11.1 Basic Properties
Definition 11.1.1 Define the following positive linear functional for f ∈ Cc
Then the measure representing this functional is Lebesgue measure.
The following lemma will help in understanding Lebesgue measure.
Lemma 11.1.2 Every open set in ℝn is the countable disjoint union of half open boxes
of the form
where ai = l2−k for some integers, l,k. The sides of these boxes are of equal length. One
could also have half open boxes of the form
and the conclusion would be unchanged.
Thus Ck consists of a countable disjoint collection of boxes whose union is ℝn. This is
sometimes called a tiling of ℝn. Think of tiles on the floor of a bathroom and you will get
the idea. Note that each box has diameter no larger than 2−k
. This is because
Let U be open and let ℬ1 ≡ all sets of C1 which are contained in U. If ℬ1,
been chosen, ℬk+1 ≡
all sets of Ck+1
Let ℬ∞ = ∪i=1∞ℬi. In fact ∪ℬ∞ = U. Clearly ∪ℬ∞⊆ U because every box of every ℬi
is contained in U. If p ∈ U, let k be the smallest integer such that p is contained in a box
from Ck which is also a subset of U. Thus
Hence ℬ∞ is the desired countable disjoint collection of half open boxes whose union is
U. The last assertion about the other type of half open rectangle is obvious. This proves
Now what does Lebesgue measure do to a rectangle, ∏
Lemma 11.1.3 Let R = ∏
i=1n[ai,bi],R0 = ∏
Proof: Let k be large enough that
for i = 1,
and consider functions gik
having the following graphs.
Then by elementary calculus along with the definition of Λ,
Letting k →∞, it follows that
This proves the lemma.
Lemma 11.1.4 Let U be an open or closed set. Then mn
Proof: By Lemma 11.1.2 there is a sequence of disjoint half open rectangles,
are also disjoint
rectangles which are identical to the Ri
but translated. From Lemma 11.1.3
It remains to verify the lemma for a closed set. Let H be a closed bounded set first.
Then H ⊆ B
large enough. First note that x
is a closed set. Thus
the last equality because of the first part of the lemma which implies
as claimed. If
bounded, consider Hm ≡B
Passing to the
limit as m →∞
yields the result in general.
Theorem 11.1.5 Lebesgue measure is translation invariant. That is
for all E Lebesgue measurable.
Proof: Suppose mn
By regularity of the measure, there exist sets G,H
such that G
is a countable intersection of open sets, H
is a countable union of
compact sets, mn
and G ⊇ E ⊇ H.
which follows from Lemma
applied to the sets which are
either intersected to form G
or unioned to form H.
and both x + H and x + G are measurable because they are either countable unions or
countable intersections of measurable sets. Furthermore,
and so by completeness of the measure, x + E is measurable. It follows
is not necessarily less than
consider Em ≡ B
by the above. Letting
it follows mn
This proves the theorem.
Corollary 11.1.6 Let D be an n × n diagonal matrix and let U be an open set.
Proof: If any of the diagonal entries of D equals 0 there is nothing to prove because
then both sides equal zero. Therefore, it can be assumed none are equal to zero. Suppose
these diagonal entries are k1,
From Lemma 11.1.2
there exist half open
having all sides equal such that
Suppose one of these is
where bj −aj
. Then DRi
0 and Ij
) if kj <
Then the rectangles, DRi
are disjoint because D
one to one and their union is DU.
and this proves the corollary.
From this the following corollary is obtained.
Corollary 11.1.7 Let M > 0. Then mn
Proof: By Lemma 11.1.4 there is no loss of generality in taking a = 0. Let D
be the diagonal matrix which has M in every entry of the main diagonal so
. Note that DB
By Corollary 11.1.6
There are many norms on ℝn. Other common examples are
any norm for
you can define a corresponding ball in terms of this
It follows from general considerations involving metric spaces presented earlier that these
balls are open sets. Therefore, Corollary 11.1.7 has an obvious generalization.
Corollary 11.1.8 Let
be a norm on ℝn. Then for M >
where these balls are defined in terms of the norm