Lecture 29 - Mar 27

Egoroff’s Theorem (Proof)

Given

there exists such that and convergence holds on .

Fix . For , define

that is,

For each fixed , , and

Hence, since ,

So for each choose such that

Set

Then

Now if , then for each and all ,

Hence uniformly on .

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Lusin’s Theorem (Version Used in Class)

Let be a Radon measure on a locally compact metric space . Suppose is Borel measurable and . Then there exists continuous such that

If is bounded, one can take

Proof sketch (as in notes)

For , define

Then . Choose so that

Define truncated function

Then (bounded on a finite-measure set).

Choose such that

Take a subsequence pointwise a.e. By Egoroff on a finite-measure set, there exists measurable such that

and convergence is uniform on .

Using inner regularity, choose compact with

Using outer regularity, choose open with

Apply the cutoff lemma: choose such that

For large , define

Then is continuous and equals on (up to the approximation step on ). The bad set is contained in

whose measure is at most .

Hence the claim follows.

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Normed Linear Spaces (Recap)

Let be a vector space over or . A norm is a map such that:

1. for all .
2. .
3. for all .
4. for all .

Define metric

If complete under this metric, is a Banach space.

Theorem (recall): is Banach for .

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Linear Functionals and Bounded Linear Maps

If is a vector space over , a linear map is called a linear functional.

Let and be normed spaces. A linear map

is bounded if there exists such that

For linear , the following are equivalent:

1. is continuous.
2. is continuous at .
3. is bounded.

Let

Define operator norm

If is Banach, then is Banach.

Dual space of is

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Duality for : The Canonical Map

Let and satisfy

Each defines a bounded linear functional on by

By Holder,

Proposition

For ,

If is semi-finite, the analogous statement also holds for :

Here is semi-finite if every with contains such that and .

Proof idea

Upper bound is Holder. For equality when , choose

so that and

For , let

Using semi-finiteness choose with and set

Then

and let .

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Main Theorem (Riesz Representation for )

For , for every

there exists such that

Remark: this also holds for under suitable -finite (or semi-finite) hypotheses.