∫ ∫
Lf ≡ ⋅⋅⋅ f (x ,⋅⋅⋅,x )dx ⋅⋅⋅dx (21.8)
ℝ ℝ 1 n i1 in
(21.8)
where
(i1,⋅⋅⋅,in)
is some permutation of
(1,⋅⋅⋅,n)
and f ∈ C_{c}
(Ω)
. These are the familiar iterated
improper Rieman integrals. Thus
∫ ∫ r ∫ a
f (x1,⋅⋅⋅,xn)dxi1 ≡ lrim→∞ f (x1,⋅⋅⋅,xn)dxi1 = f (x1,⋅⋅⋅,xn)dxi1
ℝ −r −a
for all a sufficiently large since f has compact support. The measure which results is called m_{n}. It is n
dimensional Lebesgue measure and has all the regularity properties mentioned in the Riesz representation
theorem above. In particular, m_{1} refers to ℝ and
∫ ∫
f (x)dx = f (x)dm1
for f continuous with compact support. I will continue using the dx notation in the context of iterated
integrals instead of something like the following which is arguably more consistent with the above notation
but more cumbersome to think about and write.
∫ ∫
⋅⋅⋅ fdm1 (xi1)⋅⋅⋅dm1(xin)
ℝ ℝ
Lemma 21.1.1Let E be Borel. Then if f ∈ C_{c}
(ℝn)
is nonnegative,
∫ ∫ ∫
fXEdmn = ⋅⋅⋅ fXEdxi1 ⋅⋅⋅dxin (*)
ℝn ℝ ℝ
(*)
with each iterated integral making sense. Also,
∫ ∫ ∫
ℝn XEdmn = ℝ ⋅⋅⋅ ℝ XEdxi1 ⋅⋅⋅dxin (**)
(**)
with all the iterated integrals making sense.
Proof: Let K denote sets of the form ∏_{i=1}^{n}
(ai,bi)
. Then it is clearly a π system. Also, since every
open set is the countable union of sets of this form, σ
(K)
⊇ℬ
(ℝn)
, the Borel sets because it contains the
open sets. Then ∗ holds for any set in K thanks to an approximation like that used in determining
the Lebesgue measure of a box and the monotone or dominated convergence theorem. Now
let G denote the measurable sets for which ∗ holds. It clearly contains K and is closed with
respect to complements and countable disjoint unions and so by Dynkin’s lemma, Lemma
19.3.2, G ⊇ σ
(K)
⊇ℬ
(ℝn)
. The reason is it closed with respect to complements is that if
E ∈G,
∫ ∫ ∫ ∫
⋅⋅⋅ fX dx ⋅⋅⋅dx + ⋅⋅⋅ fX Cdx ⋅⋅⋅dx
ℝ ℝ E i1 in ℝ ℝ E i1 in
The last equality follows from repeated iterations of the monotone convergence theorem for each successive
iterated integral, described more in the next major theorem. ■
be an increasing sequence of Borel measurable simple functions converging to f
pointwise. Then ∗ holds with f replaced with s_{k} by Lemma 21.1.1. Now apply the monotone convergence
theorem to obtain the result. In the right side you must apply it to the successive iterated integrals to yield
that the iterated integrals all make sense and in the limit yield the desired iterated integral. To illustrate,
note that
∫ ∫ ∫
⋅⋅⋅ sk(x1,⋅⋅⋅,xn )dxi1dxi2 ⋅⋅⋅dxin−1
ℝ ℝ ℝ
is measurable with respect to one dimensional Lebesgue measure by the above lemma. This allows moving
the limit inside the first integral from the the left in
Then the same reasoning allows one to move it inside the next integral from the left and so forth, each
time preserving the one dimensional Lebesgue measurability. Thus the above is increasing to
Thus G is closed with respect to countable disjoint unions and complements. Therefore, by Dynkin’s
lemma, G⊇ σ
(K)
but α
(K)
obviously includes all open sets so it contains the Borel sets. Since it
is a subset of the Borel sets, it is equal to the Borel sets. Now let k →∞ to conclude that
m_{n}
(E )
= m_{n}
(x+ E )
for all E Borel.
Now by regularity, if F is any measurable set, there exists E a countable union of compact sets and G a
countable intersection of open sets such that E ⊆ F ⊆ G and
mn (E ) = mn (F) = mn (G )
Then m_{n} is translation invariant for both E,G because these are Borel sets. Therefore, it is translation
invariant for F also. Specifically,
Therefore, all inequalities are actually equalities and this shows the desired result. ■
In the following proposition,
∥⋅∥
refers to a norm on ℝ^{n}, not necessarily the usual one. Actually
something better will be proved because it is more convenient to do so.
Proposition 21.1.4Let α
(n)
denote the m_{n}measure of the ball B
(0,1)
≡
{x ∈ ℝn : ∥x∥ < 1}
.Then
n
mn (B (x,r)) = α (n)r .
Proof:More generally, let G denote those Borel sets with the following property. Letting
R_{p}≡∏_{i=1}^{n}
[− p,p]
, for all r > 0,
n
mn (r (E ∩ Rp)) = r mn (E ∩ Rp).
Let K denote those sets of the form ∏_{k=1}^{n}I_{k} where I_{k} is an interval. Then if U ≡∏_{k=1}^{n}
[ak,bk]
is one of
these sets, you would have U ∩R_{p} = ∏_{k=1}^{n}
[ak,bk]
or else ∅ in which case, there is nothing to prove, and
so the condition holds because
n∏
mn (r (U ∩ Rp)) = r (bk − ak) = rnmn (U ∩ Rp)
k=1
The situation is the same if the intervals are open or half open or if they are a mixture of various kinds of
intervals. Thus K⊆G. Now if you have disjoint sets of G