Next I will compare ℋ^{n} and m_{n}. To do this, recall the following covering theorem which is a
summary of Corollary 9.3.6 found on Page 670.
Theorem 19.2.1Let E ⊆ ℝ^{n}and letℱ be a collection of balls of boundedradii such that ℱ covers E in the sense of Vitali. Then there exists a countable collectionof disjoint balls from ℱ, {B_{j}}_{j=1}^{∞}, such that m_{n}(E ∖∪_{j=1}^{∞}B_{j}) = 0.
In the next lemma, the balls are the usual balls taken with respect to the usual distance in
ℝ^{n}.
Lemma 19.2.2If m_{n}
(S )
= 0 then ℋ^{n}
(S)
= ℋ_{δ}^{n}
(S )
= 0. Also, there exists aconstant, k such that ℋ^{n}
(E )
≤ km_{n}
(E)
for all E Borel. Also, if Q_{0}≡ [0,1)^{n}, the unitcube, then ℋ^{n}
n
([0,1) )
> 0.
Proof: Suppose first m_{n}
(S)
= 0. Without loss of generality, S is bounded. Then by outer
regularity, there exists a bounded open V containing S and m_{n}
(V)
< ε. For each x ∈ S, there
exists a ball B_{x} such that
^Bx
⊆ V and δ > r
( )
^Bx
. By the Vitali covering theorem there is a
sequence of disjoint balls
{Bk}
such that
{ ^ }
Bk
covers S. Here
^
Bk
has the same center as B_{k}
but 5 times the radius. Then letting α
(n)
be the Lebesgue measure of the unit ball in ℝ^{n}
ℋn (S) ≤ ∑ β(n)r(B^ )n = β(n)5n∑ α (n)r(B )n
δ k k α(n) k k
β(n) β(n)
≤ ----5nmn (V) < ----5nε
α(n) α(n)
Since ε is arbitrary, this shows ℋ_{δ}^{n}
(S)
= 0 and now it follows ℋ^{n}
(S)
= 0.
Letting U be an open set and δ > 0, consider all balls B contained in U which have
diameters less than δ. This is a Vitali covering of U and therefore by Theorem 19.2.1,
there exists
{Bi}
, a sequence of disjoint balls of radii less than δ contained in U
such that ∪_{i=1}^{∞}B_{i} differs from U by a set of Lebesgue measure zero. Let α
(n)
be the Lebesgue measure of the unit ball in ℝ^{n}. Then from what was just shown,
Now letting E be Borel, it follows from the outer regularity of m_{n} there exists a decreasing
sequence of open sets,
{Vi}
containing E such such that m_{n}
(Vi)
→ m_{n}
(E)
. Then from the
above,
n n
ℋδ (E ) ≤ lii→m∞ ℋ δ (Vi) ≤ li→im∞ kmn (Vi) = kmn (E ).
Since δ > 0 is arbitrary, it follows that also
ℋn (E ) ≤ kmn (E ).
This proves the first part of the lemma.
To verify the second part, note that it is obvious ℋ_{δ}^{n} and ℋ^{n} are translation invariant
because diameters of sets do not change when translated. Therefore, if ℋ^{n}
n
([0,1) )
= 0, it
follows ℋ^{n}
n
(ℝ )
= 0 because ℝ^{n} is the countable union of translates of Q_{0}≡ [0,1)^{n}. Since
each ℋ_{δ}^{n} is no larger than ℋ^{n}, the same must hold for ℋ_{δ}^{n}. Therefore, there exists a
sequence of sets,
{Ci}
each having diameter less than δ such that the union of these sets
equals ℝ^{n} but
∞∑
1 > β(n)r(Ci)n.
i=1
Now let B_{i} be a ball having radius equal to diam
(Ci)
= 2r
(Ci)
which contains C_{i}. It
follows
m (B ) = α (n)2nr(C )n = α-(n)2nβ (n) r(C )n
n i i β(n) i
which implies
∞ ∞
1 > ∑ β(n)r (C )n = ∑ -β-(n)--m (B ) = ∞,
i=1 i i=1 α(n)2n n i
a contradiction. ■
Theorem 19.2.3By choosing β
(n)
properly, one can obtain ℋ^{n} = m_{n}on allLebesgue measurable sets.
Proof:I will show ℋ^{n} is a positive multiple of m_{n} for any choice of β
(n)
.
Define
k = mn-(Q0)
ℋn (Q0)
where Q_{0} = [0,1)^{n} is the half open unit cube in ℝ^{n}. I will show kℋ^{n}
(E )
= m_{n}
(E)
for any
Lebesgue measurable set. When this is done, it will follow that by adjusting β
(n)
the multiple
can be taken to be 1.
Let Q = ∏_{i=1}^{n}[a_{i},a_{i} + 2^{−k}) be a half open box where a_{i} = l2^{−k}. Thus Q_{0} is the union of
( )
2k
^{n} of these identical half open boxes. By translation invariance, of ℋ^{n} and
m_{n}
for any such half open box and by translation invariance,
for the translation of any such half open box. It follows that kℋ^{n}
(U )
= m_{n}
(U )
for all open sets. It follows immediately, since every compact set is the countable
intersection of open sets that kℋ^{n} = m_{n} on compact sets. Therefore, they are also
equal on all closed sets because every closed set is the countable union of compact
sets. Now let F be an arbitrary Lebesgue measurable set. I will show that F is ℋ^{n}
measurable and that kℋ^{n}
(F)
= m_{n}
(F )
. Let F_{l} = B
(0,l)
∩ F. Then there exists H a
countable union of compact sets and G a countable intersection of open sets such
that