Measure Theory/Riesz' representation theorem

Theorem (Riesz' representation theorem)Edit

Let   be a locally compact Hausdorff space and let   be a positive linear functional on  . Then, there exists a  -field   containing all Borel sets of   and a unique measure   such that

  1.   for all  
  2.   for all compact  
  3. If  , and   then  
  4. If  , and   then  
  5. The measure space   is complete

Proof

Recall the Urysohn's lemma:

If   is a locally compact Hausdorff space and if   is open and   is compact with   then

there exists   with   satisfying   and  . This is written in

short as  

We shall first prove that if such a measure exists, then it is unique. Suppose   are measures that satisfy (1) through (5)

It suffices to show that   for every compact  

Let   be compact and let   be given.


By (3), there exists open   with   such that  

Urysohn's lemma implies that there exists   such that  

(1) implies that  . But  , that is  . We can similarly show that  . Thus,  

Suppose   is open in  , define  

If   are open , then  


If   is a subset of   then define  

Define  

Let  

monotonicity of   is obvious for all subsets of  


Let   with  

It is obvious that   which implies that  . Hence, we have that   is complete.

Step 1Edit

Suppose   is a sequence of subsets of   then,  

Proof

Let   be open subsets of  . We wish to show that  

Given  , let   be such that   (so that  ) and  . This is possible because  .

Now by Urysohn's lemma we can find  ,   such that   and   on   with  

Thus   and   on  


As   is a linear functional   for all  

 


Thus,   for every  , i.e.  


If   is a sequence of members of  , there exist open   such that given  


 . Define  ,   is open. Let  . Then   but  


Thus  

Step 2Edit

If   is compact, then   and  


Proof

It suffices to show that   for every compact  

Let   and  , define   Then   is open and  

Then, by Urysohn's lemma, there exists   such that  , and hence,   on  

By definition   and  

As  , we have   and hence,  


Let   be

By definition, there exists open   such that  


By Urysohn's lemma, there exists   such that  , which implies that  , that is

 .

Hence,  

Step 3Edit

Every open set   satisfies

 

If   is open and  ,  , then  


Proof

Let   be open. Let   such that  . It suffices to show that there exists compact   such that  .

By definition of  , there exists   such that   and  

Let  . Obviously  .

Let   be open such that  , then   and hence  , further  


Thus,  

Step 4Edit

Suppose   is a sequence of pairwise disjoint sets in   and let  . Then,  

Proof

If  , by step 1, we are done.

If   is finite then,   and hence,   is countably additive on  


Suppose,   are compact and disjoint then  ;  


Claim:  


As   is a locally compact Hausdorff space, there exist disjoint open sets  ,   with  ,  

Hence, by Urysohn's lemma, there exists   such that   and  

Now,   and  ,  


Thus,  

Assume  . Given  , there exists compact   such that  

Let  . Obviously,   is compact.

Thus,  

and hence,   By step 1, we have  .

Thus,  

Step 5Edit

If   and   then there exists   compact and   open with   and  


Proof

  is open. As  , there exist compact   and open   such that   with

 

Now,   (by step 4)

Thus,  

Step 6Edit

  is a field of subsets of  

Proof

Let   and let   be given.

There exist compact   and open   such that  ,   with

 


Write  

As   is a closed subsetof   , it is compact


Then,  

thus,   is finite and hence,  

Now write  

and  

Step 7Edit

  is a  -field containing all Borel sets


Proof

Let   be closed

Then,   is compact for every   compact

Therefore   and hence   (by definition) and hence,   has all closed sets. In particular,  


Let  . Then,   and   and hence,  

Now let   where  

We know that   for every compact  

Let  ,  .  , but   and hence,  

Step 8Edit

 

Proof

Let  . Then   for every compact  

Now, let  ,  . Given  , there exists open   such that  ,  , that is,  . Further, there exists compact   such that  


  implies that  , that is, there exists compact   such that

 


Therefore,   implies that  

As   is arbitrary, we are done.

Step 9Edit

For  ,  


Proof

Without loss of generality, we may assume that   is real valued.

It is obviuos from the definition of   that  

Let  . Hence, as   is continuous,   is compact. and we can write   for some  . Let  . Let   be an  -fine partition of  

Let  . As   is continuous,   is compact,   is measurable for every  , and hence,  

 

Hence, we can find open sets   such that  

  for all  


We know that if compact   with   open then there exists   wiht   and   on  

Hence, there exist functions   such that   on  .

Thus,   for all  


By step 2, we have  

  on each  

Thus,  

 

 

 

 


As   is arbitrary, we have   which completes the proof.