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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 be a signed measure on x ...

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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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