be a measure space. Then thesimple functions are dense in L^{p}
(Ω)
.
Proof: Recall that a function, f, having values in ℝ can be written in the form f = f^{+}−f^{−}
where
+ −
f = max (0,f),f = max (0,− f).
Therefore, an arbitrary complex valued function, f is of the form
+ − ( + − )
f = Re f − Re f + i Im f − Im f .
If each of these nonnegative functions is approximated by a simple function, it follows f is
also approximated by a simple function. Therefore, there is no loss of generality in assuming
at the outset that f ≥ 0.
Since f is measurable, Theorem 7.1.6 implies there is an increasing sequence of simple
functions, {s_{n}}, converging pointwise to f(x).Now
|f(x)− sn(x )| ≤ |f(x)|.
By the Dominated Convergence theorem,
∫ p
0 = lni→m∞ |f(x)− sn(x)| dμ.
Thus simple functions are dense in L^{p}.
Recall that for Ωa topological space, C_{c}(Ω)is the space of continuous functions with
compact support in Ω.Also recall the following definition.
Definition 15.2.2Let (Ω,S,μ) be a measure space and suppose
(Ω,τ)
is also atopological space. Then (Ω,S,μ) is called a regularmeasure space if the σ algebra of Borel setsis contained in S and for all E ∈S,
μ(E) = inf{μ(V ) : V ⊇ E and V open}
and if μ
(E )
< ∞,
μ(E ) = sup{μ(K) : K ⊆ E and K is compact }
and μ
(K )
< ∞ for any compact set, K.
For example Lebesgue measure is an example of such a measure. More generally these
measures are often referred to as Radon measures.
Lemma 15.2.3Let Ω be a metric space in which the closed balls are compact andlet K be a compact subset of V , an open set. Then there exists a continuous functionf : Ω → [0,1] such that f(x) = 1 for all x ∈ K andspt(f) is a compact subset of V .That is, K ≺ f ≺ V.
Proof: Let K ⊆ W ⊆W⊆ V and W is compact. To obtain this list of inclusions consider
a point in K,x, and take B
(x,rx)
a ball containing x such that B
(x,rx)
is a compact
subset of V . Next use the fact that K is compact to obtain the existence of a list,
{B(xi,rxi∕2)}
_{i=1}^{m} which covers K. Then let
m ( rxi)
W ≡ ∪i=1B xi,2 .
It follows since this is a finite union that
----------
W-= ∪m B (x , rxi)
i=1 i 2
and so W, being a finite union of compact sets is itself a compact set. Also, from the
construction
It is clear that f is continuous if the denominator is always nonzero. But this is clear because
if x ∈ W^{C} there must be a ball B
(x,r)
such that this ball does not intersect K. Otherwise, x
would be a limit point of K and since K is closed, x ∈ K. However, x
∈∕
K because
K ⊆ W.
It is not necessary to be in a metric space to do this. You can accomplish the same thing
using Urysohn’s lemma.
Theorem 15.2.4Let (Ω,S,μ) be a regular measure space as in Definition15.2.2where the conclusion of Lemma 15.2.3holds. Then C_{c}(Ω) is dense in L^{p}(Ω).
Proof: First consider a measurable set, E where μ
(E)
< ∞. Let K ⊆ E ⊆ V where
μ
(V ∖ K )
< ε. Now let K ≺ h ≺ V. Then
∫ ∫
|h − XE|pdμ ≤ XVp∖Kd μ = μ (V ∖K ) < ε.
It follows that for each s a simple function in L^{p}
(Ω)
, there exists h ∈ C_{c}
(Ω )
such that
||s− h||
_{p}< ε. This is because if
m∑
s(x) = ciXEi(x)
i=1
is a simple function in L^{p} where the c_{i} are the distinct nonzero values of s each μ
(Ei)
< ∞
since otherwise s
∕∈
L^{p} due to the inequality
∫ p p
|s|dμ ≥ |ci| μ(Ei).
By Theorem 15.2.1, simple functions are dense in L^{p}