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Real Analysis HW 3 Solutions

Problem 31.   Justify the assertion in the proof of Vitali’s Theorem that it suffices toconsider the case that  E   is bounded.

Solution.  We know by a previous problem that any positive measure set contains a boundedset which is also positive measure. Therefore if we can prove Vitali’s theorem for any boundedset of positive measure, we have also proven it for any set of positive measure.  

Problem 32.   Does Lemma 16 remain true if Λ is allowed to be finite or to be uncountablyinfinite? Does it remain true if Λ is allowed to be unbounded.

Solution.  Recall Lemma 16 from the text:

Lemma 16   Let  E   be a bounded measurable set of real numbers. Suppose there is a bounded, countably infinite set of real numbers  Λ   for which the collection of translates of  E ,  (λ + E )λ∈Λ, is disjoint. Then  m(E ) = 0.

Lemma 16 does not remain true if Λ is allowed to be finite. Let  E  = [0, 1) and Λ =  {0, 1}.Then  E  + 1 and  E  are disjoint, but  E  is not measure 0.

If Λ is uncountable, then we may always pick a countable subset which satisfies the conditions

of the theorem. Hence the Lemma holds if Λ is uncountable.

The Lemma is not true if Λ is unbounded. Consider  E  = [0, 1) and Λ =  Z. Then  E  + λ  areall disjoint for all  λ ∈  Λ.  

Problem 33.   Let  E  be a non-measurable set of finite outer measure. Show that there is aGδ   set  G  that contains  E  for which

m∗(E ) = m∗(G),   while  m∗(G ∼  E ) >  0

Proof.   Define  G   in the usual way,   G   =

 ∞k=1 Ok   where  {Ok}

∞k=1  are open sets satisfying

m∗(Ok)   < m∗(E ) +   1

k. Using monotonicity and that fact that  m∗(G)  ≤  m∗(E ) + 1/k   for

every   k   ≥   1, we readily conclude that   m∗(G) =   m∗(E ). However we cannot have thatm∗(G ∼  E ) = 0, since that would imply that  E  = G ∩ (G ∼  E )C  is measurable. Thereforem∗(G ∼  E ) >  0.  

Problem 38.  Let the function  f   : [a, b] →  R be Lipschitz, that is, there is a constant  c  ≥  0such that for all [u, v] ∈  [a, b],  |f (u) − f (v)| ≤ c|u − v|. Show that  f  maps a set of measurezero onto an set of measure zero. Show that  f   maps an  F σ   set onto an  F σ   set. Concludethat  f  maps a measurable set to a measurable set.

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Note that for any   y   ∈   Range(f ) such that   |x −  y |   < δ , there exist  α0, with   y   =   f x(α0).Since   |x −  f x(α)|   is strictly increasing in   |α|   we must have   |α0|   < . This implies thatα0  =  f 

−1x   (y) = f 

−1(y) − f −1(x) and so

|f −1(y) − f −1(x)| =  |α0| < .

Problem 46.   Let  f  be a continuous function and  B  be a Borel set. Show that  f −1(B) is aBorel set.

Proof.   Let  B  denote the Borel sets. We define the set

E  = {E  ⊂  R|f −1(E ) ∈ B}

to be the collection of sets  E   for which  f −1(E ) is Borel. Note that, since f   is continuousE   contains all the open sets. Whats left is to show that  E   is a  σ-algebra. To do this, weconsider a set  E  ∈ E . We note that  f −1(E C ) =  f −1(E )C  ∈ B , therefore  E C  ∈ E . Similarly

for any collection of open sets  {E k}∞k=1 ⊂ E  we have

f −1

∞k=1

E k

=

∞k=1

f −1(E k) ∈ B 

and therefore∞

k=1 E k  ∈ E . It follows that E   is a  σ-algebra containing the open sets, and soby the definition of the Borel sets,  B ⊆ E . Therefore by the definition of  E   if  B  ∈ B ⊆ E ,then  f −1(B) ∈ B .  

Problem 47: (Not assigned) Use the preceding two problems to show that a continuousstrictly increasing function that is defined on an interval maps Borel sets to Borel sets.

Solution:   Since  f  is continuous and strictly monotone, it has a continuous inverse  g =  f −1

(see problem 45). Since g  is continuous, we have by problem 46 that if  B  is Borel,  g−1(B) =f (B) is Borel.  

Problem 4:   Suppose  f   is a real-valued function on  R  such that  f −1(c) is measurable foreach number  c. Is  f   necessarily measurable?

Solution:   No. Let V  be a non-measurable subset of (0, 1) and consider the function  f (x) =ex · (2χV   − 1), where χV   denote the idicator function on  V  . Note that f > 0 on V   and f  ≤ 0on  R ∼  V  . Since f  is one-to-one we also have that  f −1(c) is either empty or a singleton andtherefore measurable. However, by construction we have f −1((0, ∞)) = {x :   f (x) >  0} =  V  .

Therefore  f   is not measurable.   Problem 6.   Let  f  be a function with measurable domain  D. Show that  f   is measurable if and only if the function  g  defined on  R by g(x) = f (x) for  x  ∈  D  and  g(x) = 0 for  x /∈ D   ismeasurable.

Solution.   Let  c be a real number. Suppose that  g  is measurable, then

{x ∈  D  :  f (x) > c} =  {x ∈  R :  g(x) > c} ∩ D

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is measurable, since  {x ∈  R :  g(x) > c}  and  D  are both measurable, therefore  f   is measur-able. In the case that  f   is measurable, we see that if  c ≥  0, then

{x ∈  R :  g(x) > c} =  {x ∈  D  :  f (x) > c}

and if  c c} =  Dc ∪ {x ∈  D  :  f (x) > c}.

Therefore  g  is measurable since  {x ∈  D  :  f (x) > c}  and  Dc is.  

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