Measure Theory/Riesz' representation theorem

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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 1[edit]

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 2[edit]

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 3[edit]

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 4[edit]

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 5[edit]

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 6[edit]

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 7[edit]

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 8[edit]

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 9[edit]

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.