Lecture 28 - Mar 25

Recap

Lemma

Let be a locally compact metric space. Suppose , where is compact and is open. Then there exists

such that ,

and

Theorem

Suppose is a locally compact metric space and is a Radon measure. Then

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Proof (continued)

Since the space of all integrable simple functions is dense in , it is enough to prove is dense in with respect to the metric.

Given and , choose such that

If we show: given and , there exists such that

then we are done.

To prove this, it is enough to show: given and a Borel set , there exists such that

Now fix . Since is Radon, there exist compact and open such that

By the lemma, there exists with

Consider :

1. It is on .
2. It is on (since and ).
3. It is bounded by on .

Hence

so

This proves the required approximation.

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Lusin-Type Statements

1. Every measurable set is “nearly” a finite union of intervals, i.e., given and a Borel set , there exist finitely many disjoint intervals () such that

2. Every a.e. finite measurable function is “nearly” uniformly continuous.

3. Every Borel measurable function is “nearly” a continuous function.

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Egoroff’s Theorem

Let be a sequence of complex-valued measurable functions on a finite measure space . Suppose

Then, given , there exists measurable such that

Remarks

1. Egoroff’s theorem is not true if (counterexample on ).
2. Egoroff’s theorem does not give uniform convergence a.e.

Example: take

Then a.e., but uniformly a.e.