To begin with, certain kinds of functions map measurable sets to measurable sets. It was
shown earlier that Lipschitz functions do this. So do differentiable functions. It will be
assumed that U is an open set in ℝ^{p} and that h : U → ℝ^{p} satisfies
= 0. Thus multiplication by a p × p matrix satisfies 9.9.
It is convenient in the following lemma to use the norm on ℝ^{p} given by
||x|| = max {|xk| : k = 1,2,⋅⋅⋅,p} .
Thus B
(x,r)
is the open box,
∏p
(xk − r,xk + r)
k=1
and so m_{p}
(B (x,r))
=
(2r)
^{p}. Also for a linear transformation A ∈ℒ
(ℝp,ℝp)
,
||A|| ≡ ||suxp||≤1 ||Ax||.
It is interesting to consider the measurability of x →
∥Df (x)∥
. However, these
considerations are not needed in the following lemma. Note that Dh
(x)
is a matrix whose i^{th}
column is h_{xi}, the partial derivative with respect to x_{i} given by
h (x+tei) − h(x)
lit→m0 -------t-------
Each component of the above difference quotient is a continuous function and so it follows
that each component of the above limit is Borel measurable. Therefore, x →
||Dh (x)||
is also
a Borel measurable function because it equals
sup ||Dh (x )v||
||v||≤1,v∈D
where D is a dense countable subset of the unit ball. Since it is the sup of countably
many Borel measurable functions, it must also be Borel measurable. It follows the
set
Tk ≡ {x ∈ T : ||Dh (x)|| < k}
for T a measurable set, must be measurable as well. In most of what follows, the
measurability of Dh will be automatic because it will be assumed that h is C^{1}.
and let ε > 0 be given. Since T_{k} is a subset of a set of measure zero, it is measurable, but we
don’t need to pay much attention to this fact. Now by outer regularity, there exists an open
set V , containing T_{k} which is contained in U such that m_{p}
(V )
< ε. Let x ∈ T_{k}. Then by
differentiability,
h(x+ v ) = h(x)+ Dh (x)v + o(v)
and so there exist arbitrarily small r_{x}< 1 such that B
(x,5r )
x
⊆ V and whenever
||v||
≤ 5r_{x},
||o(v )||
<
1
5
||v||
. Thus
h (B (x,5rx)) ⊆ Dh (x) (B (0,5rx)) +h (x)+ B (0,rx) ⊆ B (0,k5rx)+
+B (0,rx)+ h (x) ⊆ B (h (x) ,(5k +1)rx) ⊆ B(h (x ),6krx)
From the Vitali covering theorem, there exists a countable disjoint sequence of these balls,
{B(xi,ri)}
_{i=1}^{∞} such that
{B (xi,5ri)}
_{i=1}^{∞} =
{ }
^Bi
_{i=1}^{∞} covers T_{k} Then letting m_{p}
denote the outer measure determined by m_{p},
because it is the union of two sets which are in ℱ_{p}. In general, let S_{k} = S ∩ B
(0,k)
and
observe that h
(S )
= ∪_{k}h
(Sk )
. ■
In particular, this proves most of the following theorem from a different point of view to
that done before.
Theorem 9.6.3Let A be a p × p matrix. Then if E is a Lebesgue measurableset, it follows that A
(E )
is also a Lebesgue measurable set.
Proof: By Theorem 9.1.7, there exists F a countable union of compact sets
{Ki}
, and a
set of measure zero N disjoint from F such that E = F ∪ N, with m_{p}
(N)
= 0. Since
x → Ax is a continuous map, each AK_{i} is compact and so AF is a countable union of
compact sets. Therefore, it is a Borel set and is therefore, Lebesgue measurable. Thus
A
(F ∪N )
= A
(F)
∪A
(N)
and from what was just shown, A
(N)
is of measure zero and so it
is measureable. ■
class=”left” align=”middle”(Y )9.7. CHANGE OF VARIABLES, NONLINEAR
MAPS