17. Riesz' representation theorem - Aarhus Universitet
17. Riesz' representation theorem - Aarhus Universitet
17. Riesz' representation theorem - Aarhus Universitet
Create successful ePaper yourself
Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.
<strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong><br />
Klaus Thomsen matkt@imf.au.dk<br />
Institut for Matematiske Fag<br />
Det Naturvidenskabelige Fakultet<br />
<strong>Aarhus</strong> <strong>Universitet</strong><br />
November 2005<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong>
We read the lecture notes on Riesz’ <strong>representation</strong> <strong>theorem</strong>, mostly<br />
stolen from Walter Rudins book: ’Real and complex analysis’.<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong>
The Riesz <strong>representation</strong> <strong>theorem</strong> - the setting<br />
Throughout the note X will be a fixed locally compact Hausdorff<br />
space. By Cc(X) we denote the vector space of continuous<br />
compactly supported functions on X, i.e. a continuous function<br />
f : X → C is in Cc(X) if and only if<br />
supp f = {x ∈ X : f (x) = 0}<br />
is a compact subset of X.<br />
A linear functional Λ : Cc(X) → C is said to be positive when<br />
f ∈ Cc(X), f (x) ≥ 0 ∀x ∈ X ⇒ Λ(f ) ≥ 0.<br />
Some simple examples of positive linear functionals on Cc(X) are<br />
easy to come by: For example, one may choose a couple of points<br />
x1,x2 ∈ X and two non-negative real numbers α1,α2, and define<br />
Γ : Cc(X) → C such that<br />
Γ(f ) = α1f (x1) + α2f (x2).<br />
Then Γ is a positive linear functional on Cc(X).<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong>
The Riesz <strong>representation</strong> <strong>theorem</strong> - the setting<br />
More generally: Let µ be a measure defined on a σ-algebra of<br />
subsets of X which (at least) contains the Borel sets in X. If µ is<br />
finite on every compact subset of X, i.e. if µ(K) < ∞ when K ⊆ X<br />
is compact, then any function f ∈ Cc(X) is integrable with respect<br />
to µ. (This follows from Proposition 15 ii) on p. 267 in Roydens<br />
book since |f | ≤ M1K, where M = sup {|f (x)| : x ∈ X } < ∞ and<br />
1K is the characteristic function of K = supp f .) We can therefore<br />
define Λµ : Cc(X) → C by integration with respect to µ, i.e.<br />
<br />
Λµ(f ) = f dµ,<br />
(sometimes written <br />
X f (x) dµ(x) to emphasize the variable.) It<br />
follows from Proposition 15 i)+ ii) on p. 267 in Roydens book that<br />
Λµ is then a positive linear functional on Cc(X).<br />
The main content of Riesz’s <strong>representation</strong> <strong>theorem</strong> is that every<br />
positive linear functional on Cc(X) arises in this way.<br />
X<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong>
The Riesz <strong>representation</strong> <strong>theorem</strong> for positive functionals<br />
Theorem<br />
Let X be a locally compact Hausdorff space. Let Λ : Cc(X) → C<br />
be a positive linear functional.<br />
Then there exists a σ-algebra M in X which contains all Borel sets<br />
in X, and there exists a unique positive measure µ on M which<br />
represents Λ in the sense that<br />
(a)<br />
<br />
Λ(f ) = f dµ for every f ∈ Cc(X),<br />
X<br />
and which has the following additional properties:<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong>
The Riesz <strong>representation</strong> <strong>theorem</strong> for positive functionals<br />
Theorem<br />
(b)<br />
(c) For every E ∈ M, we have<br />
(d) The relation<br />
µ(K) < ∞ when K ⊆ X is compact.<br />
µ(E) = inf {µ(V) : E ⊆ V, V open} .<br />
µ(E) = sup {µ(K) : K ⊆ E, K compact}<br />
holds for every open set E, and every E ∈ M with µ(E) < ∞.<br />
(e) If E ∈ M, A ⊆ E, and µ(E) = 0, then A ∈ M.<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong>
The proof - uniqueness<br />
The uniqueness part of the statement is that if µ1 and µ2 are both<br />
measures on M with the properties (a)-(d), then µ1 = µ2.<br />
We start by proving this. For this we introduce the following<br />
notation: When f ∈ Cc(X) is a function taking values in [0,1], and<br />
E ⊆ X we shall write<br />
f ≺ E,<br />
when supp f ⊂ E, and<br />
E ≺ f<br />
when f (x) = 1 for x ∈ E.<br />
It follows then from Urysohn’s lemma, Lemma 0.17 in<br />
’Kommentarer’ that whenever K is a compact subset of X, V an<br />
open subset of X, and K ⊆ V , then there is an f ∈ Cc(X) such<br />
that<br />
K ≺ f ≺ V.<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong>
The proof - uniqueness<br />
In particular, it follows that<br />
<br />
µ1(K) = 1K dµ1 ≤ f dµ1 =<br />
X<br />
X<br />
X<br />
<br />
f dµ2 ≤<br />
X<br />
1V dµ2 = µ2(V)<br />
(1)<br />
in this situation, i.e. when K is compact, V is open and K ⊆ V .<br />
Since µ2 satisfies (c) we deduce from (1) that µ1(K) ≤ µ2(K)<br />
when K ⊆ X is compact.<br />
Since µ1 and µ2 both have property (d) we conclude that<br />
µ1(V) ≤ µ2(V) when V ⊆ X is open.<br />
Since they also both have property (c) we conclude that<br />
µ1(E) ≤ µ2(E) for all E ∈ M. By symmetry we must also have the<br />
reversed inequality, and we may therefore conclude that µ1 = µ2.<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong>
The proof - the construction<br />
We turn now to the construction of the measure µ. First we define<br />
µ(V) when V ⊆ X is open:<br />
µ(V) = sup {Λ(f ) : f ≺ V }. (2)<br />
Since a part of the conditions in ’f ≺ V ’ is that f (X) ⊆ [0,1] it<br />
follows that µ(V) ≥ 0 because Λ is a positive linear functional.<br />
Note that it may very well be that µ(V) = ∞! Note also that<br />
when V1 and V2 are both open and V1 ⊆ V2, then it follows from<br />
(2) that µ(V1) ≤ µ(V2).<br />
Hence, when we set<br />
µ(E) = inf {µ(V) : E ⊆ V } (3)<br />
for any subset E ⊆ X, we haven’t changed the definition on open<br />
sets. In other words, with (3) we have an extension of µ from open<br />
to arbitrary sets.<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong>
The proof - the construction<br />
This extension will not in general give us a measure defined, as it<br />
is, on all subsets of X, and a major part of the proof is to identify a<br />
σ-algebra of sets in X which contains the Borel sets, and on which<br />
µ does give us a measure. (The trouble, of course, is the countable<br />
additivity of µ!)<br />
Let MF be the subsets E of X for which µ(E) < ∞ and<br />
µ(E) = sup {µ(K) : K ⊆ E compact} . (4)<br />
Let M be the subsets E of X with the property that<br />
when K ⊆ X is compact.<br />
E ∩ K ∈ MF<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong><br />
(5)
The proof - checking the properties of M and µ<br />
We prove that µ : M → [0, ∞] has the required properties:<br />
First of all, observe that µ is monotone, in the sense that<br />
since {µ(V) : B ⊆ V } ⊆ {µ(V) : A ⊆ V }.<br />
Next we establish the following<br />
A ⊆ B ⇒ µ(A) ≤ µ(B) (6)<br />
If E1,E2,E3,..., are arbitrary subsets of X,<br />
<br />
∞<br />
<br />
∞<br />
µ ≤ µ(Ei).<br />
i=1<br />
Ei<br />
i=1<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong><br />
(7)
Proof of (7)<br />
We prove first that<br />
µ (V1 ∪ V2) ≤ µ(V1) + µ(V2) (8)<br />
when V1 and V2 are open.<br />
To this end, let g ≺ V1 ∪ V2. It follows then from Theorem 0.18 in<br />
’Kommentarer’ - the <strong>theorem</strong> on partitions of unity - that there are<br />
functions hi ∈ Cc(X),i = 1,2, such that h1(x) + h2(x) = 1 when<br />
x ∈ supp g, and gi ≺ Vi,i = 1,2.<br />
It follows that g = gh1 + gh2 so the additivity of Λ shows that<br />
Λ(g) = Λ(gh1) + Λ(gh2).<br />
Since ghi ≺ Vi we conclude that Λ(g) ≤ µ(V1) + µ(V2). It follows<br />
then from (2) that µ2(V1 ∪ V2) ≤ µ(V1) + µ(V2), proving (8).<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong>
Proof of (7)<br />
By induction it follows that<br />
<br />
n<br />
µ<br />
i=1<br />
Vi<br />
<br />
≤<br />
n<br />
µ(Vi) (9)<br />
for any finite collection V1,V2,... ,Vn of open sets in X.<br />
This is then used to prove (7) in the following way: If µ (Ei) = ∞<br />
for some i, the inequality (7) is trivial, so we may assume that<br />
µ(Ei) < ∞ for all i.<br />
Let ǫ > 0. It follows from the definition, (3), that there are open<br />
sets Vi,i = 1,2,... , in X such that Ei ⊆ Vi and<br />
µ(Vi) ≤ µ(Ei) + 2 −i ǫ for all i.<br />
i=1<br />
Since ∞<br />
i=1 Ei ⊆ ∞<br />
i=1 Vi we conclude that<br />
µ<br />
∞<br />
i=1<br />
Ei<br />
<br />
≤ µ<br />
∞<br />
i=1<br />
Vi<br />
<br />
. (10)<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong>
Proof of (7)<br />
To estimate the right-hand side, let f ≺ ∞ i=1 Vi. Then<br />
Vi,i = 1,2,... , is an open cover of the compact set suppf so<br />
there is an n ∈ N such that f ≺ n i=1 Vi.<br />
Then<br />
<br />
∞<br />
<br />
Λ(f ) ≤ µ ,<br />
and by use of (9) we find that<br />
Λ(f ) ≤<br />
n<br />
µ(Vi) ≤<br />
i=1<br />
i=1<br />
i=1<br />
Vi<br />
n <br />
µ(Ei) + 2 −i ǫ n<br />
∞<br />
≤ ǫ+ µ(Ei) ≤ ǫ+ µ(Ei).<br />
Since f ≺ ∞ i=1 Vi was arbitrary we conclude (by using (2)) that<br />
µ ( ∞ i=1 Vi) ≤ ǫ + ∞ i=1 µ(Ei).<br />
Since ǫ > 0 was arbitrary, we conclude that<br />
µ ( ∞ i=1 Vi) ≤ ∞ i=1 µ(Ei).<br />
In combination with (10), this proves (7).<br />
i=1<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong><br />
i=1
Proof of (11)<br />
The next step is to establish<br />
If K ⊆ X is compact, then K ∈ MF, and<br />
µ(K) = inf {Λ(f ) : K ≺ f }.<br />
(11)<br />
If K ≺ f , and 0 < α < 1, set Vα = {x ∈ X : f (x) > α}.<br />
Then K ⊆ Vα since f (x) = 1 when x ∈ K, and when g ∈ Cc(X) is<br />
any function such that g ≺ Vα, we have that αg ≤ f .<br />
Hence Λ(αg) = αΛ(g) ≤ Λ(f ) because Λ is linear and positive.<br />
It follows that<br />
µ(K) ≤ µ(Vα) = sup {Λ(g) : g ≺ Vα} ≤ α −1 Λ(f ).<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong>
Proof of (11)<br />
In particular we see that µ(K) < ∞, and by letting α → 1 we<br />
obtain the conclusion that µ(K) ≤ Λ(f ).<br />
Since K ≺ f was arbitrary, we see that µ(K) ≤ inf {Λ(f ) : K ≺ f }.<br />
On the other hand, if we let ǫ > 0, it follows from (3) that there is<br />
an open set V ⊇ K such that µ(V) ≤ µ(K) + ǫ.<br />
By Urysohn’s lemma there is a function K ≺ f ≺ V , and for this<br />
function Λ(f ) ≤ µ(V) by (2). Since K ≺ f , we conclude that<br />
inf {Λ(f ) : K ≺ f } ≤ µ(K) + ǫ.<br />
Since ǫ > 0 was arbitrary, we have established the equality in (11).<br />
As we saw, µ(K) < ∞, so it follows from (6) and the compactness<br />
of K that K ∈ MF. - (11) is proved.<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong>
Proof of (12)<br />
Every open set satisfies (4).<br />
Hence V ∈ MF when V is open and µ(V) < ∞.<br />
Let V ⊆ X be open, and let α be a real number such that<br />
α < µ(V). By definition of µ(V), cf. (2), there is an f ∈ Cc(X)<br />
such that f ≺ V and Λ(f ) > α.<br />
Then K = supp f is a compact subset of V , and when W is an<br />
open subset containing K, viz. K ⊆ W, then f ≺ W and hence<br />
µ(W) ≥ Λ(f ).<br />
It follows from (3) that µ(K) ≥ Λ(f ) > α. Since α < µ(V) was<br />
arbitrary, we conclude that that sup {µ(K) : K ⊆ V } ≥ µ(V).<br />
The reversed inequality is trivial, thanks to (6), so (12) has been<br />
established.<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong><br />
(12)
Proof of (13)<br />
Suppose E =<br />
We prove first that<br />
∞<br />
Ei, where E1,E2,E3,... , are<br />
i=1<br />
pairwise disjoint members of MF. Then<br />
∞<br />
µ(E) = µ (Ei).<br />
i=1<br />
If, in addition, µ(E) < ∞, then also E ∈ MF.<br />
(13)<br />
µ (K1 ∪ K2) = µ(K1) + µ(K2) (14)<br />
when Ki,i = 1,2, are compact and disjoint.<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong>
Proof of (13)<br />
By Urysohn’s lemma there is an f ∈ Cc(X) such that 0 ≤ f ≤ 1,<br />
f (x) = 1,x ∈ K1, and f (x) = 0 when x ∈ K2.<br />
Let ǫ > 0. It follows from (11) that there is g ∈ Cc(X) such that<br />
K1 ∪ K2 ≺ g and Λ(g) ≤ µ(K1 ∪ K2) + ǫ.<br />
Then K1 ≺ fg and K2 ≺ (1 − f )g. Hence µ(K1) ≤ Λ(fg) and<br />
µ(K2) ≤ Λ((1 − f )g) by (11).<br />
Since Λ is linear, we find that<br />
µ(K1) + µ(K2) ≤ Λ(fg) + Λ((1 − f )g) = Λ(g) ≤ µ(K1 ∪ K2) + ǫ.<br />
Since ǫ > 0 was arbitrary we conclude that<br />
µ(K1) + µ(K2) ≤ µ(K1 ∪ K2), and then (14) follows from (7).<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong>
Proof of (13)<br />
To prove the equality in (13) we pick compact subsets Ki ⊆ Ei such<br />
that µ(Ki) ≥ µ(Ei) − 2−iǫ. This is possible since Ei ∈ MF.<br />
For each n, n i=1 Ki is a compact subset of E, and it follows from<br />
(14) that<br />
<br />
n<br />
<br />
n n<br />
µ(E) ≥ µ = µ (Ki) ≥ µ (Ei)−2 −i n<br />
ǫ ≥ µ (Ei)−ǫ.<br />
i=1<br />
Ki<br />
i=1<br />
i=1<br />
Since n and ǫ > 0 are arbitrary here, we conclude that<br />
µ(E) ≥<br />
i=1<br />
(15)<br />
∞<br />
µ (Ei) . (16)<br />
i=1<br />
Combined with (7) this yields the equality in (13).<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong>
Proof of (13)<br />
Returning to (15) we conclude then that if µ(E) < ∞,<br />
Kn = n i=1 Ki is a compact subset of E such that<br />
µ(K) ≥ n i=1 µ (Ei) − ǫ = µ(E) − ∞ j=n+1 µ (Ej) − ǫ.<br />
Since ∞ i=1 µ (Ei) = µ(E) < ∞ we have that ∞ j=n+1 µ (Ej) < ǫ if<br />
n is large enough. So for n large, Kn is a compact subset of E such<br />
that µ (Kn) ≥ µ(E) − 2ǫ.<br />
It follows that E satisfies (4) when µ(E) < ∞. Thus E ∈ MF in<br />
this case.<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong>
Proof of (17)<br />
If E ∈ MF, and ǫ > 0, there is a compact subset K ⊆ X<br />
and an open subset V ⊆ X such that K ⊆ E ⊆ V , and µ(V \K) ≤ ǫ.<br />
(17)<br />
It follows from (3) that there is an open set V ⊇ E such that<br />
µ(V) ≤ µ(E) + ǫ<br />
2<br />
and since (4) holds and µ(E) < ∞ there is a compact subset<br />
K ⊆ E such that µ(K) ≥ µ(E) − ǫ<br />
2 .<br />
Note that V = K ∪ (V \K), and that V \K is open. It follows from<br />
(12) that V \K ∈ MF since µ(V \K) ≤ µ(V) < ∞,<br />
and from (11) that K ∈ MF, so we conclude from (13) that<br />
µ(V) = µ(K) + µ(V \K),<br />
which implies that<br />
µ(V \K) = µ(V) − µ(K) ≤ µ(E) + ǫ ǫ<br />
2 − µ(E) + 2 = ǫ.<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong>
Proof of (18)<br />
If A ∈ MF, B ∈ MF, then A\B, A ∪ B, and A ∩ B belong to MF.<br />
(18)<br />
Since µ(A ∪ B) ≤ µ(A) + µ(B) < ∞ (using (7)), it follows from<br />
(6) that µ(A\B) < ∞, µ(A ∪ B) < ∞ and µ(A ∩ B) < ∞.<br />
So it remains just to show that all three sets in the statement<br />
satisfy (4).<br />
To this end, let ǫ > 0. By (17) there are compact sets Ki,i = 1,2,<br />
and open sets Vi,i = 1,2, such that K1 ⊆ A ⊆ V1, K2 ⊆ B ⊆ V2,<br />
and µ(Vi\Ki) ≤ ǫ,i = 1,2.<br />
Note that<br />
A\B ⊆ V1\K2 ⊆ (V1\K1) ∪ (K1\V2) ∪ (V2\K2) .<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong>
Proof of (18)<br />
Of the three last sets two are open and one is compact. (12) and<br />
(11) they are all in MF and hence (7) implies that<br />
µ(A\B) ≤ µ (V1\K1)+µ (K1\V2)+µ (V2\K2) ≤ ǫ+µ (K1\V2)+ǫ.<br />
Then K1\V2 is a compact subset of A\B such that<br />
µ (K1\V2) ≥ µ(A\B) − 2ǫ.<br />
Since ǫ > 0 was arbitrary, this shows that A\B ∈ MF. Once this is<br />
established it follows from (13) that A ∪ B = (A\B) ∪ B and<br />
A ∩ B = A\(A\B) are both in MF.<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong>
Proof of (19)<br />
M is a σ-algebra in X which contains all Borel sets. (19)<br />
In the following K is an arbitrary compact subset of X. Recall that<br />
K ∈ MF by Observation 11.<br />
Let A ∈ M, and consider the complement Ac = X \A. By definition<br />
of M, A ∩ K ∈ MF, and Ac ∩ K = K\(A ∩ K) is therefore the<br />
set-theoretic difference of two sets from MF. Hence Ac ∩ K ∈ MF<br />
by (18). This shows that Ac ∈ M since K was arbitrary.<br />
Consider then a sequence Ai,i = 1,2,3,... , of sets from M. Put<br />
B1 = A1 ∩ K, and Bn = (An ∩ K) \(B1 ∪ B2 ∪ · · · ∪ Bn−1),n ≥ 2.<br />
Then the Bi’s are disjoint subsets of X and<br />
K ∩ ( ∞ n=1 An) = ∞ n=1 Bn. It follows from (18) that Bi ∈ MF for<br />
all i.<br />
Since (K ∩ ( ∞ n=1 An)) ≤ µ(K) < ∞ by (6) and (11), we deduce<br />
from (13) that ∞ n=1 Bn ∈ MF. It follows that ∞ n=1 An ∈ M since<br />
K was arbitrary.<br />
It follows that M is a σ-algebra.<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong>
End of proof of (19), beginning of proof of (20)<br />
If C is a closed subset of X, the intersection C ∩ K is compact and<br />
hence an element of MF by Observation 11. Thus C ∈ M since K<br />
was arbitrary, and we see that M contains all closed subsets of X.<br />
Being a σ-algebra it must therefore contain all Borel subsets of X.<br />
MF = {A ∈ M : µ(A) < ∞} . (20)<br />
If A ∈ MF, µ(A) < ∞ by definition. Furthermore, it follows from<br />
(11) and (18) that A ∩ K ∈ MF for every compact subset K.<br />
Hence A ∈ M. This proves one of the desired inclusions.<br />
To prove the other, assume that A ∈ M and that µ(A) < ∞. To<br />
prove that A ∈ MF, let ǫ > 0.<br />
By definition of µ there is an open set V ⊇ A such that<br />
µ(V) < ∞. By (12) V ∈ MF, and by (17) we can find a compact<br />
subset K ⊆ V such that µ (V \K) ≤ ǫ.<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong>
Proof of (20)<br />
Since A ∩ K ∈ MF, there is a compact subset H ⊆ A ∩ K such that<br />
µ(H) ≥ µ(A ∩ K) − ǫ.<br />
Since A ⊆ (A ∩ K) ∪ (V \K), it follows that<br />
µ(A) ≤ µ(A ∩ K) + µ (V \K) ≤ µ(H) + ǫ + µ (V \K) ≤ µ(H) + 2ǫ.<br />
Since ǫ > 0 was arbitrary, we see from this that A ∈ MF, as<br />
desired.<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong>
Proof of (21)<br />
µ is a measure on M. (21)<br />
Let E1,E2,... be a sequence of mutually disjoint elements of M.<br />
If µ (Ei) = ∞ for some i, it is clear that<br />
µ ( ∞ i=1 Ei) = ∞ = ∞ i=1 µ (Ei).<br />
Otherwise, it follows from (20) that Ei ∈ MF and hence<br />
µ ( ∞<br />
i=1 Ei) = ∞<br />
i=1 µ (Ei) by (13).<br />
The reader is now asked to observe that the properties of µ<br />
stipulated in (b)-(e) all hold: (b) follows from (11), (c) follows from<br />
the definition of µ, (d) follows from (20) and (12), while (e) follows<br />
from (20) since the set A of (e) is in MF be definition of MF.<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong>
Proof of a)<br />
It remains to check that condition (a) holds. To this end it suffices<br />
to check that<br />
<br />
Λ(f ) ≤ f dµ (22)<br />
for every real-valued element of Cc(X).<br />
Indeed, it follows then that also Λ(−f ) ≤ <br />
X −f dµ holds, which<br />
implies first that Λ(f ) ≥ <br />
<br />
X f dµ and then that Λ(f ) = X f dµ.<br />
By linearity this yields (a).<br />
We prove (22): Choose a < b in R such that f (X) ⊆ [a,b], and let<br />
ǫ > 0. Choose y0 < y1 < y2 < · · · < yn such that yi − yi−1 < ǫ for<br />
all i, y0 < a and yn = b. Put<br />
Ei = {x ∈ R : yi−1 < f (x) ≤ yi} ∩ K,<br />
i = 1,2,... ,n, where K = supp f .<br />
Since f is continuous and hence Borel measurable, the Ei’s are<br />
Borel sets. They are mutually disjoint.<br />
X<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong>
Proof of a)<br />
It follows from the definition of µ that there are open sets Wi ⊇ Ei<br />
such that µ (Wi) < µ (Ei) + ǫ<br />
n . Set Vi = Wi ∩ f −1 (] − ∞,yi + ǫ[.<br />
Then Vi is a an open set such that Vi ⊇ Ei,f (x) < yi + ǫ,x ∈ Vi,<br />
and<br />
µ (Vi) ≤ µ (Wi) < µ (Ei) + ǫ<br />
. (23)<br />
n<br />
Note that K ⊆ n i=1 Vi. It follows from Theorem 0.18 in<br />
’Kommentarer’ that there are continuous functions hi : X → [0,1]<br />
such that hi ≺ Vi,i = 1,2,... ,n, and n i=1 hi(x) = 1,x ∈ K.<br />
Then f = n i=1 hif , and it follows from (19) that<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong>
Proof of a)<br />
µ(K) ≤ Λ<br />
n<br />
i=1<br />
hi<br />
<br />
=<br />
n<br />
Λ(hi). (24)<br />
Since hif ≤ (yi + ǫ)hi, and since yi − ǫ < f (x) on Ei, we find that<br />
Λ(f ) =<br />
=<br />
≤<br />
n<br />
Λ(hif ) ≤<br />
i=1<br />
n<br />
(yi + ǫ)Λ(hi)<br />
i=1<br />
n<br />
(|a| + yi + ǫ)Λ(hi) −<br />
i=1<br />
n<br />
(|a| + yi + ǫ)µ(Vi) −<br />
i=1<br />
i=1<br />
n<br />
|a|Λ(hi)<br />
i=1<br />
n<br />
|a|Λ(hi) (using (2))<br />
i=1<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong><br />
(25)
Proof of a)<br />
n <br />
≤ (|a| + yi + ǫ) µ(Ei) +<br />
i=1<br />
ǫ<br />
n<br />
− |a|Λ(hi)<br />
n<br />
i=1<br />
(using (23))<br />
n <br />
≤ (|a| + yi + ǫ) µ(Ei) +<br />
i=1<br />
ǫ<br />
<br />
− |a|µ(K)<br />
n<br />
(using (24))<br />
n <br />
= (yi + ǫ) µ(Ei) + ǫ<br />
n<br />
+ |a|<br />
n<br />
ǫ<br />
n<br />
n<br />
(since µ(Ei) = µ(K))<br />
=<br />
i=1<br />
i=1<br />
n<br />
(yi − ǫ)µ(Ei) + 2ǫµ(K) +<br />
i=1<br />
(since<br />
n<br />
i=1<br />
i=1<br />
(|a| + yi + ǫ) ǫ<br />
n<br />
n<br />
µ(Ei) = µ(K))<br />
i=1<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong><br />
(26)
Proof of a)<br />
<br />
≤<br />
<br />
≤<br />
X<br />
X<br />
f dµ + 2ǫµ(K) +<br />
n<br />
i=1<br />
(|a| + yi + ǫ) ǫ<br />
n<br />
(since<br />
n<br />
(yi − ǫ)1Ei ≤ f )<br />
i=1<br />
f dµ + 2ǫµ(K) + ǫ(|a| + b + ǫ) (since yi ≤ b for all i).<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong><br />
(27)
Proof of a)<br />
Since ǫ > 0 was arbitrary here, this yields (22).<br />
Klaus Thomsen <strong>17.</strong> Riesz’ <strong>representation</strong> <strong>theorem</strong>