It was shown above that if A is an n × n matrix, then AE is Lebesgue measurable whenever E
is.
Lemma 21.3.1Let A be an n × n matrix and let E be Lebesgue measurable. Then
mn (A (E)) = |det(A )|mn (E )
Proof: This is to be shown first for elementary matrices. To begin with, let A be the elementary matrix
which involves adding the i^{th} row of the identity to the j^{th} row. Thus
A (x1,⋅⋅⋅,xn) = (x1,⋅⋅⋅,xi,⋅⋅⋅,xi + xj,⋅⋅⋅,xn)
Consider what it does to B ≡∏_{k=1}^{n}
[ak,bk]
. The j^{th} variable now goes from a_{j} + x_{i} to b_{j} + x_{i}. Thus, by
Fubini’s theorem
∫ ∫
A (B )dm = X (x ,⋅⋅⋅,x )dm
ℝn n ℝn A(B) 1 n n
∫ ∫
= ⋅⋅⋅ XA (B )(x1,⋅⋅⋅,xn)dxjdxidx1 ⋅⋅⋅dxj−1dxj+1⋅⋅⋅dxn
the integration taken in an order such that the first two two are with respect to x_{i} and x_{j}. Then from what
was just observed, this reduces to
∫ ∫ ∫ ∫
X [a1,b1](x1)⋅⋅⋅ X[an,bn](xn) X [ai,bi](xi) X[aj+xi,bj+xi](xj)dxjdxidx1 ⋅⋅⋅dxn
That inner integral is still
(b − a )
j j
and so the whole thing reduces to ∏_{k}
(b − a )
k k
= m_{n}
(B)
. The
determinant of this elementary matrix is 1 because it is either upper triangular or lower triangular with all
ones down the main diagonal. Thus in this case, m_{n}
(A (B ))
=
|det(A )|
m_{n}
(B)
. In case A is the elementary
matrix which involves multiplying the i^{th} row by α≠0,
|det(A )|
=
|α |
and A
(B )
is also a
box which changes only the interval corresponding to x_{i}, making it
|α|
times as long. Thus
m_{n}
(A (B))
=
|α|
m_{n}
(B)
=
|det(A )|
m_{n}
(B)
. The remaining kind of elementary matrix A involves
switching two rows of I and it takes B to
Bˆ
where
ˆB
is the Cartesian product of the same intervals as in B
but the intervals now occur with respect to different variables. Thus the m_{n}
(A(B ))
= m_{n}
(B)
. Also,
|det(A)|
=
|− det(I)|
= 1 and so again m_{n}
(A(B ))
=
|det (A )|
m_{n}
(B )
.
Letting A be one of these elementary matrices, let G denote the Borel sets E such that
mn (A (E ∩Rm )) = |det(A)|mn (E ∩ Rm ), Rm ≡ [− m,m ]n (21.16)
(21.16)
Then if K consists of sets of the form ∏_{k=1}^{n}
[ak,bk]
, the above shows that K⊆G. Also it is clear that G is
closed with respect to countable disjoint unions and complements. Therefore, G⊇ σ
(K)
. But
σ
(K )
⊇ℬ
(ℝn)
because σ
(K)
clearly contains every open set. Now let m →∞ in 21.16 to obtain the
desired result.
Next suppose that F is a Lebesgue measurable set, m_{n}
(F)
< ∞. Then by inner regularity, there is a set
E ⊆ F such that E is the countable union of compact sets, hence Borel, and m_{n}