This section contains generalizations of the earlier change of variables formulas. First is a very
interesting proposition which says that differentiable functions map measurable sets to measurable
sets.
Proposition 9.6.1Let h : U → ℝ^{p}is continuous where U is an open subset of ℝ^{p}. Also supposeh is differentiable on H ⊆ U where H is Lebesgue measurable. Then if E is a Lebesgue measurableset contained in H, then h
(E )
is also Lebesgue measurable. Also if N ⊆ H is a set of measure zero,then h
(N)
is a set of measure zero.
Proof:Consider the second claim first. Let N be a set of measure zero contained in H and
let
N ≡ {x ∈ N : ∥Dh (x)∥ ≤ k}
k
There is an open set V ⊇ N_{k} such that m_{p}
(V )
< ε. For each x ∈ N_{k}, there is a ball B_{x} centered at x with
radius r_{x}< 1 such that B_{x}⊆ V and for y ∈ B_{x},
h(y) ∈ h (x)+ Dh (x)B (0,rx)+ B (0,εrx)
Thus
h(Bx) ⊆ h (x)+ B (0,(∥Dh (x)∥+ ε)rx)
⊆ B (h(x),(k +ε)r )
--- p x
mp (h(Bx )) ≤ (k +ε) mp (B (x,rx))
By the Besicovitch covering theorem, there are balls of this sort such that
Mp
Nk ⊆ ∪j=1 ∪ {B ∈ Gj}
where G_{j} is a countable disjoint collection of these balls. Thus,
Mp Mp
mp-(h(Nk)) ≤ ∑ ∑ mp-(h(B )) ≤ (k+ ε)p∑ ∑ mp (B)
j=1B∈Gj j=1 B∈Gj
p p
≤ (k+ ε) Mpmp (V ) ≤ ε(k + ε) Mp
Since ε is arbitrary, it follows that m_{p}
(h (Nk ))
= 0 and so in fact h
(Nk)
is measurable and has
m_{p} measure zero. Now let k →∞ to conclude that m_{p}
(h(N ))
= 0. Now the other claim is
shown as follows. By Proposition 8.1.3, if E is Lebesgue measurable, E ⊆ H, there is an F_{σ} set
F contained in E such that m_{p}
(E ∖F)
= 0. Then h
(F )
is clearly measurable because h is
continuous and F is a countable union of compact sets. Thus h
(E )
= h
(F )
∪ h
(E ∖F )
and the
second was just shown measurable while the first is an F_{σ} set so it is actually a Borel set.
■
In particular, h
(H )
is Lebesgue measurable since H is.
Now recall Lemma 3.7.1 which is stated here for convenience.
Lemma 9.6.2Let g be continuous and map B
(p,r)
⊆ ℝ^{n}to ℝ^{n}. Suppose that for all x ∈B
(p,r)
,
|g (x )− x| < εr
Then it follows that
(------)
g B (p,r) ⊇ B (p,(1− ε)r)
Now suppose that U ⊆ ℝ^{p} is open, h : U → ℝ^{p} is continuous, and m_{p}
(h(U ∖H ))
= 0 where H ⊆ U and
H is Lebesgue measurable. Suppose also that h is differentiable on H and is one to one on H. Define for
measurable E
λ(E ) ≡ mp (h (E ∩H )), μ (E) ≡ mp (E ∩ U)
Since h is one to one, this along with Proposition 9.6.1 implies that λ ≪ μ. Therefore, for measurable E, it
follows from the Radon Nikodym theorem presented above that
∫ ∫
λ(E) = mp (h (E ∩H )) = XED μλ(x)dmp = XED μλ(x)dμ
U
So what is D_{μ}λ
(x )
? Recall that this exists for μ a.e. x, is measurable and for such x,
p p
|det(Dh (x ))|(1− ε) ≤ Dmp λ(x) ≤ |det(Dh (x))|(1+ ε)
Since ε is arbitrary, this shows D_{mp}λ
(x )
=
|det(Dh (x))|
.
If Dh
(x)
^{−1} does not exist, then you have
h (B (x,r))− h(x) ⊆ Dh (x )B (0,r)+ B (0,εr)
and Dh
(x)
maps into a bounded subset of a p − 1 dimensional subspace. Therefore, the right side has
measure no more than an expression of the form Cr^{p}ε, C depending on Dh
equals a measurable function a.e. By completeness of the measure m_{p} it
follows that x →X_{E}
(h (x ))
|det(Dh (x))|
is also measurable. I am not saying that x →X_{E}
(h(x))
is
measurable. In fact this might not be so because it is X_{h−1(E)
}
(x)
and the inverse image of a measurable set
is not necessarily measurable. For a well known example, see Problem 11 on Page 476. The thing
which is measurable is the product in the integrand. Therefore, for E a measurable subset of
H
The following theorem gives a somewhat improved change of variables formula over what was presented
earlier in the sense that the mapping h only needs to be differentiable rather than C^{1} and the set over
which the integral is taken is just a measurable set.
Theorem 9.6.3Let U ⊆ ℝ^{p}be open, h : U → ℝ^{p}continuous, and m_{p}
(h(U ∖H ))
= 0 where H ⊆ U andH is Lebesgue measurable. Suppose also that h is differentiable on H and is one to one on H. Then h
(H )
is Lebesgue measurable and if f ≥ 0 is Lebesgue measurable, then
∫ ∫
f (y)dmp = f (h (x ))|det(Dh (x))|dmp (9.19)
h(H ) H
(9.19)
and all needed measurability holds.
Proof:Formula 9.18 implies that 9.19 holds for any nonnegative simple function s. Then for f
nonnegative and measurable, it is the pointwise increasing limit of such simple functions. Therefore, 9.19
follows from the monotone convergence theorem. ■
In the exercises which follow, if no measure is mentioned, assume it is Lebesgue measure.