be a measure space where Y is a closed subset of X a finite dimensional
normed vector space and ℬ
(Y)
, the Borel sets in Y . Suppose also that μ
(K )
< ∞ whenever
K is a compact set in Y . By Corollary 7.5.8, it follows μ is regular. This regularilty of μ
implies an important approximation result valid for any f ∈ L^{1}
(Y)
. It turns out that in this
situation, for all ε > 0, there exists g a continuous function defined on Y with g equal to 0
outside some compact set and
∫
|f − g|dμ < ε.
Definition 9.2.1Let f : X → Y where X is a normed vector space. Then thesupport off,denoted byspt
(f)
is the closure of the set where f is not equal to zero.Thus
spt(f) ≡ {x : f-(x-) ⁄=-0}
Also, if U is an open set, f ∈ C_{c}
(U )
means f is continuous on U andspt
(f)
⊆ U. Similarlyf ∈ C_{c}^{m}
(U )
if f has m continuous derivatives andspt
(f)
⊆ U and f ∈ C_{c}^{∞}
(U)
ifspt
(f)
⊆ U and f has continuous derivatives of every order on U.
Lemma 9.2.2Let Y be a closed subset of X a finite dimensional normed vectorspace. Let K ⊆ V where K is compact in Y and V is open in Y . Then there exists acontinuous function f : Y →
[0,1]
such thatspt
(f)
⊆ V , f
(x)
= 1 for all x ∈ K. If
(Y,ℬ (Y),μ)
is a measure space with μ
(K )
< ∞, for every compact K, then ifμ
(E )
< ∞ where E ∈ℬ
(Y )
, there exists a sequence of functions in C_{c}
(Y)
{fk}
suchthat
∫
kl→im∞ Y |fk(x)− XE (x)|dμ = 0.
Proof: For each x ∈ K, there exists r_{x} such that
B (x,rx ) ≡ {y ∈ Y : ||x − y|| < rx} ⊆ V.
Since K is compact, there are finitely many balls,
{B (xk,rxk)}
_{k=1}^{m} which cover K. Let
W = ∪_{k=1}^{m}B
(xk,rxk)
. Since there are only finitely many of these,
--- m ---------
W = ∪k=1B (x,rxk)
and W is a compact subset of V because it is closed and bounded, being the finite union of
closed and bounded sets. Now define
( C )
f (x) ≡-----distCx,W---------
dist(x,W )+ dist(x,K )
The denominator is never equal to 0 because if dist
(x,K )
= 0 then since K is
closed, x ∈ K and so since K ⊆ W, an open set, dist
( )
x,W C
> 0. Therefore, f is
continuous.
When x ∈ K, f
(x)
= 1. If x
∕∈
W, then f
(x)
= 0 and so spt
(f)
⊆W⊆ V. In the above
situation the following notation is often used.
K ≺ f ≺ V. (9.4)
(9.4)
Thus this notation is used if spt(f) is a compact subset of the open set V and if f has values
in [0,1] while f equals 1 on a compact subset of V .
It remains to prove the last assertion. By Corollary 7.5.8, μ is regular and so there exist
compact sets
{Kk}
and open sets
{Vk}
such that V_{k}⊇ V_{k+1}, K_{k}⊆ K_{k+1} for all k,
and
K ⊆ E ⊆ V ,μ (V ∖K ) < 2−k.
k k k k
From the first part of the lemma, there exists a sequence
{f}
k
such that
K ≺ f ≺ V .
k k k
Then f_{k}
(x)
converges to X_{E}
(x)
a.e. because if convergence fails to take place, then x must
be in infinitely many of the sets V_{k}∖ K_{k}. Thus x is in
Now the functions are all bounded above by 1 and below by 0 and are equal to zero off V_{1}, a
set of finite measure so by the dominated convergence theorem,
∫
lim |XE (x)− fk(x)|dμ = 0,
k→ ∞
the dominating function being X_{E}
(x)
+ X_{V 1}
(x)
. This proves the lemma. ■
Note that if ℱ⊇ℬ
(Y )
and ℱ is Borel regular in the sense that whenever F ∈ℱ there
exist H,G Borel sets such that H ⊆ F ⊆ G and μ
(G ∖H )
= 0 which is the case for Lebesgue
measure for example, then you could replace ℬ
(Y )
with ℱ in all of the above. Thus assume
this kind of Borel regularity holds or simply assume the measure is regular on ℱ in what
follows.
With this lemma, here is an important major theorem.
Theorem 9.2.3Let Y be a closed subset of X a finite dimensional normed vectorspace. Let
(Y,ℱ, μ)
be a measure space with ℱ⊇ℬ
(Y)
and μ
(K)
< ∞, for every compact Kin Y . Let f ∈ L^{1}
(Y)
and let ε > 0 be given. Then there exists g ∈ C_{c}
(Y )
suchthat
∫
|f (x)− g (x )|dμ < ε.
Y
Proof: By considering separately the positive and negative parts of the real
and imaginary parts of f it suffices to consider only the case where f ≥ 0. Then
by Theorem 7.1.6 and the monotone convergence theorem, there exists a simple
function,