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NYU MATH-AD 231 - Real Analysis Oral Exam Notes

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Real Analysis Oral Exam Notes 2008Chaitanya Ekanadham1 Basic properties of R1.1 Definitionssup/inf, open/closed sets, Gδand Fσsets, borel sets, Lindelof property, finiteintersection property, open cover, subcover.1.2 Useful TheoremsTheorem 1. (R satisfies the Lindelof property)[αGα⊇ A ⇒ ∃{αj} s.t.[jGαj⊇ ADepends on: axiom of choice, countability of rationalsProof idea: For each x ∈ A, take an interval Ixwith rational endpointsthat contains x and is in some Gαfrom the open cover. Use countability.Theorem 2. (Heine Borel Theorem)C ⊆ R closed and bounded ⇒ every open covering {Gα} of C contains afinite sub cover.Depends on: existence of sup/infProof idea: Do first for C = [a, b]. Let A = {x ∈ C : (a, x] has a finitesub cover}. Show that sup A = b. Extend to arbitrary sets by adding Ccto{Gα}, and applying to some interval I ⊇ C.Theorem 3. (R satisfies finite intersection property.)Let {Cα} be a collection of closed sets in R containing at least o ne boundedset C. Thenn\j=1Cj6=  ∀n ⇒\αCα6= Depends on: Heine Borel (#2)Proof idea: Since no finite union from {Ccα} can cover C, {Ccα} cannotcover C by Heine-Borel1Theorem 4. Every open set O ⊆ R is a disjoint countable union of openintervals[jIjDepends on: def. of open set, sup/inf, countability of ratio nalsProof idea: ∀x ∈ O, let Ixbe the biggest open interval containing x s.t.Ix⊆ O. Then any two intervals in {Ix} are disjoint or the same, and O =[xIx. {Ix} is countable since each Ixcontains a distinct rational.1.3 Important examples1. Something not Lindelof2. Spaces in which compactness is not equivalent to closed and bounded2 Measure theory2.1 Definitionsfield, sigma field, semi-ring, (finite/sigma-finite) measure, counta ble additiv-ity, outer measure, subadditivity, monotone class2.2 Useful TheoremsTheorem 1. (Equivalent definitions of countable additivity)1. ∀(Aj) disjoint, µ([jAj) =Xjµ(Aj) ⇔2. ∀(Aj) increasing/decreasing, µ([jAj/\jAj) = limj∞µ(Aj) ⇔3. Aj↓  ⇒ limj∞µ(Aj) = 0Depends on: set theory, limits of sumsProof idea: Reinterpret increasing/decreasing sequences of sets as count-able disjoint unions and vice versa.2Theorem 2. (Caratheodory extension theorem)If F is a field of subsets of X and µ c.a. on (X, F) ⇒ ∃ an extension µ∗ofµ that is c.a. on (X, σ(F)).Depends on: properties of countable additity, definition of sigma algebra,monotone class theorem (for uniqueness only)Proof idea: 5 steps:1. Introduce outer measure µ∗(defined on 2X) and show it is nonneg,monotonic, agrees with µ on F (easy) and is subadditive (use 2−jǫtrick and def. of outer measure).2. Define S = {E ∈ 2X: ∀A ∈ 2X, µ∗(A) = µ∗(A ∩ E) + µ∗(A ∩ Ec)} tobe the class of measurable sets. Show S is a σ-field (show finite unionswhich also implies finite a dditivity of µ on S, then take limit and usesubadditivity to show countable unions).3. Show µ∗is countably additive on S (take limit with finite additivity,and then use subadditivity f or the reverse inequality)4. Show t hat S ⊇ F (use the definition of outer measure and a pply count-able additivity of µ on F).5. Use monotone class theorem to show uniqueness (show that the class ofsubsets on which 2 extensions agree is a monotone class and thereforecontains σ(F)).Theorem 3. A monotone class M that is a field is a σ-algebra.Depends on: definitionsProof idea: Take An∈ M disjoint. Then since each partial union is inM, and since monotone classes are closed under increasing limits, the resultfullows.Theorem 4. Monotone class theoremLet F be a field and let M be the smallest monotone class containing F.Then F ⊇ σ(F).Depends on: Previous theorem3Proof idea: STS that M is an algebra. Define for each A ∈ F the setMA= {B ∈ M : A ∩ B ∈ M, A ∩ Bc∈ M, Ac∩ B ∈ M}. Then one canshow that MAis a monotone class and is thus equal to M. But this meansthat ∀B ∈ M, MB= M ⇒ M is an algebra.Theorem 5. Existence of Lebesgue measure∃ a c.a. measure λ on (R, B) that is translation-invariant and agr ees withthe length of intervals.Depends on: Caratheodory extension, Heine BorelProof idea: Define λ((a, b]) = b − a for a ll intervals (a, b] and then showthat t he extension theorem applies. STS λ defined on the field F generatedby these intervals is countably additive. We do this in 2 steps:1. Show on the semi-ring of semi-closed intervals. Suppose I = (a, b] =∞\j=1Ij. Clearly, λ(I) ≥Pλ(Ij). To show the reverse, make t he Ij’s2−j(ǫ/2) bigger and open, and make I ǫ/2 smaller and closed. Thenthe stretched Ij’s are an open cover for the shrunken I, and then wecan use Heine Borel to conclude thatPλ(Ij) ≥ λ( I) − ǫ, ∀ǫ > 0.2. Show on the field. Suppose A =N[j=1Ij=∞[j=1Ajfor some A, Aj∈ Fwith Ajdisjoint. Then show that λ(A) =∞Xj=1λ(Aj) by commutingsums and applying the previous step (just manipulations).2.3 Important examples1. item12. item243 Measurable sets and functions3.1 Definitionsmeasurable set (Caratheodory definition), measurable function, a.e. conver-gence, uniform convergence3.2 Useful TheoremsTheorem 1. (Measurable sets are like open/closed sets)1. E measurable ⇔2. ∀ǫ > 0 ∃O ⊇ E open s.t. µ(O ∩ Ec) < ǫ ⇔3. ∀ǫ > 0 ∃C ⊆ E closed s.t. µ(E ∩ Cc) < ǫ ⇔4. ∀ǫ > 0 ∃ a Gδset G ⊇ E s.t. µ(G ∩ Ec) < ǫ ⇔5. ∀ǫ > 0 ∃ a Fσset F ⊆ E s.t. µ(E ∩ Fc) < ǫDepends on: def. of outer measure, set theoryProof idea: Finish this proof!1. (1)  ( 2): use def. of o uter measure to find a set of open intervals(tightly) containing E, use def. of E measurable to bound µ∗(O ∩ Ec)2. (2)  (4): Take ǫn= 1/n in (2) to get open set On. Take G =\nOn3. (4)  (???)Theorem 2. Alternative definitions of real- valued measurable f unctions:f : (X, Ω)  (R, B) measurable ⇔∀a ∈ R, {x ∈ X : f(x) <, >, ≤, ≥, = a} ∈ ΩDepends on: commuting of preimage and count able unions, def. of sigmaalgebra5Proof idea: ⇒ is clear.⇐ : NTS ∀A ∈ B, f−1(A) ∈ Ωf−1(Ij) ∈ Ω ⇒ f−1([jIj) ∈ Ω since preimage and countable unions com-mute. ≥ is the complement of <, and ≤ is the countable union of <.Remark: It is easily shown that measurable functions are closed under{+, ∗, sup, inf, lim inf, lim sup, lim}Theorem 3. (Measurable fns on finite intervals a re almost simple/step/cnts)Let f : [a, b]  R be measurable s.t. f = ±∞ on a set of measure 0. Then:1. ∀ǫ > 0, ∃ a step fn g : [a, b]  R and a


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