14.5 Representations for Positive Linear Functionals on C_{c}
(ℝp)
In this section I will show that positive linear functionals defined on C_{c}
(ℝp)
(Think iterated integral of a
continuous function which is zero off a compact set.) are represented by a unique Radon measure. This is
another topic which holds in far greater generality than ℝ^{p}. It is really no harder to include the more
general situation which consists of a locally compact Hausdorff space in place of ℝ^{p}. However, I am trying
to avoid esoterica, especially since the main interest is in ℝ^{p} anyway. Nevertheless, I will dispense with
using bold face to indicate points in ℝ^{n} because it is not customary to use this in the more general
setting.
Definition 14.5.1Define C_{c}
(ℝp)
to be the functions which have complex values and compactsupport. Thismeansspt
(f)
≡
{x ∈ ℝp : f (x) ⁄= 0}
is a compact set. Then L : C_{c}
(ℝp)
→ ℂ iscalled a positive linear functionalif it is linear and if, whenever f ≥ 0, then L
(f)
≥ 0 also. Whenf is a continuous function andspt
(f )
⊆ V an open set, we say f ∈ C_{c}
(V)
.
The following definition gives some notation.
Definition 14.5.2If K is a compact subset of an open set, V , then K ≺ ϕ ≺ V if
p
ϕ ∈ Cc(V ),ϕ(K ) = {1},ϕ(ℝ ) ⊆ [0,1],
where ℝ^{p}denotes the whole topological space considered. Also for ϕ ∈ C_{c}(ℝ^{p}), K ≺ ϕ if
ϕ(ℝp) ⊆ [0,1] and ϕ(K) = 1.
and ϕ ≺ V if
p
ϕ (ℝ ) ⊆ [0,1] and spt(ϕ) ⊆ V.
Next is a useful theorem.
Theorem 14.5.3Let H be a compact subset of an open set U in ℝ^{p}. Then there exists an open set Vsuch that
H ⊆ V ⊆ ¯V ⊆ U
withV compact. There also exists ψ such that H ≺ f ≺ V , meaning that f = 1 on H andspt
(f)
⊆V .
Proof: Consider h →dist
( )
h,UC
. This continuous function achieves its minimum at some
h_{0}∈ H because H is compact. Let δ ≡dist
( )
h0,U C
. The distance is positive because U^{C} is
closed. Now H ⊆∪_{h∈H}B
(h,δ)
. Since H is compact, there are finitely many of these balls
which cover H. Say H ⊆∪_{i=1}^{k}B
(hi,δ)
≡ V. Then, since there are finitely many of these balls,
let
-- k ------- k
V ≡ ∪i=1B (hi,δ),V ≡ ∪ i=1B (hi,δ)
V is a compact set since it is a finite union of compact sets.
To obtain f, let
( C )
f (x) ≡ -----distC-x,V--------
dist (x,V )+ dist(x,H )
Then f
(x)
≤ 1 and if x ∈ H, its distance to V^{C} is positive and dist
(x,H )
= 0 so f
(x)
= 1. If x ∈ V^{C},
then its distance to H is positive and so f
(x)
= 0. It is obviously continuous because the denominator is a
continuous function and never vanishes. Thus H ≺ f ≺ V . ■
Theorem 14.5.4(Partition of unity)Let K be a compact subset of ℝ^{p}and suppose
K ⊆ V = ∪ni=1Vi,Vi open.
Then there exist ψ_{i}≺ V_{i}with
∑n
ψi(x) = 1
i=1
for all x ∈ K. If H is a compact subset of V_{i}for some V_{i}there exists a partition of unity such thatψ_{i}
(x)
= 1 for all x ∈ H
Proof: Let K_{1} = K ∖∪_{i=2}^{n}V_{i}. Thus K_{1} is compact and K_{1}⊆ V_{1}. Let K_{1}⊆ W_{1}⊆W_{1}⊆ V_{1}with
W_{1}compact. To obtain W_{1}, use Theorem 14.5.3 to get f such that K_{1}≺ f ≺ V_{1} and let
W_{1}≡
{x : f (x) ⁄= 0}
.Thus W_{1},V_{2},
⋅⋅⋅
V_{n} covers K and W_{1}⊆ V_{1}. Let K_{2} = K ∖ (∪_{i=3}^{n}V_{i}∪W_{1}). Then
K_{2} is compact and K_{2}⊆ V_{2}. Let K_{2}⊆ W_{2}⊆W_{2}⊆ V_{2}W_{2} compact. Continue this way finally obtaining
W_{1},
⋅⋅⋅
,W_{n}, K ⊆ W_{1}∪
⋅⋅⋅
∪ W_{n}, and W_{i}⊆ V_{i}W_{i} compact. Now let W_{i}⊆ U_{i}⊆U_{i}⊆ V_{i},U_{i} compact.
PICT
By Theorem 14.5.3, let U_{i}≺ ϕ_{i}≺ V_{i},∪_{i=1}^{n}W_{i}≺ γ ≺∪_{i=1}^{n}U_{i}. Define
∪_{i=1}^{n}U_{i}. Consequently γ(y) = 0 for all y near x and so
ψ_{i}(y) = 0 for all y near x. Hence ψ_{i} is continuous at such x. If ∑_{j=1}^{n}ϕ_{j}(x)≠0, this situation
persists near x and so ψ_{i} is continuous at such points. Therefore ψ_{i} is continuous. If x ∈ K, then
γ(x) = 1 and so ∑_{j=1}^{n}ψ_{j}(x) = 1. Clearly 0 ≤ ψ_{i}
(x)
≤ 1 and spt(ψ_{j}) ⊆ V_{j}. As to the last claim,
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.■
Now with this preparation, here is the main result called the Riesz representation theorem for positive
linear functionals.
Theorem 14.5.5(Riesz representation theorem)Let L be a positive linear functional on C_{c}(ℝ^{p}). Thenthere exists a σ algebra S containing the Borel sets and a unique measure μ, defined onS, such that
μ is complete, (14.16)
μ(K ) < ∞ for all K compact, (14.17)
μ(F ) = sup{μ(K) : K ⊆ F,K compact},
for all F ∈S ,
μ(F) = inf{μ(V) : V ⊇ F,Vopen}
for all F ∈S, and
∫
fdμ = Lf for all f ∈ C (ℝp ). (14.18)
c
(14.18)
The plan is to define an outer measure and then to show that it, together with the σ algebra of sets
measurable in the sense of Caratheodory, satisfies the conclusions of the theorem. Always, K will be a
compact set and V will be an open set.
Definition 14.5.6μ(V ) ≡ sup{Lf : f ≺ V } for V open, μ(∅) = 0.μ(E) ≡ inf{μ(V ) : V ⊇ E} forarbitrary sets E.
Lemma 14.5.7μ is a well-defined outer measure.
Proof: First it is necessary to verify μ is well defined because there are two descriptions of it on open
sets. Suppose then that μ_{1}
(V)
≡ inf{μ(U) : U ⊇ V and U is open}. It is required to verify that
μ_{1}
(V )
= μ
(V )
where μ is given as sup{Lf : f ≺ V }. If U ⊇ V, then μ
(U)
≥ μ
(V)
directly from the
definition. Hence from the definition of μ_{1}, it follows μ_{1}
(V)
≥ μ
(V)
. On the other hand, V ⊇ V and so
μ_{1}
(V )
≤ μ
(V )
. This verifies μ is well defined. ■
It remains to show that μ is an outer measure. Let V = ∪_{i=1}^{∞}V_{i} and let f ≺ V . Then
spt(f) ⊆∪_{i=1}^{n}V_{i} for some n. Let ψ_{i}≺ V_{i},∑_{i=1}^{n}ψ_{i} = 1 on spt(f).
Now let E = ∪_{i=1}^{∞}E_{i}. Is μ(E) ≤∑_{i=1}^{∞}μ(E_{i})? Without loss of generality, it can be assumed
μ(E_{i}) < ∞ for each i since if not so, there is nothing to prove. Let V_{i}⊇ E_{i} with μ(E_{i}) + ε2^{−i}> μ(V_{i}).
Since ε was arbitrary, μ(E) ≤∑_{i=1}^{∞}μ(E_{i}) which proves the lemma. ■
Lemma 14.5.8Let K be compact, g ≥ 0,g ∈ C_{c}(ℝ^{p}), and g = 1 on K. Then μ(K) ≤ Lg. Alsoμ(K) < ∞ whenever K is compact.
Proof: Let α ∈ (0,1) and V_{α} = {x : g(x) > α} so V_{α}⊇ K and let h ≺ V_{α}.
PICT
Then h ≤ 1 onV_{α} while gα^{−1}≥ 1 on V_{α}and so gα^{−1}≥ h which implies L(gα^{−1}) ≥ Lh and that
therefore, since L is linear,
Lg ≥ αLh.
Since h ≺ V_{α} is arbitrary, and K ⊆ V_{α},
Lg ≥ αμ(Vα) ≥ αμ(K ).
Letting α ↑ 1 yields Lg ≥ μ(K). This proves the first part of the lemma. The second assertion follows from
this and Theorem 14.5.3. If K is given, let
K ≺ g ≺ ℝp
and so from what was just shown, μ
(K )
≤ Lg < ∞. ■
For two sets A,B, let
dist(A, B) ≡ inf{|a− b| : a ∈ A, b ∈ B}.
Lemma 14.5.9If A and B are disjoint subsets of ℝ^{p}, withdist
(A,B)
> 0 then μ(A ∪ B) =
μ(A) + μ(B).
Proof: There is nothing to show if μ
(A∪ B )
= ∞. Let δ ≡dist
(A,B )
> 0. Then let
U_{1}≡∪_{a∈A}B
( δ)
a,3
,V_{1}≡∪_{b∈B}B
( δ)
b,3
. It follows that these two open sets have empty intersection.
Also, there exists W ⊇ A ∪ B such that μ
(W )
− ε < μ
(A ∪ B)
. let U ≡ U_{1}∩ W,V ≡ V_{1}∩ W.
Then
μ (A ∪ B) + ε > μ(W ) ≥ μ(U ∪V )
Now let f ≺ U,g ≺ V such that Lf + ε > μ
(U)
,Lg + ε > μ
(V )
. Then
μ(U ∪ V) ≥ L (f + g) = L (f)+ L (g)
> μ (U )− ε+ (μ(V )− ε)
≥ μ (A )+ μ(B )− 2ε
It follows that
μ(A ∪B )+ ε > μ (A )+ μ(B )− 2ε
and since ε is arbitrary, μ
(A ∪B )
≥ μ
(A )
+ μ
(B )
≥ μ
(A ∪ B)
. ■
It follows from Theorem 5.9.2 that the measurable sets S contains the Borel σ algebra ℬ
(ℝp)
. Since
closures of balls are compact, it follows from Lemma 14.5.8 that μ is finite on every ball. Corollary 5.4.7
implies that μ is regular for every E a Borel set. That is,
μ(E ) = sup{μ(K ) : K ⊆ E },
μ(E ) = inf{μ (V) : V ⊇ E}
In particular, μ is inner regular on every open set V . This is obtained immediately. In fact the same thing
holds for any F ∈S in place of E in the above. The second of the two follows immediately from the
definition of μ. It remains to verify the first. In doing so, first assume that μ
(F)
is contained in a closed
ball B. Let V ⊇
(B ∖ F)
such that
μ (V ) < μ (B ∖ F)+ ε
Then μ
(V ∖ (B ∖ F))
+ μ
(B ∖ F)
= μ
(V)
< μ
(B ∖F)
+ ε and so μ
(V ∖ (B ∖ F))
< ε. Now consider
V^{C}∩ F. This is a closed subset of F. To see that it is closed, note that V^{C}∩ F = V^{C}∩ B which is a
closed set. Why is this so? It is clear that V^{C}∩ F ⊆ V^{C}∩ B. Now if x ∈ V^{C}∩ B, then since
V ⊇
(B ∖ F)
, it follows that x ∈ V^{C}⊆ B^{C}∪ F and so either x ∈ B^{C} which doesn’t occur, or x ∈ F
and so this must be the case. Hence, V^{C}∩ B is a closed, hence compact subset of F. Now
( ( )) ( ( ))
μ F ∖ V C ∩B = μ F ∩ V ∪BC
≤ μ(V ∖B ) ≤ μ (V ∖ (B ∖ F)) < ε
It follows that μ
(F)
< μ
( C )
V ∩ B
+ ε which shows inner regularity in case F is contained in some closed
ball B. If this is not the case, let B_{n} be a sequence of closed balls having increasing radii and let
F_{n} = B_{n}∩ F. Then if l < μ
(F)
, it follows that μ
(Fn)
> l for all large enough n. Then picking one of
these, it follows from what was just shown that there is a compact set K ⊆ F_{n} such that also
μ
(K )
> l.
Thus S contains the Borel sets and μ is inner regular on all sets of S.