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REAL ANALYSIS II HOMEWORK 1 ˙ CIHANBAHRAN The questions are from Folland’s text. Section 3.1 1. Prove Proposition 3.1. Proposition 1. Let ν be a signed measure on (X,M). If {Ej} is an increasing sequence S in M, then ν( ∞E ) = lim ν(E ). If {E } is a decreasing sequence in M and ν(E ) 1T j j→∞ j j 1 is finite, then ν( ∞E ) = lim ν(E ). 1 j j→∞ j S Proof. Assume that {E } is an increasing sequence in M. Write E = ∞E,E =∅ j r 1 j 0 and for every j ≥ 1 define F = E E . Then j j j−1 • {Fj} is a disjoint sequence in M, S • E = ∞Fj, and 1 • EN =SNFj for every N ≥ 1. 1 Therefore Ñ é ∞ N ν(E) = Xν(F ) = lim Xν(F) = lim ν(E ). j N→∞ j N→∞ N j=1 j=1 Now assume that {E } is a decreasing sequence in M with |ν(E )| < ∞. Write E = j 1 T∞E. Then{ErE }isanincreasingsequenceinMsuchthatS∞(E rE )=E rE. 1 j 1 j 1 1 j 1 Therefore by the first part ν(E rE)= lim ν(E rE ). 1 j→∞ 1 j Note that for every j, ν(E ) = ν(E )+ν(E rE ) is finite, so ν(E ) is finite. Similarly 1 j 1 j j ν(E) is finite. Thus the above equality can be rewritten as ν(E )−ν(E)= lim (ν(E )−ν(E )) . 1 j→∞ 1 j As ν(E ) is finite, we get 1 ν(E) = lim ν(Ej). j→∞ 2. If ν is a signed measure, E is ν-null iff |ν|(E) = 0. Also, if ν and µ are signed measures, ν ⊥ µ iff |ν| ⊥ µ iff ν+ ⊥ µ and ν− ⊥ µ. Assume |ν|(E) = 0. Then given F ⊆ E, since |ν| is a measure we have |ν|(F) = 0. Thus ν+(F) = 0 = ν−(E) = 0 and so ν(F) = 0. 1 REAL ANALYSIS II HOMEWORK 1 2 Conversely, assume that E is ν-null. Since ν+ ⊥ ν− there exists disjoint sets A,B ⊆ X such that A∪B = X and ν+ is null on B and ν− is null on A. Observe that 0 = ν(E ∩A) = ν+(E ∩A)−ν−(E ∩A)=ν+(E∩A)=ν+(ErB)=ν+(E) and 0 = ν(E ∩B) = ν+(E ∩B)−ν−(E∩B)=−ν−(E∩B)=−ν−(ErA)=ν−(E). Thus |ν|(E) = ν+(E)+ν−(E) = 0. To keep track of the equivalences, let’s write (1) ν ⊥ µ. (2) |ν| ⊥ µ. (3) ν+ ⊥ µ and ν− ⊥ µ. (1) ⇒ (2): There exists A,B ⊆ X such that X is the disjoint union of A and B, and ν is null on B and µ is null on A. By above, |ν|(B) = 0. Being a positive measure, so |ν| is null on B. Hence |ν| ⊥ µ. (2) ⇒ (3): There exists A,B ⊆ X such that X is the disjoint union of A and B, and |ν| is null on B and µ is null on A. In particular ν+ and ν− are null on B, so ν+ ⊥ µ and ν− ⊥ µ. (3) ⇒ (1): There exists A,B ⊆ X such that X is the disjoint union of A and B, and ν+ is null on B and µ is null on A. And there exists C,D ⊆ X such that X is the disjoint union of C and D, and ν− is null on D and µ is null on C. So µ is null on A∪C. Moreover, given E ⊆ Xr(A∪C) = B∩D, we have ν+(E) = 0 and ν−(E) = 0 so |ν|(E) = 0. Thus by above, ν is null on B ∩ D. Thus ν ⊥ µ. 3. Let ν be a signed measure on (X,M). a. L1(ν) = L1(|ν|). b. If f ∈ L1(ν), |R fdν| ≤ R |f|d|ν|. c. If E ∈ M, |ν|(E) = sup{|RE fdν| : |f| ≤ 1}. a. By definition L1(ν) = L1(ν+) ∩ L1(ν−). Also since |ν| = ν+ + ν−, we also have L1(|ν|) = L1(ν+)∩L1(ν−). b. We have Z Z Z + − fdν = fdν − fdν Z Z + − ≤ fdν + fdν ≤Z |f|dν++Z |f|dν− = Z |f|d|ν|. c. If |ν|(E) = ∞, then either ν+(E) or ν−(E) is ∞, so in any case ∞ = |ν(E)| = | RE 1dν|. Thus the equality holds. Next, assume |ν|(E) < ∞. Then every ν-measurable f with |f| ≤ 1 satisfies fχE ∈ L1(|ν|) = L1(ν). Thus by (b), Z Z Z Z fdν = fχ dν ≤ |fχ |d|ν| = |f|χ d|ν| ≤ |ν|(E). E E E E Thus |ν|(E) is an upper bound for the given set, hence we get the “≥” part of the desired inequality. For the other direction, let X = P ∪N be a Hahn decomposition of REAL ANALYSIS II HOMEWORK 1 3 ν and consider f = χ −χ ∈L1(ν). Since P ∩N = ∅, |f| ≤ 1 and we have P∩E N∩E Z Z Z fdν = fdν − fdν E P∩E N∩E =|ν(P ∩E)+ν(N ∩E)| + − = ν (E)+ν (E) =|ν|(E) which yields “≤”. R 6. Suppose ν(E) = Efdµ where µ is a positive measure and f is an extended µ-integrable function. Describe the Hahn decompositions of ν and the positive, negative, and total variations of ν. Write (X,M,µ) for the measure space. Let P = {x ∈ X : f(x) ≥ 0} and N = {x ∈ X : f(x) < 0}. Clearly X is the disjoint union of P and N. Moreover for every E∈Mcontained in P we have ν(E) = R fdµ ≥ 0 so P is ν-positive and similarly N E is ν-negative. Thus X = P ∪N is a Hahn decomposition of ν. Therefore for every E ∈ M Z Z ν+(E) = ν(P ∩E) = fdµ= f+dµ P∩E E and Z Z Z ν−(E) = −ν(N ∩E) = − fdµ=− −f−dµ= f−dµ. N∩E E E Finally, |ν|(E) = ν+(E)+ν−(E) = Z (f+ +f−)dµ. E Section 3.2 8. ν ≪ µ iff |ν| ≪ µ iff ν+ ≪ µ and ν− ≪ µ. Assume ν ≪ µ. Let X = P ∪N be a Hahn decomposition. Then given E ∈ M with µ(E) = 0, we have µ(P ∩E) = 0 so ν+(E) = ν(P ∩E) = 0 and similarly ν−(E) = 0. Thus ν+ ≪ µ and ν− ≪ µ. Assume ν+ ≪ µ and ν− ≪ µ. Then given E ∈ M with µ(E) = 0, we have |ν|(E) = ν+(E)+ν−(E)=0+0=0. Thus |ν|≪µ. Assume |ν| ≪ µ. Then given E ∈ M with µ(E) = 0, we have ν+(E)+ν−(E) = 0 so ν+(E) = ν−(E) = 0 since ν+, ν− are positive measures. Thus ν(E) = 0 − 0 = 0, so ν ≪µ. 9. Suppose {νj} is a sequence of positive measures, If νj ⊥ µ for all j, then P P ∞ν ⊥µandif ν ≪µfor all j, then ∞ν ≪µ. 1 j j 1 j P Write ν = ∞ν . Assume ν ⊥ µ for all j. So for every j there exists A ,B ∈ M such 1 j j j j that • Aj ∩Bj = ∅, • Aj ∪Bj = X, REAL ANALYSIS II HOMEWORK 1 4 • ν is null on B , j j • µ is null on A . j Let A = S∞Aj and B = Tj Bj. Then 1 1 • A∩B=∅, • A∪B=X, • µ is null on A, since µ is null on every Aj, • ν is null on B. Thus ν ⊥ µ. Now assume ν ≪ µ for every j. So given E ∈ M with µ(E) = 0, we have ν (E) = 0 j j for every j and hence ∞ ν(E) = Xνj(E) = 0, 1 hence ν ≪ µ. 10. Theorem 3.5 may fail when ν is not finite. (Consider dν(x) = dx/x and dµ(x) = dx on (0,1), or ν = counting measure and µ(E) = P 2−n on N.) n∈E Consider the measure space (N,P(N),µ) where µ:P(N)→R X −n E7→n∈E2 . Clearly µ is a measure. Now if we let ν to be the counting measure, then if µ(E) = 0 for some E ⊆ N, that means E = ∅ hence ν(E) = 0. Thus we have ν ≪ µ. However, suppose the conclusion of Theorem 3.5 holds for our µ and ν. So there exists P −n δ > 0 such that ν(E) < 1 whenever µ(E) < δ. Since n∈N2 converges, there exists N∈Nsuchthat X 2−n < δ. n≥N So picking E = {n ∈ N : n ≥ N}, we have µ(E) < δ. However ν(E) = ∞. 11. Let µ be a positive measure. A collection of functions {f } ⊆L1(µ)iscalled α α∈A R uniformly integrable if for every ε > 0 there exists δ > 0 such that | f dµ| < ε E α for all α ∈ A whenever µ(E) < δ. 1 a. Any finite subset of L (µ) is uniformly integrable. b. If {f } is a sequence in L1(µ) that converges in the L1 metric to f ∈ L1(µ), n then {f } is uniformly integrable. n Write (X,M,µ) for the measure space at hand. Let’s first show why singletons are uniformly integrable. Given f ∈ L1(µ) the map ν : M → R E7→Z fdµ E
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