Chapter 21 Measures From Positive Linear Functionals
Rudin does it this way and I really don’t know a better way to do it. In this chapter, we will only consider
the measure space to be a complete separable metric space in which the measure of balls is finite. Rudin
does this for an arbitrary locally compact Hausdorff space, but all examples of interest to me are metric
spaces. In fact, the one of most interest is ℝ^{n}.
Definition 21.0.1Let Ω be a Polish space (complete separable metric space). Define C_{c}
(Ω)
to be the functions which have complex values and compact support. Thismeansspt
(f)
≡
{x ∈ Ω : f (x) ⁄= 0}
is a compact set. Then L : C_{c}
(Ω)
→ ℂ is called a positive linear functional ifit is linear and if, whenever f ≥ 0, then L
(f )
≥ 0 also.
The following definition gives some notation.
Definition 21.0.2If K is a compact subset of an open set, V , then K ≺ ϕ ≺ V if
ϕ ∈ Cc(V),ϕ(K ) = {1},ϕ(Ω ) ⊆ [0,1],
where Ω denotes the whole topological space considered. Also for ϕ ∈ C_{c}(Ω), K ≺ ϕ if
ϕ(Ω ) ⊆ [0,1] and ϕ(K ) = 1.
and ϕ ≺ V if
ϕ(Ω) ⊆ [0,1] and spt(ϕ) ⊆ V.
Now we need some lemmas.
Theorem 21.0.3Let H be a compact subset of an open set U in a metric space having the property thatthe closures of balls are compact. Then there exists an open set V such 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,U C)
. This continuous function achieves its minimum at some h_{0}∈ H
because H is compact. Let δ ≡dist
(h ,U C)
0
. 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
(h ,δ)
i
≡ V. Then, since there are finitely many of these balls, V = ∪_{i=1}^{k}B
(h ,δ)
i
which is a
compact set since it is a finite union of compact sets.
≤ 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 21.0.4(Partition of unity)Let K be a compact subset of a Polish space in which the closuresof balls are compact 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 21.0.3 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 21.0.3, 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
^V
j
≡ 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 21.0.5(Riesz representation theorem)Let (Ω,τ) be a Polish space for which theclosures of the balls are compact and let L be a positive linear functional on C_{c}(Ω). Then thereexists a σ algebra S containing the Borel sets and a unique measure μ, defined onS, such that
μ is complete, (21.1)
μ(K ) < ∞ for all K compact, (21.2)
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 ∈ Cc(Ω). (21.3)
(21.3)
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 21.0.6μ(V ) ≡ sup{Lf : f ≺ V } for V open, μ(∅) = 0.μ(E) ≡ inf{μ(V ) : V ⊇ E} forarbitrary sets E.
Lemma 21.0.7μ 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 21.0.8Let K be compact, g ≥ 0,g ∈ C_{c}(Ω), 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 21.0.3. If K is given, let
K ≺ g ≺ Ω
and so from what was just shown, μ
(K )
≤ Lg < ∞. ■
Lemma 21.0.9If A and B are disjoint subsets of Ω, withdist
(A, B)
> 0 then μ(A ∪ B) =
μ(A) + μ(B).
Proof: There is nothing to show if μ
(A ∪ B)
= ∞. Thus we can 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 19.5.2 that the measurable sets S contains the Borel σ algebra ℬ
(Ω)
. Since
closures of balls are compact, it follows from Lemma 21.0.8 that μ is finite on every ball. Corollary 19.4.8
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.