Definition 1Let . Definewhere .

Then is called d-dimensional Hausdorff measure in .

Remark 1Hausdorff measure is a Borel outer measure, it is not Radon for because is not finite with respect to . And it is a measure if restricted on Lebesgue measurable sets (by Caratheodory condition: We say a set satisfies Caratheodory condition, if for any set , ). Moreover, note that gives the volume of unit ball in dimension if is an integer, we naturally have (not trivially) , where is the n-dimensional Lebesgue measure.

Here are some basic properties of Hausdorff measures:

Theorem 2on , for all . And hence on .

*Proof:* Note that , we have for .

Denote by for . Then

So for all , hence on .

Theorem 3on for .

*Proof:* Let be a unit cube in , we can divide it into cubes with side-length , diameter . So

Send we will have , hence on .

Theorem 4Let , , then

- If , then .
- If , then .

*Proof:* Note that the second statement is the contraposition of the first one. To prove the first statement, fix a , there exists a cover of , with , such that

Then we have

Send we will have .

Having this theorem, we know that for any set there exists only one such that is neither nor . It is natural to define this to be its dimension.

Definition 5We say the Hausdorff dimension of a set is

Note that the Hausdorff dimension need not be an integer. Even if is an integer, it doesn’t mean is a -dimensional surface in any sense.

Another important and related measure is called capacity, and it is defined as follows.

Definition 6Let , we define the p-capacity of by

Obviously, by density we can consider instead of .

It’s not hard to see is an outer measure, by know that given a sequence of functions , denote by and , then -a.e. As we know, given an outer measure, one can construct a -algebra by Caratheodory condition such that it is a measure restricted to this -algebra. But in this case, we are not that interested in it because the -algebra only contains sets of capacity 0 or . And here are some properties of capacities that one can easily prove.

Theorem 7 (Properties of )

- open,
- , for .
- .
- .

From the scaling, we can see that the capacity does not measure the “volume” of the set. And according this scaling, it is natural to look for the relations between and .

Theorem 8Assume , if , then .

Note that this is a refinement of the third statement of last theorem for $latex {1

*Proof:* First we claim that there exists a constant such that for any neighborhood containing , there exists an open set and such that

To proof the claim, let be a neighborhood of and set . Since , we can find a sequence of ball with such that , and

Set

Then . Set and , then and

Now we use the claim inductively. We find a sequence of , from the claim, and relate them by . Set and . Then and on . Therefore

since are discrete and .

The next important relation is that, given , we will have some information on its Hausdorff dimension. Before proving it, we need a Lemma.

Lemma 9Let , suppose and defineThen .

*Proof:* From Lebesgue differential theorem one can see . Then by the absolute continuity of Lebesgue integral, there exists a and such that whenever . Fix a and an , we define a set

Since , we can fix an open set with , and consider a family of balls

Then this family of balls cover , and by Vitali’s covering theorem, there exists a sequence of disjoint balls such that . Hence

Then we can take to yield the result.

Theorem 10Let and . If , then for all .

*Proof:* Since , we can choose a sequence of functions such that . Define , by GNS inequality we know is well defined in , and for any . So , the average of on , goes to 0 as if . Now I claim that

Assume by contradiction, there exists an such that , by Poincare inequality, we know that

for small enough. Then

for some . This means converges as , which leads to a contradiction. Therefore, , and by the previous lemma we know since .

For case, we can say more about the relation between capacity and Hausdorff measure since we will have more geometric insights. For instance, we can take advantage of isoperimetric inequality and co-area formula to prove the following result.

Theorem 11Let , then if and only if .

*Proof:* From Theorem 8 we know that implies . Assume , for any , we can find an such that and . By co-area formula, we know

This implies and for some . Fix this , by isoperimetric inequality, one can see . Obviously we know , then for each , we can find an such that

Since is a small portion in , by relative isoperimetric inequality we have

Since these balls cover , by Vitali’s covering theorem, there exists a sequence of disjoint balls such that . From the estimates we get above,

And we know that for each . Hence . Send we will get the result.

Remark 2In the previous theorem, we actually prove the smallness of is equivalent to the smallest of , which we don’t have for .

The next important result is fine property of Sobolev functions. Comparing to Lebesgue integrable functions, Sobolev functions have a smaller singular set, and their representatives enjoy a better continuity result. To show it, we first need a Chebyshev type inequality.

Lemma 12Let and , define the setThen

*Proof:* First note that, without loss of generality, we can assume by homogeneity. For each , we consider a ball such that . This collection of balls cover , and by Besicovitch covering theorem, there exist and countable collections of disjoint balls , such that

Denote by the element in for . For each , consider the function , which is in , on . By extension theorem we know this function can be extended to , say , such that

for all , where depends only on . Then we have in for all . Define , then since is open. Therefore

Now we are ready to prove the fine properties of Sobolev functions.

Theorem 13Let . Then there exists a Borel set such that , and exists for any . In addition,for all . Furthermore, given any , there exists an open set such that is continuous restricted on .

*Proof:* For each , choose an such that . Define the set

From Lemma 14, we know . Consider the set

then by Theorem 9 and Lemma 10, we have . For any , by Poincare’s inequality we have

Then for any , we have

Define , then for , we have . This implies converges uniformly to a continuous function in . So

as , which implies . Define , then

And we know that exists for all . In addition,

For the continuity statement, given any , we can find an such that . From theorem 8, we can find an open set such that . Then we are done since we know is continuous restricted on .