B Y ) measurable and φ (E) is open in Y, we have f (φ (E)) M. Fact. If V is an open set in R 2, then there exists a family {R j } j= of open rectangles in R2 satisfying. R j V for all j =, 2, j= R j = V Proposition 5. Let (, M) be a measurable space. Let u, v : R be (M, B R ) measurable. If φ : R 2 R is continuous, then h : R defined by h(x) = φ(u(x), v(x)) is (M, B R ) measurable. Proof. Define f : R 2 by f(x) = (u(x), v(x)). Since h = φ f and φ is continuous, by Theorem 3 it is enough to show that f is (M, B R 2) measurable. First we show that if R R 2 is an open rectangle, then f (R) M. Let (a, b), (c, d) R be open intervals such that R = {(y, z) R 2 a < y < b, c < z < d}. If (u(x), v(x)) R, then u(x) (a, b) and v(x) (c, d) implies x u ((a, b)) v ((c, d)). Hence f (R) = u ((a, b)) v ((c, d)). Since u, v are (M, B R ) measurable, we see f (R) M as M is closed under countable intersections. Now let V be an open ( set in R 2. By the above fact, there is a family {R j } j= of open rectangles such that ) R j = V. So f (V) = f R j = f (R j ) M. Thus f is (, B R 2) measurable. By Theorem 3, h = φ f is (, B R ) measurable. Proposition 6 (p45). Let (, M) be a measurable space. If c R and f, g : R are (M, B R ) measurable, then. cf is (M, B R ) measurable 2. f + g is (M, B R ) measurable 3. fg is (M, B R ) measurable Proof.. Define φ : R R by φ(y) = cy. Then φ f = cf and by Theorem 3, cf is measurable. 2. Define φ : R 2 R by φ(y, z) = y + z. Then φ(f, g) = f + g and since φ is continuous, by Proposition 5, f + g is measurable. 3. Define φ : R 2 R by φ(y, z) = yz. Then φ(f, g) = fg and since φ is continuous, by Prop 5, fg is measurable. Note. In the above proposition, points () and (2) imply its a vector space and adding on point (3) implies it is an algebra. Also, the proposition is true if we consider f, g : R. Proposition 7. If {f j } j= is a sequence of R valued measurable functions on (, M) then g (x) = sup j f j (x) g 2 (x) = inf j f j (x) g 3 (x) = lim sup j f j (x) g 4 (x) = lim inf j f j (x) are all (M, B R ) measurable. Moreover, if f(x) = lim j f j(x) exists for all x, then f is measurable. Proof. Let a R be given. Then {x : g (x) > a} = since f j is measurable. Thus g is measurable. Also {x : g 3 (x) > a} = j= k=j {x : f j (x) > a} implies g ((a, ]) = f j ((a, ]) M j= j= k=j {x : f k (x) > a} implies g ((a, ]) = f k ((a, ]) M. So g 3 is measurable. Since g 2 (x) = inf j f j (x) = sup j f j (x) and g 4 (x) = lim sup f j (x), we see g 2 and g 4 are measurable. Corollary. If f, g : R are measurable functions, then max{f, g} and min{f, g} are also measurable. 3

7 Corollary 2. If f : R is measurable, then so are f + = max{f, 0} and f = min{f, 0}. Corollary 3. If f : R is measurable, then so is f = f + + f..2 Simple Functions (Generalized Step Functions) Recall that for E, the characteristic function of E is if x E, χ E (x) = 0 if x / E. Definition (p46). A simple function on is a measurable function whose range consists of a finite number of values in R. If φ is a simple function with range {a,..., a n }, then for all j =, 2,..., n, the set E j = φ (a j ) is measurable. The n standard representation for φ is φ(x) = a j χ Ej (x). j= Theorem 4 (p47). Let (, M) be a measurable space.. If f : [0, ] is a measurable function, then there exists a sequence {φ n } n=0 of simple functions such that 0 φ 0 φ... f φ n (x) f(x) for all x φ n converges uniformly to f on the sets where f is uniformly bounded. 2. If f : C is measurable, there is a sequence {φ n } of simple functions such that 0 φ φ 2 f, φ n f pointwise and φ n f uniformly on any set on which f is bounded. Proof. (of ) For all n = 0,, 2,... and k = 0,,..., 2 2n, set En k = f ((k2 n, (k + )2 n ]) and F n = f ((2 n, ]). Define φ n (x) = 2 2n k=0 k2 n χ E k n + 2 n χ Fn (x). We see that ( En k = f ((k2 n, k + ) ]) ((( 2 n f k + ) )) 2 n, (k + ) 2 n = En+ 2k En+ 2k On the set En, k we see φ n = k2 n χ E k n and φ n+ = (2k)2 n χ E 2k n+ +(2k +)2 n χ E 2k+ = k2 n χ n+ E 2k n+ +(k + 2 )2 n χ E 2k+. n+ So φ n+ φ n on each En. k Also, we can see φ n+ φ n on F n. Therefore φ n+ φ n. Since φ n f, on each En k we see 0 f φ n (k + )2 n k2 n. It follows that φ n f and on the sets where f is bounded, it converges uniformly (as these sets fall into some E k n.)

8 Definition. Let (, M) be a measurable space.. A positive measure on M is a function µ : M [0, ] with the properties µ( ) = 0 and if {E j } j= M is a sequence of mutually disjoint sets, then µ = µ(e j ). To avoid trivialities, we assume µ(e) < for some E M. Usually, we refer to a positive measure as just a measure. 2. A measure space is a triple (, M, µ) where µ is a measure on M. Theorem 5. Let (, M, µ) be a measure space. Then. µ( ) = 0 2. (monotonicity) If E, F M and E F, then µ(e) µ(f ). 3. (subadditivity) If {E j } j= M, then µ E j µ(e j ). j= E k j= j= j= 4. (continuity from above) If {E j } j= M and E E 2..., then µ( E j ) = lim j µ(e j ). 5. (continuity from below) If {E j } j= M and E E 2... and µ(e ) <, then µ( E j ) = lim j µ(e j ). Proof.. Since there exists E M such that µ(e) <, we see µ(e) = µ(e ) = µ(e) + µ( ) since E and are disjoint. Now, subtracting µ(e) from both sides, we see µ( ) = Let E, F M such that E F. Then F = E (F \ E). Since E and F \ E are disjoint, we see µ(f ) = µ(e (F \ E)) = µ(e) + µ(f \ E) µ(e). 3. Use Lemma 4. Let {E j } j= M satisfying E E 2 with E 0 =. Define F j = E j \ j k= E k = E j \ E j. By Lemma, j= F j = j= E j. Thus as µ(e 0 ) = Similar µ( E j ) = µ( F j ) = µ(f j ) = µ(e j \ E j ) = µ(e j ) µ(e j ) = lim n µ(e n) Definition.. Let (, M, µ) be a measure space. Then a µ null set, or simply null set, is a set in M that has measure If some statement P is true for all points in except possibly those points in a null set, then we say P holds almost everywhere (a.e.) or we may say P holds for almost every x or P holds µ a.e..3 Integration Let (, M, µ) be a measure space. We set L + = {f : [0, ] : f is measurable}. Definition. Let φ L + be a simple function. Then there exists {a, a 2,..., a n } [0, ) and {E j } n j= M such that φ = n j= a jχ Ej. We define the Lebesgue Integral of φ with respect to µ by φdµ := n j= a jµ(e j ). More generally, if A M is measurable, then we define the Lebesgue Integral of φ over A with respect to µ as A φdµ := φχ Adµ = n j= a jµ(e j A).

9 Definition. Let f L + be any function. Then the Lebesgue Integral of f with respect to µ is { } fdµ = sup φdµ 0 φ f, φ L +, φ is simple. Also, if A M is measurable, then the Lebesgue Integral of f over A with respect to µ is given by A fdµ = fχ Adµ. Proposition 8. Let f, g L + and c [0, ]. Then. If f g, then fdµ gdµ. 2. If A, B M and A B then A fdµ B fdµ. 3. If A M, then A cfdµ = c A fdµ. 4. If f(x) = 0 for all x A M, then fdµ = 0. A 5. If A M and µ(a) = 0, then fdµ = 0. A Proposition 9. Let φ L + be a simple function. Define λ : M [0, ] by λ(e) = φdµ. Then λ is a measure on M. E Proof. Since φ is simple, there exists {a, a 2,..., a n } [0, ) and {E j } n j= M such that φ = n j= a jχ Ej. Let {A k } M be mutually disjoint sets. Then ( ) λ A k k= = = = = = = = φdµ k= A k φχ k= A dµ k ( ( n )) a j µ E j A k ( k= n ) a j µ (E j A k ) j= j= n j= a j k= j= k= k= k= µ(e j A k ) n a j µ(e j A k ) A k φdµ = λ(a k ) k= Theorem 6 (Monotone Convergence Theorem). Let {f n } n= L + be given. Suppose that. f j f j+ for all j =, f(x) = lim n f n (x) for all x Then f L + and fdµ = lim n f ndµ. Proof. Since f(x) = sup n f n (x), by Prop 7, f L +. By Prop 8(), we see { f ndµ} n= [0, ] is a nondecreasing sequence of real numbers and thus by the MCT for R, there exists M [0, ] such that lim n f ndµ = M. Since f n f for all n, Prop 8() also tells us f ndµ fdµ. Thus M fdµ. Thus we just need to show M fdµ. Let α (0, ) and φ L + be a simple function such that 0 φ f. Set E n := {x f n (x) αφ(x)}. Since f j f j+, we see E E 2. Since αφ f, we also have n=e n =. Thus f n dµ f n dµ α φdµ. () E n E n

10 By Prop 9 and Thm 5(4), lim n E n φdµ = φdµ. Thus, taking the limit of Equation M = lim n f ndµ α φdµ. By definition of the Lebesgue Integral for f, taking the sup over φ and α gives us M fdµ. Proposition 0. Let φ, ψ L + be simple functions. Then (φ + ψ)dµ = φdµ + ψdµ. Proof. Let n j= a jχ Ej and m k= b kχ Fk be the standard representations for φ and ψ respectively. Clearly, E j = m k= (E j F k ) for each j and F k = n j= (E j F k ) for each k. So (φ + ψ)dµ = = = = = n m (a j + b k )µ(e j F k ) j= k= n m a j j= j= µ(f k E j ) + k= m n b k k= j= µ(f k E j ) k= ( n m ) m n a j µ F k E j + b k µ F k E j n m a j µ(e j ) + a k µ(f k ) j= k= φdµ + ψdµ k= j= Theorem 7. If {f n } n= L + and f(x) = f n (x) for all x, then fdµ = f n dµ. n= n= Proof. First we will show for a sum of two functions, then n functions, then an infinite series of functions. So let f, f 2 L +, then by Theorem 4, there exists {φ j } j=, {ψ j} j= L+ such that φ j, ψ j are simple with 0 φ φ 2... f and 0 ψ ψ 2... f 2 lim φ j = f and lim ψ j = f 2. From these it follows that 0 φ + ψ φ 2 + ψ 2... f + f 2 lim φ j + ψ j = f + f 2. By the Monotone Convergence Theorem and Proposition 0, (f + f 2 )dµ = lim j (φ j + ψ j )dµ = lim φ j dµ + ψ j dµ = f dµ + f 2 dµ j Using Induction, we can show for n functions. To show for an infinite series, note that 0 lim f n n= N n= 2 f n... n= N f n (x) = f(x) n= Thus, applying the Monotone Convergence Theorem again, we see f n fdµ = lim N N f n dµ = lim n= N N n= f n dµ = n= f n dµ. Lemma (Fatou s Lemma- P.52). If {f n } n= L +, then (lim inf f n)dµ lim inf f ndµ.

11 Proof. Define g k (x) = inf f n(x) for all k and for all x. By Proposition 7, g k L + for all k. Also (g k ) k= is a monotone n k sequence with g(x) := lim g k(x) = lim inf f n(x). By the Monotone Convergence Theorem, k n We also see lim g k dµ = k lim k g k dµ = lim inf k lim g kdµ = k lim inf n f ndµ. g k dµ lim inf f k dµ k since g k f k for all k. Combining these two equations, we get what we wanted. Proposition (P 5). If f L +, then fdµ = 0 if and only if f = 0 a.e. Proof. First, we will show for simple functions. Let φ L + be a simple function and say φ = n j= a jχ Ej. Suppose φ = 0 a.e. Then either a j = 0 or µ(e j ) = 0 for all j =,..., n. Thus φdµ = n j= a jµ(e j ) = 0. Now suppose φdµ = 0. Then a j µ(e j ) = 0 for all j =,..., n, which implies either a j or µ(e j ) = 0 for all j. Thus φ = 0 a.e. Now let f L +. Suppose f = 0 a.e. Then for all simple φ L + such that 0 φ f, φ = 0 a.e. Then φdµ = 0 and by the definition of a Lebesgue Integral, fdµ = sup{ φdµ : φ L+, 0 φ f, and φ is simple} = 0. Now suppose f 0 a.e. Then for sufficiently large n, µ({x : f(x) > n }) > 0. Set E = {x : f(x) > n }. Then µ(e) > 0. Consider the simple functions n χ E. We see 0 n χ E f. By Proposition 8(), fdµ n χ Edµ = µ(e) > 0. n Remark. This shows for f L +, the Lebesgue Integral does not see values of f on the null sets. Corollary 4. If {f n } n= L + and lim inf f n(x) f(x) a.e. with f L +, then fdµ lim inf n n Proof. Set E = {x : lim inf f n (x) < f}. By hypothesis, µ(e) = 0. Thus we have lim inf n f nχ E C fχ E C for all x, and ( ) fχ E = 0 a.e. implies fχ E dµ = 0. Using these together with Fatou s Lemma, we see ( ) lim inf f ndµ lim inf f nχ E C dµ by Proposition 8() lim inf f nχ E C dµ by Fatou fχ EC dµ by Proposition 8() and ( ) = fχ E C dµ + fχ Edµ by Prop = fχ E C + fχ Edµ by ( ) = fdµ. f n dµ. Definition. Let (, M, µ) be a measure space. Then the measure µ is complete if whenever E M is a nullset, we find F M for all F E. Note. If µ is complete, then for E M with µ(e) = 0 and F E, we must have µ(f ) = 0. Theorem 8. Suppose (, M, µ) is a measure space. Set N = {N N µ(n) = 0} and M = {E F E M, F N for some N N }. Then M is a σ algebra and there exists a unique extension of µ to a measure µ on M. Say µ is the completion of µ.

12 Proof.. Clearly, M. 2. Let G M. Want to show G C M. Find E M and F N N such that G = E F. Define N = N \ E and F = F \ E. Then G = E F and E N = E F =. Also F N N. Then E F = E F = (E N C ) F = ((E N) N C ) ((E N) F ) = (E N) (N C F ). So G C = (E F ) C = [(E N) (N C F )] C = (E N) C (N C F ) C = (E N) C (N F C ). Now E N M which implies (E N) C M. Also N F C N. So G C M by definition. 3. If {G j } j= M, then there exists {E j} j= M and {N j} j= M and {F j} j= such that F j N j and G j = E j F j. Then G j = E j F j =. Notice that F j N j and µ( N j ) µ(n j ) = 0. So j= j= F j N N. So G j M. j= j= E j j= F j Definition. Define µ : M [0, ] by µ(e) = µ(e) for all E M and µ(e F ) = µ(e) for all E M and F N N. Notes.. µ defines a measure. (prove) 2. µ is well-defined and unique. Well-defined: Suppose E F = E 2 F 2 with E, E 2 M and F N N, F 2 N 2 N. Then µ(e F ) = µ(e ) µ(e 2 N 2 ) = µ(e 2 ) = µ(e 2 F 2 ). Similarly,. So µ(e F ) = µ(e 2 F 2 ). Unique: Suppose ν : M [0, ] is another completion. Let E F M. Then ν(e F ) ν(e N) = µ(e N) = µ(e) = µ(e F ) = µ(e) = ν(e) ν(e F ). Thus ν(e F ) = µ(e F ). Definition. Let (, M, µ) be a measure space. If E = σ finite. j= E j j= j= with {E j } j= M and µ(e j) < for all j, then E is Proposition 2 (p. 52). If f L + and fdµ <, then {x f(x) = } is a null set and {x f(x) > 0} is a σ finite set. Proof. Set E = {x f(x) = }. Then > fdµ fdµ = µ(e), which implies µ(e) = 0. Also for all j, set E E j = {x f(x) > j }. Then {x f(x) > 0} = E j and > fdµ E j fdµ > j µ(e j), which says µ(e j ) <. Definition. Let (, M, µ) be a measure space. Define L (, M, µ) to be the collection of all measurable functions f : R such that f dµ <. Note. If f is measurable, so is f L + and f = f + + f. Thus f ± dµ f dµ < if f L. Definition. If f L (, M, µ), then f is integrable and define fdµ = f + dµ f dµ. Proposition 3 (p. 53). If f L (, M, µ), then fdµ f dµ. Proof. By Theorem 7, fdµ = f + dµ f dµ f + dµ + f dµ f dµ = f dµ. Proposition 4 (p. 54). If f, g L (, M, µ), then TFAE. E fdµ = gdµ for all E M. E 2. f g dµ = 0

13 3. f = g a.e. Proof. (2) (3) By Prop (3) (2) If f = g a.e., then f g = 0 a.e. which implies f g = 0 a.e. and thus f g dµ = 0 by Prop. (2) () If f g dµ = 0, then for all E M Thus E fdµ = E gdµ. E fdµ gdµ = (f g)χ E dµ E f g χ E dµ f g dµ = 0. () (3) Contrapositive. Then µ({x f(x) g(x) 0}) > 0. Define E = {x f(x) g(x) > 0} and E 2 = {x f(x) g(x) < 0}. Then either µ(e ), µ(e 2 ) or both are > 0. Suppose µ(e ) > 0. Then (f g)χ E L + and so E (f g)dµ > 0. This implies E fdµ > E gdµ and so() does not hold. Similarly for µ(e 2 ) > 0. Note. We say f and g are related if f = g µ a.e. This defines an equivalence relation between functions in L (, M, µ). Definition. Let (, M, µ) be a measure space. Define L (, µ) to be the collection of all equivalence classes of integrable functions with respect to the relation just described. Notation. If we write f L (µ), then we really mean f is a representative for its equivalence class. Proposition 5 (p.47). Suppose µ is a complete measure. Then. If f is measurable and g = f µ a.e., then g is measurable. 2. If {f n } n= is a sequence of measurable functions and lim n f n (x) = f(x) a.e., then f is measurable. Proposition 6 (p.48). Let (, M, µ) be a measure space and (, M, µ) be its completion. If f : R is (M, B R ) measurable, then there exists an (M, B R ) measurable function g such that f = g µ a.e. Note. We identify L (µ) with L (µ). Definition. Let (, M, µ) be a measure space. Define ρ : L (µ) L (µ) [0, ) by ρ (f, g) = f g dµ where f, g L (, M, µ) are representatives for the equivalence classes f and g. Proposition 7. The function ρ is a metric on L (µ). Proof. Clearly, ρ (f, g) = ρ (g, f). Also if f, g, h L (µ), then ρ (f, g) = f g dµ = f h + h g dµ f h + h g dµ = ρ (f, h) + ρ (h, g). Finally, let f, g L (µ) and f, g L (, M, µ) be representatives for f and g. Then ρ (f, g) = f g dµ = 0 if and only if f = g a.e. which happens if and only if f, g are in the same equivalence class. Definition. If {f n } n= L (µ) and f L (µ) satisfies lim n ρ (f n, f) = 0, then we write f n f in L (µ) and say f n converges (strongly) to f in L (µ). Theorem (Lebesgue s Dominated Convergence Theorem). Let {f n } n= L (µ) be a sequence such that lim f n(x) = n f(x) µ a.e. and there exists g L (µ) such that f n (x) g µ a.e. for all n. Then f L (µ) and fdµ = lim f n dµ. n

14 Proof. By Propositions 5 and 6, we may assume f is measurable. By hypothesis, we see f(x) g(x) µ a.e. which implies f dµ gdµ <. So f L (µ). Since f n (x) < g(x) µ a.e., we also see that g +f n 0 and g f n 0 µ a.e. for all n. Notice that since lim f n = f µ a.e., lim inf (g + f n)(x) = g(x) + f(x) µ a.e. and lim inf (g f n)(x) = g(x) f(x) n n n µ a.e. By the corollary to Fatou s Lemma (Corollary 4), (g +f)dµ lim inf (g +f n )dµ = gdµ+lim inf f n dµ and n n (g f)dµ lim inf (g f n )dµ = gdµ lim sup n n is obvious, we see they are all = and thus fdµ = lim n f ndµ. f n dµ. Thus lim sup n f n dµ fdµ lim inf n f n dµ. Since Corollary 5. Suppose {f n } n= L (µ) satisfies the hypotheses of the LDC Theorem. Then f n f in L (µ), that is, lim n f n f dµ = 0. Proof. Notice lim n f n (x) f(x) = 0 µ a.e. f n (x) f(x) 2g(x) µ a.e. for all n. Then by the LDC Theorem, f n f dµ = 0dµ = 0. Theorem 9 (p 55). Suppose {f n } n= L (µ) satisfies function f L (µ) and f n dµ = f n dµ. n= n= Proof. Define g(x) = n= f n(x) for all x. By Theorem 7 and our hypotheses g(x)dµ = n= n= f n dµ = f n dµ <. Then n= f n converges µ a.e. to some n= f n dµ <. Then g L (µ). By Proposition 2, n= f n(x) < µ a.e. Hence f n convergence absolutely µ a.e. So we may put f(x) = n= f n(x) for those x where the series converges and f(x) = 0 everywhere else (i.e., on a null set). Moreover, n= f n(x) n= f n(x) g(x) µ a.e. By the LDC Theorem, f L (µ) and lim N N f n (x) = Types of Convergence f n dµ. f n f pointwise if lim n f n(x) = f(x) for all x. f n f a.e. if lim n f n(x) = f(x) µ a.e. fdµ = lim N n= f n f uniformly if for all ɛ > 0 there exists N ɛ such that for all n > N ɛ we have f n f < ɛ for all x. f n f in L if lim f n f dµ = 0. (strong convergence) n f n f in measure if for all ɛ > 0 we have lim n µ({x f n(x) f(x) ɛ}) = 0. Definition. We say that {f n } n= L (µ) is Cauchy in measure if for all ɛ > 0 we have f m (x) ɛ}) = 0. Proposition 8. Suppose {f n } n= L (µ) and f L (µ). If f n f in L, then f n f in measure. Proof. Let ɛ > 0 be given. Set E n = {x f n (x) f(x) ɛ}. Now Thus µ(e n ) = 0. 0 = ɛ lim f n f dµ lim f n f dµ lim ɛdµ = lim n n ɛ E n n ɛ µ(e n) 0. E n n N f n (x) = lim µ({x f n(x) m,n

15 Theorem 0. Suppose that {f n } n= is a sequence of measurable functions that are Cauchy in measure. Then there exists a measurable function f such that f n f in measure. Proof. Choose {g j } j= = {f n j } j= {f n} j= such that for all j we have µ({x : g j (x) g j+ (x) 2 j }) 2 j. }{{} E j Set F k = j=k E j. Then µ(f k ) j=k µ(e j) 2 j = 2 k. For x F k, we have for all i j k g j (x) g i (x) i l=j g l+(x) g l (x) i l=j 2 l 2 j. It follows that {g j } j= is pointwise Cauchy on F k C for all k. Then there exists f : R such that g j f on Fk C for all k, that is, g j f pointwise on \ ( F k) and f = 0 on the null set. Since µ(f ) = µ(e j ) 2 j = and F F 2 F 3, we find that 0 µ( F k) = lim µ(f k ) lim 2 k = 0. Thus µ( F k) = 0. Thus g j f µ a.e. and by Proposition 5, f is measurable. For each x F j we see g j (x) f(x) lim g j (x) g i (x) + lim g i (x) f(x) lim i i i i l=j i g l+ (x) g l (x) lim i l=j 2 l 2 j. (We know lim g i (x) f(x) = 0 as x F j implies x F i.)thus g j f in measure. Observe f n (x) f(x) f n (x) g j (x) + g j (x) f(x). So if f n (x) f(x) ɛ then either f n (x) g j (x) ɛ 2 g j (x) f(x) ɛ 2. Thus ({ µ ({x : f n (x) f(x) ɛ}) µ x x : f n (x) g j (x) ɛ }) ({ + µ x : g j (x) f(x) ɛ }). 2 2 or So taking the limit of both sides as n, j, we get lim µ({x : f n(x) f(x) ɛ} = 0 n since lim µ ({ x : f n (x) g j (x) ɛ 2}) = 0 for fn is Cauchy in measure and lim µ ({ x : g j (x) f(x) ɛ 2}) = 0 for f n converges in measure. Theorem. Suppose {f n } n= is a sequence of measurable functions such that f n f in measure with f measurable. Then there exists {f nj } j= {f n} n= such that f nj f µ a.e. Proof. Choose a subsequence {f nj } j= such that µ({x C : f n j f 2 j }) }{{} 2 j. Setting F k = j=k E j, µ(f k ) 2 k. E j For x F k and j k we see that f nj (x) f(x) 2 j. It follows that f nj f pointwise in \ k= F k. Thus f nj fµ a.e. since µ( k= F k) = 0. Theorem 2. Suppose {f n } n= is a sequence of measurable functions and f n f and f n g in measure for some measurable f and g. Then f = g µ a.e. Corollary 6. If {f n } n= L (µ) and f L (µ) with f n f in L, then there exists a subsequence {f nj } j= f nj f µ a.e. such that Examples. Take = N, M = P(N), µ(e) = the number of elements in E. (that is, the counting measure). If f L + (µ), then fdµ = f(k) If f L (µ), then k= f(k) = N f dµ <. So k= f(k) is absolutely convergent. Suppose that f n (k) = k n. Then f n(k) 0 pointwise (and thus µ a.e.), but not uniformly (as for all ɛ > 0, k n ɛ when k nɛ) and not in measure (as µ({k N : k n ɛ}) = ) Suppose that f n (k) = n. Then f n(k) 0 pointwise, uniformly, in measure, and µ a.e., but not in L.

16 k n for k n, Suppose that f n (k) = 2 0 otherwise. Then f n (k) 0 pointwise, uniformly, µ a.e., and in measure, but not in L. Theorem (Egoroff s Theorem). Suppose µ() < and f, f 2,..., f are complex valued and measurable functions on such that f n fa.e. Then for all ɛ > 0 there exists E such that µ(e) < ɛ and f n f uniformly on \ E. Proof. Let N be the set of all points where f n (x) f(x). So µ(n) = 0. For each k, n N, define E n,k = m=n{x \ N : f m (x) f(x) k }. Observe for all k that E n+,k E n,k and n=e n,k =. Since µ(e,k ) µ() <, we may use Theorem 5 to conclude that 0 = µ( n=e n,k ) = lim µ(e n,k ). So for all k there exists n k such that µ(e nk,k) < 2 k ɛ. Set E = N ( k= E n k,k). Then µ(e) µ(n) + µ( E nk,k) ɛ k= 2 k < ɛ. If x E, then for all n > n k, f n (x) f(x) < k, that is, f n f uniformly on \ E..4 L p Spaces Definition. A function F : (a, b) R is convex on (a, b) R if F (λx + ( λ)y) λf (x) + ( λ)f (y) for all λ [0, ] and x, y (a, b). Theorem 3. If F is convex on (a, b) R, then for all [x, y] (a, b) with x < y, we find that there exists M < such that F (s) M for all s [x, y]. Proof. Suppose there does not exist M <. Then for all n N there exists s n [x, y] such that F (s n ) < n. Since [x, y] is compact, there exists a subsequence (call it s n for simplicity) such that s n s for some s [x, y]. Let s [x, y] \ {s } be given. For each λ [0, ), we have F (λs + ( λ)s n ) λf (s) + ( λ)f (s n ) < λf (s) + ( λ)( n). It follows that F (λs + ( λ)s ) = for all λ [0, ). So F (s) = for all s [x, y] \ {s }, which contradicts the fact that F : (a, b) R. Theorem 4. If F is convex on (a, b) R, then F is continuous on (a, b). Proof. We will first prove a claim. Claim: For each x, y, z (a, b) satisfying x < y < z, we have Proof: Let y = λx + ( λ)z with λ = z y z x. Then F (y) F (x) y x < F (z) F (y) z y. as F is convex. This implies and thus Thus F (y) z y z x F (x) + y x z x F (z) F (x) z x z y F (y) y x z y F (z) y x F (x) z y F (z) z x (y x)(z y) F (y). F (y) F (x) y x y x F (y) + z y F (z) z x (y x)(z y) F (y) = y z F (y) + F (z) F (y) z y F (z) = z y. Let [x, y] (a, b) with x < y be given. Then F is uniformly bounded from below by Theorem 3. Let s (x, y) and t (s, y). So x < s < t < y. Then F (s) F (x) s x F (t) F (s) t s F (y) F (t) y t

17 which implies t s s x t s [F (s) F (x)] + F (s) F (t) y t [F (y) F (t)] + F (s). Since F is uniformly bounded, the RHS does not blow up, so as t s we see F (t) F (s). Similarly for t (x, s). Thus lim t s F (t) = F (s). Theorem 5 (Jensen s Inequality). Suppose that (, M, µ) is a measure space with µ() <. If F is a convex function on R and f L (µ), then F ( ) fdµ F fdµ. µ() µ() Proof. Since f L (µ), f dµ <. By Proposition 2, {x : f = + } is a nullset. So WLOG we may assume f is R valued (just redefine it to be 0 on the nullset). Put t = µ() fdµ. For each s (, t) and u (t, ), the claim above gives us F (t) F (s) F (u) F (t). t s u t F (t) F (s) Let β = sup s. Then β F (t) F (u) t u t s F (u) F (t) u t which implies F (u) F (t) + β(u t) for u (t, ). If u (, t), then β by definition of supremum. Thus F (t) F (u) + β(t u) which implies F (u) F (t) + β(u t). Let u = f(x). Then F (f(x)) is measurable and F (f(x)) F (t) + β(f(x) t) which implies F (f(x))dµ ( ) ( ) F (t)dµ + β f(x)dµ tdµ = F (t)µ() + β f(x)dµ tµ(). Note that if F (f(x)) is not in L then it integrates to, in which case this inequality is still true. Substituting the value for t, we see F (f(x))dµ F ( µ() ) fdµ µ(). Let = [n], M = P(), µ(k) = a k where n k= a k = and a k > 0. So fdµ = f(k)a k. Put F = e t, which is convex on R. Then by Jensen s Inequality, since µ() =, we have ( ) ( ) exp f(k)ak = exp fdµ e f dµ = a k exp(f(k)). Put y k = e f(k), that is, f(k) = ln y k. Then exp( ln y a k k ) a k exp(ln y k ) which implies n k= y a k n k = a k y k. Theorem (Young s Inequality). Let p + q = with p, q >. Then ab p a p + q b q. Proof. Use the above with α = p, α 2 = q, y = a p, y 2 = b p. Theorem (Hölder s Inequality). Let p + q = with p, q >. Let f, g L+. Then k= ( fgdµ ) /p ( f p dµ g q dµ) /q. Proof. If f p dµ = 0, then f = 0 a.e. which implies fg = 0 a.e. and thus fgdµ = 0. Similarly if g q dµ = 0. So assume f p dµ, g q dµ > 0. If f p dµ = or g q dµ =, the inequality is clear. So assume 0 < f p dµ, g q dµ <.Put F = f ( f p dµ ) /p and G = g ( gq dµ ) /q.

18 Observe F p dµ = f p dµ f p dµ =. Similarly, G q dµ =. Using Young is Inequality, So F Gdµ fg ( f p dµ) /p ( g q dµ) /q dµ which implies p F p dµ + q Gq dµ = p F p dµ + q G q dµ = p + q =. ( fgdµ ) /p ( f p dµ g q dµ) /q. Theorem (Minkowski s Inequality). Suppose p. Let f, g L + be given. Then ( /p ( (f + g) dµ) p /p ( f dµ) p + g p dµ) /p. Proof. If p =, then it is trivial. So assume p >. Then (f + g) p dµ = f(f + g) p dµ + g(f + g) p dµ ( f p dµ) /p ( (f + g) p dµ) p /p + ( g p dµ) /p ( (f + g) p dµ) p /p. If (f + g) p dµ = 0, clear. If it is then (f + g) p = 2 p ( 2 f + 2 g)p 2 p f p + 2 p g p = 2 p (f p + g p ) (since x p is convex) which implies one of f p and g p is. Thus we can divide by ( (f + g) p dµ) p /p to get Minkowski s Inequality. Definition. For each p [, ) and each measurable function f, define f p = ( f p dµ) /p and f = ess sup x f(x) = {a 0 : µ({x : f(x) > a}) = 0} (where inf =. This is called the essential supremum. Definition. For each p [, ] define L p (, µ) = {f L (µ) : f p < }..5 Normed Vector Spaces Let K denote R or C. Recall that a vector space is a set of elements with addition and scalar multiplication. By a subspace, we mean a vector subspace of. If x, denote by Kx the subspace {kx : k K}. If M and N are subspaces of, then M N = {x + y : x M, y N }. Definition. A seminorm on is a function : [0, ) such that x + y x + y for all x, y. λx = λ x for all x and λ K. If also satisfies x = 0 if and only if x = 0 then is called a norm on. A pair (, ) is called a normed vector space. Examples. R n is a VS and the function x p = ( n k= x k p ) /p for p [, ) is a norm. So is x = max{ x, x 2,..., x n }. For each p [, ], the space L P (µ) is a VS and the function p is a norm on L p. Fact. If (, ) is a NVS, then ρ (x, y) = x y for x, y is a metric on. The topology induced by this metric is called the norm (or strong) topology.

19 Definition. If ρ is a metric on a set, the topology induced by ρ is generated by E = {U : there exists ɛ > 0, x such that ρ(y, x) < ɛ for all y U}. (In Euclidean Space, these are the open balls of radius ɛ.) If E P(), then the smallest topology T (E) containing E is called the topology generated by E. Note. Each set in E is open in the topology generated by E by definition. Definition. Two norms and are equivalent if there exists constants 0 < c, c 2 < such that c x x c 2 x for all x. Examples. If = R n, then for all p, q [, ], the norms p and q are equivalent. If = R N (that is, the space of infinite sequences of real numbers), then for each p q [, ), the norms p and q are not equivalent. Definition. If (, ) is a NVS that is complete with respect to ρ, then we say that (, ), or just, is a Banach Space. Definition. Let {x n } n= be given. The series n= x N n converges to x if lim N n= x n = x (i.e., lim N N n= x n x = 0). The series x n is absolutely convergent if n= x n <. Theorem 6 (p. 52). A NVS (, ) is complete if and only if every absolutely convergent series is convergent. Proof. ( ) Suppose (, ) is complete. Let {x n } n= such that n= x n <. Then for all N we can define S N = N n= x n. Want to show S N is Cauchy. Let M > N be given. Then S M S N = M N+ x n M N+ x n 0 as M, N. Thus {S N } N= is a Cauchy Sequence in and since is complete there exists x such that lim N S N x = 0. Thus n= x n converges to x. ( ) Suppose every absolutely convergent series converges. Let {x n } n= be a Cauchy Sequence. Select a subsequence {x nj } j= such that for all j and n, m n j, we have x n x m < 2 j. Put y = x n and y j = x nj x nj for all j >. Then x nk = k j= y j. Also, y j = y + y j = x n + x nj x nj x n + 2 j x n + <. j= j=2 j=2 J Then, by hypothesis, there exists x such that lim J j= y j = x. Then lim J x nj = x. Of course x n x x n x nj + x nj x and for n n j, x n x nj 2 j and x nj x 0. Thus x n x 0. Corollary 7. If (, M, µ) is a measure space, then (L (µ), ) is a Banach Space. Theorem 7 (p. 83). For p [, ], the space (L P (µ), p ) is a Banach Space. Proof. Case : p [, ). Suppose that {f k } k= Lp (µ) satisfying k= f k <. Put A p = f k p p and G n = n f k, with G = f k. Clearly, G p < Gp 2 < < Gp and by Minkowski s inequality G n p p = By the Monotone Convergence Theorem, ( n p f k ) dµ n f k p dµ = j=2 n f k p p A p. ( n p lim G ( n p ( ) p n p p = lim f k ) dµ = lim f k ) dµ = f k dµ = G p p A p < n n

20 since G n p p A p for all n. This implies that ( f k ) p < a.e. and so f k < a.e. Thus for almost every N x there exists F (x) < such that F (x) = lim N f k(x). Define F (x) = 0 for those x where the sum is infinite. We need to show F L p (µ) and lim F N f k = 0. We have p F p dµ = f k (x) dµ ( ) p f k dµ = G p p A p <. Thus F L p (µ). Finally, for all n, F n f k p = n+ f k p ( f k ) p = G L. Also lim n F (x) f k (x) = 0 a.e. So by the Dominated Convergence Theorem, Thus by Theorem 6, L p (µ) is complete. lim F n f k p p = lim n n F n f k p dµ = 0. Case 2: p =. Let {f k } L (µ) satisfying f k <. For each k, set A k = {x : f k (x) > f k }. By the definition of, each A k is a null set. Also A = A k is a null set. For each x \ A, f k(x) f k <. So there exists F such that F (x) = f k(x) < for all x \ A. Put F (x) = 0 for all x A. N So F = lim N f k(x) µ a.e. Now for x \ A, F (x) = f k(x) f k(x) f k <. So F L as µ(a) = 0. Finally since f k <. lim F n f k = lim f k lim n n n+ n n+ f k = 0 Proposition 9. Let S = {simple functions on }. For each p [, ], the set S L P is dense in L P. Proof. The case p = is covered by Theorem 4. Suppose p [, ) and let f L p (µ) be given. Want to find a sequence {f n } n= S L p such that lim n f n f p = 0. By Theorem 4 (applied to f + and f ), there exists a sequence {f n } n= S such that f n f and lim n f n (x) = f(x) for all x. Since f L p, we find that f n p dµ f p dµ < which implies f n L p for all n. So {f n } n= S L p. Moreover, f n f p ( f n + f ) p 2 p f p L and f n f p 0 for all x. By the LDC, lim n f n f p dµ = 0 which implies lim f n f p = 0. Proposition 20. If p q r, then L p L r L q and f q f λ p f λ q where q = λ p + λ r. Proof. If p = q = r =, trivial. If p < and q = r = then clearly L p L L and if we take λ = 0, then we see f q f r = f 0 p f r. So suppose p, q <. If r =, then ( /q f dµ) q = ( ( q p q = f = f p/q p ) /q f p f q p dµ f p f q p dµ ( f p dµ f p/q. Let λ = p q. If f Lp L, then f p, f <, so f q f λ p f λ ) /q ) /q <. Thus f L q. Now suppose r <. Note

21 that λq < p. ( /q f dµ) q = = ( ) /q f λq f ( λ)q dµ ( ( f λq ) ) q p/λq ( λq p ) ( ( dµ f ( λ)q) ( ) λ/p ( f p dµ ) λ f r r dµ p p λq ) q ( p λq p ) dµ by Holder s Inequality = f λ p f λ r. By the same argument as above, f L P L implies f L q. Proposition 2. If µ() <, then for all p q, we have L q L p and f p f q µ() p q. Proof. If q <, then ( /p ( ) /p f dµ) p = f p dµ ( ) p ( f p ) q/p q ( p ) ( dµ = f q µ() p q. ) q f q p dµ p ( q p q ) 2 Measure Theory The Lebesgue Measure on R n. Suppose (R n, M, m n ) is a measure space where the measure m n : M [0, ] with M P(R n ) is the unique measure such that ( n ) m n (a k, b k ) = (b k a k ) k= where (a k, b k ) R for all k. What can we say about M if we want to measure all the open boxes (a k, b k )? Since any open set is a countable union of open boxes, all open sets in the usual topology must be in M. The smallest σ algebra M must be the Borel σ algebra. So we want to somehow extend m n from the boxes to all of B R n. Definition. Let. A family of sets C P() is a semialgebra if., C 2. If E, E 2 C, then E E 2 C (and thus all finite intersections are in C). 3. If E C, then there exists a finite sequence {E i } k i= C with E i E j = for all i j such that E C = k i= E i. Examples. The following are semialgebras: I = { open, half-open, closed intervals on R}. I n = {crossproduct of any n elements of I}. Notation. Denote any interval with endpoints a and b by I(a, b). Definition. Let. A family of sets F P() is an algebra if., F 2. If E, E 2 F, then E E 2 F (and thus all finite intersections are in F).

22 3. If E F, then E C F. Note that by 2 and 3, we are only allowing finite unions to be in F, unlike in a σ algebra. Examples. The following are algebras F(I) = {E R E = l k= I k, I k I, I j I k = for j k}. F(I) n = {E R n E = l k= I k, I k I n, I j I k = for j k}. In general, if C is a semialgebra, then is an algebra. F(C) = { E E = } l E k, E k C, E j E k = for j k k= Definition. Let C P(). A set function µ : C [0, ] is called monotone if for all A, B C satisfying A B, we have µ(a) µ(b). finite additive if {E k } l k= C such that E j E k = and l k= E k C implies µ( l k= E k) = l k= µ(e k). countably additive if {E k } k= C such that E j E k = and k= E k C implies µ( k= E k) = k= µ(e k). countably subadditive if {E k } k= C such that k= E k C implies µ( k= E k) k= µ(e k). st Goal: Given a monotone countably additive set function µ defined on a semialgebra C, we want to extend µ to a monotone countably additive function µ defined on an algebra F(C) generated by C. Proposition 22. Let C P(). Then there exists a unique algebra F(C) P() such that C F(C) and if A P() is an algebra such that C A, then F(C) A. So F(C) is the smallest algebra containing C. Proof. Define F(C) = {A C A P(), A is an algebra}. Definition. Given C P(), the algebra F(C) provided by Prop 22 is called the algebra generated by C. Proposition 23. If C is a semialgebra, then the algebra generated by C is F(C) := {E : E = l k= E k, E j E k =, j k, E k C}. Example. Recall I was a semialgebra. What kind of properties does m : I [0, ] defined by m(i(a, b)) = b a have? It is monotone, finitely additive, countably additive (2 cases: if the union is an interval which is finite or infinite), countably subadditive (by monotonicity, countable additivity and Lemma ). Theorem 8. Suppose µ is a finitely additive and countable subadditive set function on a semialgebra C such that µ( ) = 0. Then there exists a unique countably additive set function µ on F(C) such that µ(e) = µ(e) for all E C. Proof. For all E F(C), by Prop 23, there exists {E k } n k= C such that E = n k= E k and E j E k = if j k. Define µ(e) = n k= µ(e k). Claim : µ is well-defined. Proof : Let E F(C) and suppose there exists {E k } n k= and {F l} m l= C such that E j E k = for j k and F j F l = for j l and n k= E k = E = m l= F l. Then for all l =, 2,..., m, F l = F l E = F l ( n k= E k) = n k= (F l E k ) and for all k =, 2,..., n, E k = E k E = E k ( F l ) = m l= (E k F l ). So µ(e) = n k= µ(e l) = n k= µ( m l= E k F l ) = n k= m l= µ(e k F l ) = m l= n k= µ(e k F l ) = m l= µ( n k= E k F l ) = m l= µ(f l).

23 Claim 2: µ is finitely additive on F(C). Proof : Suppose {E k } n k= F(C) with E k E j = for j k, then n k= E k F(C) since F(C) is an algebra. By Prop 23, there exists {G r } s r= C such that s r=g r = n k= E k. Also, for all k =, 2,..., n, there exist mutually disjoint {F k,l } m k l= C such that E k = m k l= F k,l. Then for all k =,..., n ( n ) ( s ) E k = E k E k = E k G r = k= r= ( n ) ( n Also for all r =,.., s, G r = G r E k = G r ( n ) µ E k k= Claim 3: µ is countably subadditive. k= ( s ) = µ G r = = s r= m k r= k= l= ( mk l= m k k= l= s µ(g r ) = r= n µ(g r F k,l ) = ) ( s ) F k,l G r = F k,l ) = Proof : Same as above, except replace n with and change the = to. n r= m k k= l= m k l= r= G r F k,l. Now ( s n ) m k µ G r F k,l r= ( k= l= n s ) m k µ G r F k,l = k= r= l= Note that the countable additivity of µ follows from the next theorem (Theorem 9) s (F k,l G r ). n µ(e k ). Theorem 9. Let F be an algebra of sets on and µ : F [0, ] be a set function such that µ( ) = 0. Then µ is countably additive if and only if it is both finitely additive and countably subadditive. Proof. First note that if µ is finitely additive, then (since F is an algebra) for A, B F with A B, we see µ(b) = µ(a B \ A) = µ(a) + µ(b \ A) µ(a). Thus µ is monotone. ( :) Suppose µ is countably additive. Clearly µ is finitely additive as µ( ) = 0. To show subadditive, let {E k } k= F such that k= E k F. By Lemma, there exists a sequence {F k } k= of mutually disjoint sets such that F k = E k. Using countable additivity and monotonicity, we see µ( E k ) = µ( F k ) = µ(f k ) µ(e k ). ( ) Suppose µ is finitely additive and countably subadditive. Let {E k } F be mutually disjoint sets such that E k F. Since µ is countably subadditive, µ( E k ) µ(e k). To show the opposite inequality, we use finite additivity and monotonicity to conclude µ( E k ) µ( n E k ) = n µ(e k) for all n. Taking the limit as n, we get µ( E k ) µ(e k). Thus µ( E k ) = µ(e k). Definition. Suppose A P() is an algebra. A function µ : A [0, ] is called a premeasure if µ( ) = 0 and µ is countably additive. Theorem 8 shows how to construct a premeasure on an algebra, generated from a semialgebra, from a finitely additive countably subadditive function on that semialgebra. Notation. Define Ĩ := {(a, b] : a < b R} {(, b] : b R} {(a, ) : a R} {(, )} { }. Note the σ algebra generated by Ĩ is B R. Also Ĩ is a semialgebra. Proposition 24. Let F : R R be an increasing function. Define µ F : Ĩ [0, ] by µ F ((a, b]) = F (b) F (a), µ F ((, b]) = F (b) lim F (x), µ F ((a, )) = lim F (x) F (a), µ F ((, )) = lim F (x) lim F (x), µ F ( ) = 0. Then µ F is welldefined, finitely additive and monotone. Moreover, if F is right continuous, then µ F is countably x x x x subadditive. Proof. It is clear that µ F is well-defined. Suppose {I k } n k= Ĩ are disjoint. First suppose each I k is of the form (a k, b k ] and n k= (a k, b k ] = (a, b) Ĩ. Then WLOG, assume a = a < b = a 2 < b 2 =... = a n < b n = b. So k= n µ F (I k ) = n µ F ((a k, b k ]) = n F (b k ) F (a k ) = F (b) F (a) = µ F ((a, b]).

24 Now suppose n k= I k = (, b] Ĩ. WLOG, assume I = (, b ] and I k = (a k, b k ] with b = a 2 < b 2 =... = a n < b n = b. So n µ F (I k ) = µ F (I ) + n 2 µ F ((a k, b k ]) = F (b ) lim x F (x) + F (b n ) F (a 2 ) = F (b n ) lim x F (x) = µ F ((, b]). Similarly, the other cases hold. Thus µ F is finitely additive. Monotonicity follows. Thus we need only to show countable subadditivity in the case that F is right continuous. Suppose I = (a, b] k= I k with {I k } k= Ĩ. Let ɛ (0, b a). For each k, define (a k, b k + δ k ) if I k = (a k, b k ], I k = (, b k + δ k ) if I k = (, b k ], otherwise, I k where δ k satisfies F (b k + δ k ) F (b k ) < ɛ2 k. Now {I k } k= is an open cover for [a + ɛ, b]. Since compact, Heine Borel says there exists a finite subcover, call it {I k }n k= for simplicity. WLOG, assume a + ɛ I. If b I, then b + δ < b and thus [b + δ, b] n k=2 I k. WLOG, assume b + δ I 2. Then a 2 < b + δ. If b I 2, then b 2 + δ 2 < b and so [b 2 + δ 2, b] n k=3 I k. Continue to find m < n such that a < a + ɛ < b + δ < b 2 + δ 2 < < b < b m + δ m, that is, b I m. Note that this also says a i+ < b i + δ i. Now F (b) F (a) = F (b) F (a + ɛ) + F (a + ɛ) F (a) Letting ɛ 0 +, the right continuity of F yields F (b m + δ m ) F (a ) + F (a + ɛ) F (a) = m k= ((F (b k+ + δ k+ ) F (b k δ k )) + F (b + δ ) F (a ) + F (a + ɛ) F (a) m k= ((F (b k+ + δ k+ ) F (a k+ )) + F (b + δ ) F (a ) + F (a + ɛ) F (a) = m k= ((F (b k + δ k ) F (a k )) + F (a + ɛ) F (a) = m k= ((F (b k + δ k ) F (b k ) + F (b k ) F (a k )) + F (a + ɛ) F (a) m k= ɛ2 k + m k= µ F (I k ) + F (a + ɛ) F (a) ɛ + F (a + ɛ) F (a) + m k= µ F (I k ) µ F (I) = F (b) F (a) µ F (I k ). Now suppose I is an infinite interval. If I = (, b], then for each M >, the same argument shows that µ((m, b]) = F (b) F (M) k= µ F (I k ). Now letting M, we see µ F ((, b]) = F (b) lim M F (M) k= µ F (I k ). Similarly for the other cases. Note. It is also the case that µ F is countably additive, but we don t prove that here. For reference, Folland refers to this as µ 0. This is similar to Prop.5 in Folland. Proposition 25. Let F : R R be increasing and right continuous. Define µ F : Ĩ [0, ] as in Proposition 24. Then µ F : F(Ĩ) [0, ] defined by µ F ( n I k ) = n µ F (I k ) whenever {I j } n k= Ĩ satisfies I j I k = for j k is a premeasure on F(Ĩ). Proof. Follows from Prop 23, Thm 8, and Prop 24. Remark. If F = x, then µ F is the usual length of an interval. Proposition 26. Suppose µ : Ĩ [0, ] is finitely additive and µ((a, b]) < for each a, b R. Then there exists an increasing function F : R R such that µ((a, b]) = F (b) F (a) for all (a, b] R. If µ is also countably additive on Ĩ, then F is right continuous and µ F = µ.

25 µ((0, x]) if x > 0 Proof. For all x R, define F (x) = 0 if x = 0. We want to verify that µ((a, b]) = F (b) F (a) for all (a, b] R. µ((x, 0]) if x < 0 Since µ is finitely additive, if 0 < a < b µ((a, b]) = µ((0, b] \ (0, a]) = µ((0, b]) µ((0, a]) = F (b) F (a). Similarly for a 0 < b and a < b 0. To show F is increasing, note that for 0 < a < b, F (b) F (a) = µ((a, b]) 0. Similarly for the other two cases. Now, suppose µ is countably additive. We want to show F is right continuous. Let x R and {x k } k= (x, ) such that x k x as k and {x k } k= is decreasing. Notice {F (x k)} k= is decreasing and bounded below by F (x), so it converges. Case : Let x > 0. Then Thus F (x) = lim N F (x N+ ). Similarly if x = 0. F (x ) = µ((0, x ]) = µ((0, x]) + µ( k= (x k+, x k ]) Case 2: Let x < 0. Then for some m N we find x m < 0. Then Thus F (x) = lim n F (x n+ ). = F (x) + lim N N µ((x k+, x k ]) = F (x) + lim N N F (x k) F (x k+ ) = F (x) + lim N F (x ) F (x N+ ). F (x) = µ((x, 0]) = (µ((x m, 0]) + µ((x, x m ])) = F (x m ) µ( k=n (x k+, x k ]) = F (x m ) lim n k=n F (x k) F (x k+ ) = F (x m ) F (x m ) + lim F (x n+ ). To show µ = µ F, we need only to compare µ(i) and µ F (I) on infinite intervals. Suppose I = (, ). Then I = k= (( k, k + ] (k, k]). Since µ(i) 0 we have µ(i) = lim n = lim n n µ(( k, k + ]) + µ((k, k]) n F ( k + ) F ( k) + lim n = lim F ( n) + lim F (n) = µ F (I). n n n F (k) F (k ) 2nd Goal: Given a premeasure on an algebra A, we want to extend a measure on the σ algebra generated by A. Intermediate Goal: Approximate measure of any subset of a nonempty using the premeasure on an algebra A. Idea: Recall m : I [0, ] was given by m(i(a, b)) = b a (where I(a, b) is any interval with endpoints a and b) for a, b R. The algebra generated by I is F(I), which is the collection of all finite unions of disjoint intervals in I. The extension of m to m on F(I) is m(e) = n k= (b k a k ) for E = n k= I(a k, b k ), with {I(a k, b k )} n k= mutually disjoint. Now, we want to extend m to a set function that measures any subset of R. Suppose E R. Then we can find at least one countable family {I k } k= F(I) such that E k= I k (take I k = R for all k). Since E k= I k, we expect the measure of E to be m( k= I k) m(i k). So, in general, we want measure of E m(i k ). k=