242 CHAPTER 10. CHANGE OF VARIABLES

Theorem 10.1.4 Let L be a linear transformation which is invertible. Then for anyBorel F, L(F) is Borel and

mp (L(F)) = |det(L)|mp (F)

More generally, if L is an arbitrary linear transformation, then for any F ∈Fp,

L(F) ∈Fp

and the above formula holds.

Proof: From linear algebra, there are Li each elementary such that L = L1 ◦L2 ◦· · ·◦Ls.By Proposition 10.1.2, each Li maps Borel sets to Borel sets. Hence, using Lemma 10.1.3,

mp (L(F)) = |det(L1)|mp (L2 ◦ · · · ◦Ls (F))

= |det(L1)| |det(L2)|mp (L3 ◦ · · · ◦Ls (F))

= · · ·=s

∏i=1|det(Li)|mp (F) = |det(L)|mp (F)

the last claim from properties of the determinant.Next consider the general case. First I clam that if N has measure 0 then so does L(N)

and if F ∈Fp, then so is L(F)∈Fp for any linear L. This follows from Proposition 10.1.2since L is differentiable.

By Proposition 8.3.2, if E ∈Fp, then for L invertible, there is an Fσ set F and a Gδ setG such that mp (G\F) = 0 and F ⊆ E ⊆ G. Then for L invertible,

mp (L(F))≤ mp (L(E))≤ mp (L(G))

and so, since F,G are Borel,

|det(L)|mp (F) ≤ mp (L(E))≤ |det(L)|mp (G)

= |det(L)|mp (F) = |det(L)|mp (E)

and so all the inequalities are equal signs. Hence, mp (L(E)) = |det(L)|mp (E).If L−1 does not exist and E ∈Fp, then there are elementary matrices Lk such that L1 ◦

L2 ◦ · · · ◦Lm ◦L maps Rp to{

x ∈ Rp : xp = 0}, a set of mp measure zero. By completeness

of Lebesgue measure, L1 ◦L2 ◦ · · · ◦Lm ◦L(E) and L(E) are both measurable and

m

∏i=1|det(Li)|mp (L(E)) = mp (L1 ◦L2 ◦ · · · ◦Lm ◦L(E)) = 0

so in this case, mp (L(E)) = 0 = |det(L)|mp (0). Thus the formula holds regardless. ■For A,B nonempty sets in Rp, 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 estimateon the measure of a set. It is illustrated by the following picture.