Integration: More Exercises
			
Arun Ram 
Department of Mathematics and Statistics 
University of Melbourne 
Parkville, VIC 3010 Australia 
aram@unimelb.edu.au
			
Last updates: 18 March 2011
Integration: More Exercises
-  If  is a complex measure on a σ-algebra ,
and if , define
the supremum being taken over all finite partitions 
 of .
Does it follow that ?
- 
Prove that the example given at the end of [Ru, Sec. 6.10] has the stated properties.
-  Prove that the vector space  of
all complex regular Borel measures on a locally compact Hausdorff space 
is a Banch space if 
.  Hint:
Compare [Ru, Ch. 5, Exercise 8]. [That the difference of any two members of 
 is in 
 was used in the first paragraph of the proof 
of [Ru, Theorem 6.19]; supply a proof of this fact.]
- Suppose  and 
 is the exponent conjugate to .  Suppose 
is a positive σ-finite measure and  is a measurable function such that
 for every
.
Prove that then
.
- Suppose  consists of two points  and ;
define ,
, and 
.  Is it true, for this
 that 
is the dual space of 
?
- Suppose  and 
 is the exponent conjugate to .  Prove that 
 is the dual space
of 
even if  is not σ-finite.
- Suppose  is a complex Borel measure on 
 (or on the unit circle
), and define the Fourier coefficients of  by
| . |  |  
 Assume that 
as 
and prove that then
as .
Hint:  The assumption also holds with 
 in 
place of  if  is any trigonometric polynomial,
hence if  is continuous, hence if  is any bounded Borel function,
hence if  is replaced by 
.
-  In the terminology of Exercise 7, find all  such that
 is periodic, with period .
[This means that 
 for all integers ; of course,  is also assumed to be 
an integer.]
- Suppose that   is a sequence
of positive continuous functions on , that 
is a positive Borel measure on , and that 
- (a) ,
- (b) ,
   for all ,
- (c),
   for every .
 Does it follow that ?
- 
Let 
be a positive measure space.  Call a set 
uniformly integrable if to each 
corresponds a  such that
| , |  |  
 whenever 
and .
- (a)  
Prove that every finite subset of 
is uniformly integrable.
- (b)    
Prove the following convergence theorem of Vitali:
If 
- (i)  
,
- (ii)  
  is uniformly integrable,
- (iii)  
 a.e. 
as ,
- (iv)  
a.e.,
 then
 and| . |  |  
 Suggestion: Use Egoroff's theorem.
- (c)  
Show that (b) fails if  is Lebesgue measure on , even if 
 is assumed to be bounded.  Hypothesis (i) can therefore not be omitted in (b).
- (d)    
Show that hypothesis (iv) is redundant in (b) for some  ( for instance, for Lebesgue
measure on a bounded interval), but that there are finite measures for which the omission of (iv)
would make (b) false.
- (e)    Show that Vitali's theorem implies Lebesgue's dominated convergence theorem, for
finite measure spaces.  Construct an example in which Vitali's theorem applies although the 
hypotheses of Lebesgue's theorem do not hold.
-  (f)   Construct a sequence ,
say on , so that
,
but
 is not uniformly
integrable (with respect to Lebesgue measure).
- (g)   However, the following converse of Vitali's theorem is true:  If
,
 and
 exists for every ,
then 
 is uniformly
integrable.
 Prove this by completing the following outline.
 Define .  Then  is a complete metric space
(modulo sets of measure ), and  is continuous for each .  If ,
there exist 
(Exercise 13, Chapt. 5) so that| ,
   if  
,
   
. | (*) |  
 If , (*) holds with
and
in place of .  Thus (*) holds with  in place of 
 and  in place of .
Now apply (a) to :
There exists  such that| ,
   if  
,
   
. |  |  
 
- Suppose that  is a positive measure on ,
,
 for ,  a.e., and there exists  and  such that
 for all .  Prove that
| . |  |  
 Hint:   is uniformly integrable.
- Let  be the collection of all sets  in the 
unit interval  such that either
 or its complement is at most countable.  Let  
be the counting measure on this σ-algebra .  If 
 for 
, show that 
is not -measurable, although the mapping
makes sense for every  and defines a bounded linear functional on 
.
Thus  in this situation.
- Let ,
where  is Lebesgue measure on .  Show that there is 
a bounded linear functional  on 
 that is  on 
, and that therefore there is no 
that satisfies  for every .
Thus .
Notes and References
These exercises are taken from [Ru, Chapt. 6] for a course in "Measure Theory" at the Masters level at University of Melbourne. 
References
 [Ru]  
W. Rudin,
Real and complex analysis, Third edition, McGraw-Hill, 1987.
MR0924157.
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