As an important application of the area formula I will give a general version of the
divergence theorem. It will always be assumed n ≥ 2. Actually it is not necessary to
make this assumption but what results in the case where n = 1 is nothing more
than the fundamental theorem of calculus and the considerations necessary to
draw this conclusion seem unneccessarily tedious. You have to consider ℋ^{0}, zero
dimensional Hausdorff measure. It is left as an exercise but I will not present
it.
It will be convenient to have some lemmas and theorems in hand before beginning the
proof. First recall the Tietze extension theorem on Page 453. It is stated next for
convenience.
Theorem 27.2.1Let M be a closed nonempty subset of a metric space
(X, d)
andlet f : M →
[a,b]
be continuous at every point of M. Then there exists a function,g continuous on all of X which coincides with f on M such that g
(X )
⊆
[a,b]
.
The next topic needed is the concept of an infinitely differentiable partition of
unity.
Definition 27.2.2Let ℭ be a set whose elements are subsets of ℝ^{n}.^{1}Then ℭ is said to be locally finiteif for every x ∈ ℝ^{n}, there exists an open set, U_{x}containing x such that U_{x}has nonempty intersection with only finitely many setsof ℭ.
Lemma 27.2.3Let ℭ be a set whose elements are open subsets of ℝ^{n}and suppose∪ℭ ⊇ H, a closed set. Then there exists a countable list of open sets,
{Ui}
_{i=1}^{∞}suchthat each U_{i}is bounded, each U_{i}is a subset of some set of ℭ, and ∪_{i=1}^{∞}U_{i}⊇ H.
Proof: Let W_{k}≡ B
(0,k)
,W_{0} = W_{−1} = ∅. For each x ∈ H ∩W_{k} there exists an
open set, U_{x} such that U_{x} is a subset of some set of ℭ and U_{x}⊆ W_{k+1}∖W_{k−1}. Then
since H ∩W_{k} is compact, there exist finitely many of these sets,
{ k}
Ui
_{i=1}^{m(k)
} whose
union contains H ∩W_{k}. If H ∩W_{k} = ∅, let m
(k)
= 0 and there are no such sets
obtained.The desired countable list of open sets is ∪_{k=1}^{∞}
{ k}
Ui
_{i=1}^{m(k)
}. Each open set
in this list is bounded. Furthermore, if x ∈ ℝ^{n}, then x ∈ W_{k} where k is the
first positive integer with x ∈ W_{k}. Then W_{k}∖W_{k−1} is an open set containing
x and this open set can have nonempty intersection only with with a set of
{ k}
Ui
_{i=1}^{m(k)
}∪
{ k− 1}
Ui
_{i=1}^{m(k−1)
}, a finite list of sets. Therefore, ∪_{k=1}^{∞}
{ k}
Ui
_{i=1}^{m(k)
} is
locally finite.
The set,
{Ui}
_{i=1}^{∞} is said to be a locally finite cover of H. The following lemma gives
some important reasons why a locally finite list of sets is so significant. First of
all consider the rational numbers,
The set of rational numbers is definitely not locally finite.
Lemma 27.2.4Let ℭ be locally finite. Then
--- {-- }
∪ℭ = ∪ H : H ∈ ℭ .
Next suppose the elements of ℭ are open sets and that for each U ∈ℭ, there exists adifferentiable function, ψ_{U}havingspt
(ψU)
⊆ U. Then you can define the following finitesum for each x ∈ ℝ^{n}
∑
f (x ) ≡ {ψU (x ) : x ∈ U ∈ ℭ} .
Furthermore, f is also a differentiablefunction^{2}and
∑
Df (x ) = {D ψU (x) : x ∈ U ∈ ℭ}.
Proof:Let p be a limit point of ∪ℭ and let W be an open set which intersects only
finitely many sets of ℭ. Then p must be a limit point of one of these sets. It follows
p ∈∪
{H- : H ∈ ℭ}
and so ∪ℭ⊆∪
{H- : H ∈ ℭ}
. The inclusion in the other direction is
obvious.
Now consider the second assertion. Letting x ∈ ℝ^{n}, there exists an open set, W
intersecting only finitely many open sets of ℭ, U_{1},U_{2},
⋅⋅⋅
,U_{m}. Then for all
y ∈ W,
∑m
f (y ) = ψUi (y)
i=1
and so the desired result is obvious. It merely says that a finite sum of differentiable
functions is differentiable. Recall the following definition.
Definition 27.2.5Let K be a closed subset of an open set, U. K ≺ f ≺ U if f iscontinuous, has values in
[0,1]
, equals 1 on K, and has compact support containedin U.
Lemma 27.2.6Let U be a bounded open set and let K be a closed subset ofU. Then there exist an open set, W, such that W ⊆W⊆ U and a function,f ∈ C_{c}^{∞}
(U)
such that K ≺ f ≺ U.
Proof: The set, K is compact so is at a positive distance from U^{C}. Let
Since W is compact it is at a positive distance from W_{1}^{C} and so h is a well defined
continuous function which has compact support contained in W_{1}, equals 1 on W, and has
values in
[0,1]
. Now let ϕ_{k} be a mollifier. Letting
k− 1 < min (dist(K,W C ),2−1dist(W-1,U C)),
it follows that for such k,the function, h∗ϕ_{k}∈ C_{c}^{∞}
(U)
, has values in
[0,1]
, and equals
1 on K. Let f = h ∗ ϕ_{k}.
The above lemma is used repeatedly in the following.
Lemma 27.2.7Let K be a closed set and let
{Vi}
_{i=1}^{∞}be a locally finite list ofbounded open sets whose union contains K. Then there exist functions, ψ_{i}∈ C_{c}^{∞}
(Vi)
such that for all x ∈ K,
∑∞
1 = ψi(x)
i=1
and the function f
(x)
given by
∑∞
f (x) = ψi(x )
i=1
is in C^{∞}
n
(ℝ )
.
Proof: Let K_{1} = K ∖∪_{i=2}^{∞}V_{i}. Thus K_{1} is compact because K_{1}⊆ V_{1}. Let W_{1} be an
open set having compact closure which satisfies
K1 ⊆ W1 ⊆ W-1 ⊆ V1
Thus W_{1},V_{2},
⋅⋅⋅
,V_{n} covers K and W_{1}⊆ V_{1}. Suppose W_{1},
⋅⋅⋅
,W_{r} have been defined
such that W_{i}⊆ V_{i} for each i, and W_{1},
⋅⋅⋅
,W_{r},V_{r+1},
⋅⋅⋅
,V_{n} covers K. Then
let
K ≡ K ∖ ((∪∞ V )∪ (∪r W )).
r+1 i=r+2 i j=1 j
It follows K_{r+1} is compact because K_{r+1}⊆ V_{r+1}. Let W_{r+1} satisfy
--- ---
Kr+1 ⊆ Wr+1 ⊆ W r+1 ⊆ Vr+1,W r+1 is compact
Continuing this way defines a sequence of open sets,
{Wi }
_{i=1}^{∞} having compact closures
with the property
--- ∞
Wi ⊆ Vi,K ⊆ ∪i=1Wi.
Note
{Wi }
_{i=1}^{∞} is locally finite because the original list,
{Vi}
_{i=1}^{∞} was locally finite.
Now let U_{i} be open sets which satisfy
--- -- --
W i ⊆ Ui ⊆ U i ⊆ Vi,Ui is compact.
Similarly,
{Ui}
_{i=1}^{∞} is locally finite.
PICT
Since the sets,
{Wi}
_{i=1}^{∞} are locally finite, it follows ∪_{i=1}^{∞}W_{i} = ∪_{i=1}^{∞}W_{i} and so
it is possible to define ϕ_{i} and γ, infinitely differentiable functions having compact support
such that
Note how the sum makes sense because of local finiteness. If x is such that
∑_{j=1}^{∞}ϕ_{j}(x) = 0, then x
∕∈
∪_{i=1}^{∞}U_{i} because ϕ_{i} equals one on U_{i}. Consequently, for
each i, γ_{i}
(y)
= 0 for all y near x thanks to the fact that U_{i} is closed and so ψ_{i}(y) = 0 for
all y near x. Hence ψ_{i} is infinitely differentiable at such x. If ∑_{j=1}^{∞}ϕ_{j}(x)≠0, this
situation persists near x because each ϕ_{j} is continuous and by local finiteness, so is the
sum. So ψ_{i} is infinitely differentiable at such points also thanks to Lemma 27.2.4.
Therefore ψ_{i} is infinitely differentiable. If x ∈ K, then γ_{i}
(x)
= 1 for each i and so
∑_{j=1}^{∞}ψ_{j}(x) = 1. Clearly 0 ≤ ψ_{i}
(x)
≤ 1 and spt(ψ_{j}) ⊆ V_{j}. This proves the
theorem.
The functions,
{ψ }
i
are called a C^{∞} partition of unity.
The method of proof of this lemma easily implies the following useful corollary.
Corollary 27.2.8If H is a compact subset of V_{i}for some V_{i}there exists apartition of unity such that ψ_{i}
(x)
= 1 for all x ∈ H in addition to the conclusionof Lemma 27.2.7.
Proof:Keep V_{i} the same but replace V_{j} with
^Vj
≡ V_{j}∖H. Now in the proof above,
applied to this modified collection of open sets, if j≠i,ϕ_{j}
(x )
= 0 whenever x ∈ H.
Therefore, ψ_{i}
(x )
= 1 on H.
Lemma 27.2.9Let Ω be a metric space with the closed balls compact and supposeμ is a measure defined on the Borel sets of Ω which is finite on compact sets.Then there exists a unique Radon measure, μwhich equals μ on the Borel sets. Inparticular μ must be both inner and outer regular on all Borel sets.
Proof: Define a positive linear functional, Λ
(f )
= ∫fdμ. Let μ be the Radon
measure which comes from the Riesz representation theorem for positive linear
functionals. Thus for all f continuous,
∫ ∫ --
f dμ = f dμ.
If V is an open set, let
{fn}
be a sequence of continuous functions which is
increasing and converges to X_{V } pointwise. Then applying the monotone convergence
theorem,
∫ ∫
XV dμ = μ (V ) = XVdμ-= μ(V )
and so the two measures coincide on all open sets. Every compact set is a countable
intersection of open sets and so the two measures coincide on all compact sets. Now
let B
(a,n)
be a ball of radius n and let E be a Borel set contained in this
ball. Then by regularity of μ there exist sets F,G such that G is a countable
intersection of open sets and F is a countable union of compact sets such that
F ⊆ E ⊆ G and μ
(G ∖F )
= 0. Now μ
(G)
=μ
(G)
and μ
(F)
=μ
(F)
. Thus
μ(G ∖F )+ μ(F) = μ-(G)
= μ (G) = μ (G∖ F)+ μ (F)
and so μ
(G ∖ F)
=μ
(G ∖F )
. Thus
μ(E) = μ(F) = μ(F) = μ(G) = μ(E).
If E is an arbitrary Borel set, then
μ(E ∩ B (a,n)) = μ-(E ∩ B (a,n))
and letting n →∞, this yields μ
(E )
=μ
(E )
.
One more lemma will be useful.
Lemma 27.2.10Let V be a bounded open set and let X be the closed subspace ofC
(-)
V
, the space of continuous functions defined on V, which is given by thefollowing.
X = {u ∈ C (V-) : u (x) = 0 on ∂V }.
Then C_{c}^{∞}
(V )
is dense in X with respect to the norm given by
--
||u || = max {|u(x)| : x ∈ V }
Proof:Let O ⊆O⊆ W ⊆W⊆ V be such that dist
(O,V C)
< η and let ψ_{δ}
(⋅)
be a
mollifier. Let u ∈ X and consider X_{W}u ∗ ψ_{δ}. Let ε > 0 be given and let η be
small enough that |u
(x)
| < ε∕2 whenever x ∈ V ∖O. Then if δ is small enough
|X_{W}u ∗ ψ_{δ}
(x)
− u
(x)
| < ε for all x ∈O and X_{W}u ∗ ψ_{δ} is in C_{c}^{∞}
(V )
. For
x ∈ V ∖O,|X_{W}u ∗ ψ_{δ}
(x )
|≤ ε∕2 and so for such x,
|XW u ∗ψ δ(x )− u(x)| ≤ ε.
This proves the lemma since ε was arbitrary.
Definition 27.2.11A bounded open set, U ⊆ ℝ^{n}is said to have a Lipschitz boundaryand to lie on one side of its boundary if the following conditions hold. There exist openboxes, Q_{1},
⋅⋅⋅
,Q_{N},
∏n ( )
Qi = aij,bij
j=1
such that ∂U ≡U∖U is contained in their union. Also, for each Q_{i}, there exists k and aLipschitz function, g_{i}such that U ∩ Q_{i}is of the form
( k−1
{ ∏ ( i i)
(x : (x1,⋅⋅⋅,xk−1,xk+1,⋅⋅⋅,xn) ∈ j=1 aj,bj ×
)
n∏ (i i) i }
aj,bj and ak < xk < gi(x1,⋅⋅⋅,xk−1,xk+1,⋅⋅⋅,xn))(27.2.28)
j=k+1
or else of the form
(
{ k∏−1( )
x : (x1,⋅⋅⋅,xk−1,xk+1,⋅⋅⋅,xn) ∈ aij,bij ×
( j=1
n )}
∏ (ai,bi) and gi(x1,⋅⋅⋅,xk−1,xk+1,⋅⋅⋅,xn) < xk < bi (.27.2.29)
j=k+1 j j j)
The function, g_{i}has a derivative on A_{i}⊆∏_{j=1}^{k−1}
Also, there exists an open set, Q_{0}such that Q_{0}⊆Q_{0}⊆ U and U⊆ Q_{0}∪Q_{1}∪
⋅⋅⋅
∪Q_{N}.
Note that since there are only finitely many Q_{i} and each g_{i} is Lipschitz, it follows
from an application of Lemma 27.0.1 that ℋ^{n−1}
(∂U )
< ∞. Also from Lemma 27.2.9ℋ^{n−1} is inner and outer regular on ∂U.
Lemma 27.2.12Suppose U is a bounded open set as described above. Then there existsa unique function in L^{∞}
( n−1)
∂U,ℋ
^{n}, n
(y )
for y ∈ ∂U such that
|n(y)|
= 1,n is ℋ^{n−1}measurable, (meaning each component of n is ℋ^{n−1}measurable) and for everyw ∈ℝ^{n}satisfying
|w|
= 1, and for every f ∈ C_{c}^{1}
n
(ℝ )
,
∫ f (x+ tw)− f (x) ∫
ltim→0 --------t-------dx = f (n ⋅w)dℋn −1
U ∂U
Proof: Let U⊆ V ⊆V⊆∪_{i=0}^{N}Q_{i} and let
{ψi}
_{i=0}^{N} be a C^{∞} partition of unity on
V such that spt
(ψi)
⊆ Q_{i}. Then for all t small enough and x ∈ U,
f-(x+-tw)-− f-(x) 1∑N
t = t ψif (x + tw )− ψif (x).
i=0
Thus using the dominated convergence theorem,
∫
lim f-(x+-tw)-− f-(x)dx
t→0 U t
∫ ( 1∑N )
= lim - ψif (x+ tw) − ψif (x ) dx
t→0 U t i=0
∫ N n
∑ ∑
= Ui=0j=1Dj (ψif)(x )wjdx
∫ ∑n ∑N ∫ ∑n
= Dj (ψ0f )(x )wjdx+ Dj (ψif )(x) wjdx (27.2.30)
U j=1 i=1 U j=1
(27.2.30)
Since spt
(ψ0)
⊆ Q_{0}, it follows the first term in the above equals zero. In the second term,
fix i. Without loss of generality, suppose the k in the above definition equals n and
27.2.28 holds. This just makes things a little easier to write. Thus g_{i} is a function
of