Let A : ℝ^{p}→ ℝ^{p} be a one to one linear transformation and let
∥⋅∥
be a norm on ℝ^{p}. Let
∥|x ∥|
≡
∥Ax∥
.
Then
∥|⋅∥|
is a norm defined on ℝ^{p}. By Theorem 2.5.4, these norms are equivalent. Say
δ∥|x∥| ≤ ∥x∥ ≤ Δ ∥|x∥|
If x_{p}→x, then the left inequality shows that Ax_{p}→ Ax. Thus A is continuous. Also, if Ax_{p}→ Ax, the
right inequality shows that x_{p}→ x or in other words, A^{−1}
(Ax )
p
→ A^{−1}
(Ax)
. Thus both
A,A^{−1} are continuous. In particular, since A^{−1} is continuous, A =
(A−1)
^{−1} sends open sets to
open sets. This all follows from Corollary 2.2.14 and Theorem 2.2.24. This shows the following
lemma.
Lemma 8.2.1Let A : ℝ^{p}→ ℝ^{p}be linear and invertible. Then A maps open sets to open sets.
Corollary 8.2.2Let A : ℝ^{p}→ ℝ^{p}be linear and invertible. Then A maps Borel sets to Borel sets.
Proof: Let the pi system be K the open sets. Then let G be those Borel sets E such that A
(E )
is Borel.
Then it is clear that G contains K and is closed with respect to complements and countable disjoint unions.
By Dynkin’s lemma, G = ℬ
(ℝp)
= σ
(K)
. This last equality holds by definition of the Borel sets ℬ
(ℝp)
.
■
From Linear algebra, if A is such an invertible linear transformation, it is the composition of finitely
many invertible linear transformations which are of the following form.
( )T ( )T
x1 ⋅⋅⋅ xr ⋅⋅⋅ xs ⋅⋅⋅ xp → x1 ⋅⋅⋅ xr ⋅⋅⋅ xs ⋅⋅⋅ xp
( )T ( )T
x1 ⋅⋅⋅ xr ⋅⋅⋅ xp → x1 ⋅⋅⋅ cxr ⋅⋅⋅ xp ,c ⁄= 0
( ) ( )
x ⋅⋅⋅ x ⋅⋅⋅ x ⋅⋅⋅ x T → x ⋅⋅⋅ x ⋅⋅⋅ x + x ⋅⋅⋅ x T
1 r s p 1 r s r p
where these are the actions obtained by multiplication by elementary matrices. Denote these special linear
transformations by E
The other linear transformation which represents a sheer is a little harder. However,
∫
m (E (s → s + r)(R )) = dm
p E(s→s+r)(R) p
∫ ∫ ∫ ∫
= ⋅⋅⋅ XE(s→s+r)(R)dxsdxrdxp1 ⋅⋅⋅dxpp−2
ℝ ℝ ℝ ℝ
Now recall Theorem 6.8.5 which says you can integrate using the usual Riemann integral when the
function involved is continuous. Thus the above becomes
∫ bpp−2 ∫ bp1∫ br∫ bs+xr
a ⋅⋅⋅ a a a +x dxsdxrdxp1 ⋅⋅⋅dxpp−2
pp−2 p1 r s r
= mp (R) = |det(E(s → s+ r))|mp (R )
Recall that when a row (column) is added to another row (column), the determinant of the resulting matrix
is unchanged.
Lemma 8.2.3Let L be any of the above elementary linear transformations. Then
mp (L (F)) = |det(L )|mp (F)
for any Borel set F. Also L
(F )
is Borel if F is Borel.
Proof: Let R_{k} = ∏_{i=1}^{p}
(− k,k)
. Let G be those Borel sets F such that
mp (L(F ∩ Rk)) = |det(L )|mp (F ∩ Rk) (8.2)
(8.2)
Letting K be the open rectangles, it follows from the above discussion that the pi system K is in G. It is
also obvious that if F_{i}∈G the F_{i} being disjoint, then
= 0. Using what was proved above, that L sends sets of measure zero to 0,
( ( ˆ) ( ˆ ))
0 ≤ mp (L(Ek)) = mp L Ek ∪ L Ek ∖ Ek
≤ m (L (ˆE )) = 0 = |det(L)|m ( ˆE ) = |det(L)|m (E )
p k p k p k
Now let k →∞ and use properties of measures. ■
For A,B nonempty sets in ℝ^{p},A + B denotes all vectors of the form a + b where a ∈ A and b ∈ B. Thus
if Q is a linear transformation,
Q (A + B) = QA + QB
The following proposition uses standard linear algebra to obtain an interesting estimate on the measure
of a set. It is illustrated by the following picture.
PICT
In the above picture, the slanted set is of the form B + D where B is a ball and the un-slanted
version is obtained by doing the linear transformation Q to the slanted set. The reason the two
look the same is that the Q used will preserve all distances. It will be an orthogonal linear
transformation.
Proposition 8.2.5Let the norm be the standard Euclidean norm and let V be a k dimensional subspaceof ℝ^{p}where k < p. Suppose D is a F_{σ}subset of V which has diameter d. Then
mp (D + B (0,r)) ≤ 2p(d+ r)p− 1r
Proof:Let
{v1,⋅⋅⋅,vk}
be an orthonormal basis for V . Enlarge to an orthonormal basis of all of ℝ^{p}
using the Gram Schmidt process to obtain
{v1,⋅⋅⋅,vk,vk+1,⋅⋅⋅,vp}
. Now define an orthogonal
transformation Q by Qv_{i} = e_{i}. Thus Q^{T}Q = I and Q preserves all lengths. Thus also det
(Q )
= 1.
Then
Q (D + B(0,r)) = QD + B(0,r)
where the diameter of QD is the same as the diameter of D and QB
(0,r)
= B
(0,r)
because Q
preserves lengths in the Euclidean norm. This is why we use this norm rather than some other.
Therefore, from the definition of the Lebesgue measure and the above result on the magnification
factor,
mp (D + B (0,r)) = det(Q )mp (D + B(0,r)) = mp (QD + B (0,r))