Lecture 27 - Mar 24

Recap

Let be a measure space.

Define

Theorem (from previous lecture): is dense in for .

Exercise: prove the case.

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Definition (Support and )

Suppose is a metric space and is continuous.

The support of is

Define

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Radon Measure

Suppose is a metric space. A measure

is called a Radon measure if:

1. (Local finiteness) for every compact set .
2. (Inner regularity)

3. (Outer regularity)

Examples

1. If is -compact and satisfies the above regularity properties, then is Radon.
2. Any finite measure on with for every compact is Radon.
3. In particular, Lebesgue measure is Radon.

Also,

for Radon measures, since compact supports have finite measure.

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Lemma (Cutoff Function)

Let be a locally compact metric space. Suppose is compact and where is open.

Then there exists a continuous function such that

Proof

For each , choose open such that

(possible by local compactness).

Since is compact, choose with

Then

For any subset , the map is continuous.

Define

Since and are disjoint closed sets, denominator is for every , so is continuous.

By construction:

1. for (because ).
2. for (because ).

Hence

and is compact. Therefore .

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Theorem

Let be a locally compact metric space and let be a Radon measure on . Then is dense in for .

Start of proof

Take . Then

where

So for every .