This is on the Riesz representation theorem for positive linear functionals. It is a really marvelous result. It produces measures on locally compact Hausdorff spaces. Thus this doesn’t help a lot in producing measures on infinite dimensional spaces but it works great on ℝ^{n} or closed subsets of ℝ^{n} and so forth.
Definition 21.6.1 Let (Ω,τ) be a topological space. L : C_{c}(Ω) → ℂ is called a positive linear functional if L is linear,

and if Lf ≥ 0 whenever f ≥ 0.
Theorem 21.6.2 (Riesz representation theorem) Let (Ω,τ) be a locally compact Hausdorff space and let L be a positive linear functional on C_{c}(Ω). Then there exists a σ algebra S containing the Borel sets and a unique measure μ, defined on S, such that

for all F open and for all F ∈S with μ(F) < ∞,

for all F ∈S, and
 (21.8) 
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 21.6.3 μ(V ) ≡ sup{Lf : f ≺ V } for V open, μ(∅) = 0.μ(E) ≡ inf{μ(V ) : V ⊇ E} for arbitrary sets E.
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}
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).

Hence

since f ≺ V is arbitrary. 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 21.6.5 Let K be compact, g ≥ 0,g ∈ C_{c}(Ω), and g = 1 on K. Then μ(K) ≤ Lg. Also μ(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}.
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,

Since h ≺ V _{α} is arbitrary, and K ⊆ V _{α},

Letting α ↑ 1 yields Lg ≥ μ(K). This proves the first part of the lemma. The second assertion follows from this and Theorem 21.1.22. If K is given, let

and so from what was just shown, μ
Proof: By Theorem 21.1.22, there exists h ∈ C_{c}
From Lemma 21.6.5 μ(A ∪ B) < ∞ and so there exists an open set, W such that

Now let U = U_{1} ∩ W and V = V _{1} ∩ W. Then

Let A ≺ f ≺ U,B ≺ g ≺ V . Then by Lemma 21.6.5,

Since ε > 0 is arbitrary, this proves the lemma.
From Lemma 21.6.5 the following lemma is obtained.
Proof: Let V ⊇ spt(f) and let spt(f) ≺ g ≺ V . Then Lf ≤ Lg ≤ μ(V ) because f ≤ g. Since this holds for all V ⊇ spt(f),Lf ≤ μ(spt(f)) by definition of μ.
At this point, the conditions of Lemma 21.5.1 have been verified. Thus S contains the Borel sets and μ is inner regular on sets of S having finite measure.
It remains to show μ satisfies 21.8.
Proof: Let f ∈ C_{c}(Ω),f realvalued, and suppose f(Ω) ⊆ [a,b]. Choose t_{0} < a and let t_{0} < t_{1} <
 (21.9) 
Note that ∪_{i=1}^{n}E_{i} is a closed set, and in fact
 (21.10) 
since Ω = ∪_{i=1}^{n}f^{−1}((t_{i−1},t_{i}]). Let V _{i} ⊇ E_{i},V _{i} is open and let V _{i} satisfy
 (21.11) 

By Theorem 21.4.3 there exists h_{i} ∈ C_{c}(Ω) such that

Now note that for each i,

(If x ∈ V _{i}, this follows from 21.11. If x




From 21.10 and 21.9, the first and last terms cancel. Therefore this is no larger than

Since ε > 0 is arbitrary,
 (21.12) 
for all f ∈ C_{c}(Ω),f real. Hence equality holds in 21.12 because L(−f) ≤−∫ fdμ so L(f) ≥∫ fdμ. Thus Lf = ∫ fdμ for all f ∈ C_{c}(Ω). Just apply the result for real functions to the real and imaginary parts of f. This proves the Lemma.
This gives the existence part of the Riesz representation theorem.
It only remains to prove uniqueness. Suppose both μ_{1} and μ_{2} are measures on S satisfying the conclusions of the theorem. Then if K is compact and V ⊇ K, let K ≺ f ≺ V . Then

Thus μ_{1}(K) ≤ μ_{2}(K) for all K. Similarly, the inequality can be reversed and so it follows the two measures are equal on compact sets. By the assumption of inner regularity on open sets, the two measures are also equal on all open sets. By outer regularity, they are equal on all sets of S. ■
Example 21.6.9 Let L
Here is a nice observation.
Proposition 21.6.11 In Example 21.6.10 the order of integration is not important. The same functional is obtained in any order.
Proof: Let spt
