Download Math 641 Lecture #10 ¶2.9,2.10,2.11,2.12,2.13,2.15

Math 641 Lecture #10
¶2.9,2.10,2.11,2.12,2.13,2.15
Definitions (2.9). Let X be a topological space.
The support of an f : X → C is
supp(f ) = {x ∈ X : f (x) 6= 0}.
Let Cc (X) denote the collection of continuous f : X → C for which supp(f ) is compact
in X.
Elementary Property. Cc (X) is a complex vector space.
[Here supp(αf +βg) ⊂ supp(f )∪supp(g), and the union of two compact sets is compact.]
Theorem (2.10). If X and Y are topological spaces and f : X → Y is continuous, then
f (K) is compact for any compact K.
Proof: Let {Vα } be an open cover of f (K).
Since f is continuous, the collection {f −1 (Vα )} is an open cover of K.
Hence there are finitely many α1 , . . . , αn such that
K ⊂ f −1 (Vα1 ) ∪ · · · ∪ f −1 (Vαn ) = f −1
n
[
!
Vαi
.
i=1
Thus f (K) ⊂
Sn
i=1
Vαi .
Corollary. If f ∈ Cc (X), then f (X) is compact.
Proof. If f ≡ 0, then f (X) = {0} is compact.
Suppose f 6≡ 0.
Then K = supp(f ) 6= ∅ and compact, and f (K) is compact by Theorem 2.10.
If K C = ∅, i.e., K = X (so X is compact), then f (X) = f (K) is compact.
If K C 6= ∅, i.e., there is an x ∈ X such that f (x) = 0, then f (X) = f (K) ∪ {0} is
compact.
Definition (2.11). If f ∈ Cc (X) with f (X) ⊂ [0, 1], K compact in X, and V open in
X, then
a) K is a domain of unity for f , written K ≺ f , if f = 1 on K, and
b) f is subordinate to V , written f ≺ V , if supp(f ) ⊂ V .
Remark. Rudin does not “name” these notations; the name in a) is mine; that in b) is
Folland’s.
Urysohn’s Lemma (2.12). If X is a LCH space, V open in X, and K a compact
subset of V , then there exists an f ∈ Cc (X) such that K ≺ f ≺ V .
Proof: Let r0 = 0, r1 = 1, and r2 , r3 , r4 , . . . be an enumeration of Q ∩ (0, 1).
[These are numbered differently from the book: Rudin starts with r1 .]
By Theorem 2.7 there are open sets V0 (∼ r0 ) and V1 (∼ r1 ) such that V0 and V1 are
compact and
K ⊂ V1 ⊂ V1 ⊂ V0 ⊂ V0 ⊂ V.
Suppose that for n ≥ 1, the sets Vr0 , . . . , Vrn have been chosen so that ri < rj implies
Vrj ⊂ Vri .
Among r0 , . . . , rn is the largest one smaller than rn+1 ; call it ri .
Another of r0 , . . . , rn is the smallest one larger than rn+1 ; call it rj .
Choose Vrn+1 so that Vrn+1 is compact and Vrj ⊂ Vrn+1 ⊂ Vrn+1 ⊂ Vri .
This process gives a countable collection of open sets {Vr : r ∈ Q ∩ [0, 1]} for which
K ⊂ V1 , V0 ⊂ V , Vr compact for each r ∈ Q ∩ [0, 1], and s > r =⇒ Vs ⊂ Vr .
Define
(
r if x ∈ Vr ,
fr (x) =
0 otherwise,
(
1 if x ∈ Vs ,
gs (x) =
s otherwise.
Then each fr is lsc, and each gs is usc.
[The function gs is usc because compact sets are closed by Corollary a) of Theorem 2.5.]
Hence f = supr fr is lsc, and g = inf gs is usc.
The range of f is contained in [0, 1] because r ∈ [0, 1], K ≺ f because f = 1 on V1 ⊃ K,
and f ≺ V because supp(f ) ⊂ V0 ⊂ V .
Continuity of f follows by showing that f = g, i.e., that f is both lsc and usc.
Suppose fr (x) > gs (x) for some x.
Then r > s, x ∈ Vr , and x ∈
/ Vs .
But r > s implies Vr ⊂ Vs .
Hence fr ≤ gs for all r and s; so f ≤ g.
Suppose f (x) < g(x) for some x.
Then there are r, s ∈ Q ∩ [0, 1] such that f (x) < r < s < g(x).
Here, f (x) < r implies x ∈
/ Vr , and s < g(x) implies x ∈ Vs .
But r < s implies Vs ⊂ Vr .
This contradiction implies that f (x) = g(x), and thus f ∈ Cc (X) with K ≺ f ≺ V .
Definition. A partition of unity on a compact K subordinate to a cover {V1 , . . . , Vn }
of K is a collection hi ∈ Cc (X), i = 1, . . . , n for which hi ≺ Vi and h1 (x) + · · · + hn (x) = 1
for all x ∈ K.
Theorem (2.13). If X is a LCH space, then for any finite cover {V1 , . . . , Vn } of a compact
K there exists a partition of unity on K subordinate to the cover {V1 , . . . , Vn }.
Proof: Since each x ∈ K belongs to one of the Vi ’s, there is by Theorem 2.7 an open set
Wx such that Wx is compact and x ∈ Wx ⊂ Wx ⊂ Vi .
By the compactness of K, there are finitely many points x1 , . . . , xm in K for which
K ⊂ W x1 ∪ · · · ∪ W xm .
S
For each i = 1, 2, . . . , n, set Hi = {Wxj : Wxj ⊂ Vi }; each Hi is compact.
Then K ⊂ H1 ∪ · · · ∪ Hn .
By Urysohn’s Lemma, there are gi ∈ Cc (X) such that Hi ≺ gi ≺ Vi .
Define h1 = g1 , h2 = (1 − g1 )g2 , . . . , hn = (1 − g1 )(1 − g2 ) · · · (1 − gn−1 )gn .
Then hi ≺ Vi because
supp(h1 ) = supp(g1 ) ⊂ V1
supp(h2 ) = {x ∈ X : g1 (x) = 1}C ∩ supp(g2 ) ⊂ V2
etc.
Since
h1 + h2 = g1 + (1 − g1 )g2 = g1 + g2 − g1 g2 = 1 − (1 − g1 )(1 − g2 ),
induction gives
h1 + · · · + hn = 1 − (1 − g1 ) · · · (1 − gn ).
If x ∈ K, then x ∈ Hi for some i, so that gi (x) = 1, implying that
h1 (x) + · · · + hn (x) = 1.
Thus h1 , . . . , hn ∈ Cc (X) is the desired partition of unity on K subordinate to the cover
{V1 , . . . , Vn }.
Definition(2.15). For a LCH space X, a positive measure µ on BX is called a Borel
measure.
Definition (Folland p. 205). If µ is a Borel measure, then
a) µ is outer regular on an E ∈ BX if µ(E) = inf{µ(U ) : U ⊃ E, U open},
b) µ is inner regular on an E ∈ BX if µ(E) = sup{µ(K) : K ⊂ E, K compact},
c) µ is outer regular if it is outer regular for all E ∈ BX ,
d) µ is inner regular if it is inner regular for all E ∈ BX ,
e) µ is regular if it is both outer and inner regular.
Remark. Sometimes, regularity is a bit too much to ask of a Borel measure.
Definition (Folland p. 205). A Borel measure µ is called a Radon measure if
(a) µ(K) < ∞ for all compact K,
(b) µ is outer regular, and
(c) µ is inner regular on all open sets.
Remarks. Not every Borel measure is Radon because of a), and not every Radon
measure is regular because of c).