1 Outer measures and metric outer measures
Suppose that is a set. A function is said to be an outer measure if (i) , (ii) when , and, (iii) for any countable collection ,
We say that a subset of is -measurable if
Here, instead of taking a -algebra as given and then defining a measure on this -algebra (namely, on the measurable sets), we take an outer measure as given and then define measurable sets using this outer measure. Carathéodory’s theorem11 1 Gerald B. Folland, Real Analysis, second ed., p. 29, Theorem 1.11. states that the collection of -measurable sets is a -algebra and that the restriction of to is a complete measure.
Suppose that is a metric space. An outer measure on is said to be a metric outer measure if
We prove that the Borel sets are -measurable.22 2 Gerald B. Folland, Real Analysis, second ed., p. 349, Proposition 11.16. That is, we prove that the Borel -algebra is contained in the -algebra of -measurable sets.
If is a metric outer measure on a metric space , then every Borel set is -measurable.
Because is an outer measure, by Carathéodory’s theorem the collection of -measurable sets is a -algebra, and hence to prove that contains the Borel -algebra it suffices to prove that contains all the closed sets. Let be a closed set in , and let be a subset of . Because is an outer measure,
In the case , certainly . In the case , for each let
which satisfies . Because , the fact that is a metric outer measure tells us that
Because is closed, for any we have , and hence
hence for each , using this and (2) we have
To prove that , it now suffices to prove that
Let . For and satisfying , we have
which implies that . Thus,
For any , using (4) and the fact that is a metric outer measure,
But so , and hence each of the series and converges to a value . Thus the series converges to a value . But for any ,
Because the series converges, the sum on the right-hand side of the above tends to as , so
the last inequality is due to (3), which tells us . Therefore,
which completes the proof. ∎
We shall use the following.33 3 Gerald B. Folland, Real Analysis, second ed., p. 29, Proposition 1.10.
Let be a metric space. Suppose that satisfies and that satisfies . Then the function defined by
is an outer measure.
We remark that if there is no covering of a set by countably many elements of then is an infinimum of an empty set and is thus equal to .
2 Hausdorff measure
Suppose that is a metric space and let , . Let be the collection of those subsets of with diameter together with the set , and define . By Lemma 2, the function defined by
is an outer measure. If then , from which it follows that for each , as tends to , tends to some element of . We define and show that this is a metric outer measure.44 4 Gerald B. Folland, Real Analysis, second ed., p. 350, Proposition 11.17.
Suppose that is a metric space and let . Then defined by
is a metric outer measure.
First we establish that is an outer measure. It is apparent that . If , then, using that is a metric outer measure,
If is countable then, using that is a metric outer measure,
Hence is an outer measure.
To obtain that is a metric outer measure, we must show that if then . Let and let be the collection of those subsets of with diameter together with the set . If there is no covering of by countably many elements of , then . Otherwise, let be a covering of . For each , because , it follows that does not intersect both and . Write
where and . Then and , so
This is true for any covering of by countably many element of , so
The above inequality is true for any , and taking yields
completing the proof. ∎
We call the metric outer measure in the above theorem the -dimensional Hausdorff outer measure. From Theorem 1 it follows that the restriction of to the Borel -algebra of a metric space is a meausure. We call this restriction the -dimensional Hausdorff measure, and denote it also by .
It is straightforward to verify that if is an isometric isomorphism then . In particular, for , is invariant under translations.
We will use the following inequality when talking about Hausdorff measure on .55 5 Gerald B. Folland, Real Analysis, second ed., p. 350, Proposition 11.18.
Let be a set and be a metric space. If satisfy
then for any ,
Take and . There are countably many sets that cover each with diameter and such that
Let . There is some with , so and then . Therefore the sets cover . For , there are with . Because ,
Since the sets cover and each has diameter ,
This is true for all , so taking ,
This is true for all , so taking ,
3 Hausdorff dimension
If then for all .
Let . Then Let be countably many sets each with diameter such that and
This gives us
This is true for any and , so taking we obtain . ∎
For , we define the Hausdorff dimension of to be
If the set whose infimum we are taking is empty, then the Hausdorff dimension of is .
4 Radon measures and Haar measures
Before speaking about Hausdorff measure on , we remind ourselves of some material about Radon measures and Haar measures. Let be a locally compact Hausdorff space. A Borel measure on is said to be a Radon measure if (i) it is finite on each compact set, (ii) for any Borel set ,
and (iii) for any open set ,
It is a fact that if is a locally compact Hausdorff space in which every open set is -compact, then every Borel measure on that is finite on compact sets is a Radon measure.66 6 Gerald B. Folland, Real Analysis, second ed., p. 217, Theorem 7.8.
Suppose that is a locally compact group. A Borel measure on is said to be left-invariant if for all and ,
A left Haar measure on is a nonzero left-invariant Radon measure on . It is a fact that if and are left Haar measures on then there is some such that .77 7 Gerald B. Folland, Real Analysis, second ed., p. 344, Theorem 11.9.
5 Hausdorff measure in Rn
Let denote Lebesgue measure on .
If is a Borel set in , then
Let and let be countably many closed sets that cover and such that
The isodiametric inequality (which one proves using the Brunn-Minkowski inequality) states that if is a Borel set in , then
Using this gives
But because the sets cover we have , so we get
This expression does not involve the sets (which depend on ), and since this expression is true for any , taking yields
For any , the cube can be covered by cubes of side length . Let and let . The distance from the center of to one of the vertices of is
Inscribe in a closed ball with the same center as and radius . These balls cover . Hence
Taking gives .
On the other hand, by Lemma 6,
There is some constant such that
is a locally compact Hausdorff space in which every open set in is -compact. Therefore, to show that is a Radon measure it suffices to show that is finite on every compact set. If is a compact subset of , there is some such that . By Lemma 4 and Lemma 7 we get , so . Therefore is a Radon measure.
Because , is not the zero measure. Any translation is an isometric isomorphism , so is invariant under translations. Thus is a left Haar measure on . But Lebesgue measure is also a left Haar measure on , so there is some such that
proving the claim. ∎