17. Riesz' representation theorem - Aarhus Universitet

17. Riesz’ representation theorem 

Klaus Thomsen matkt@imf.au.dk 

Institut for Matematiske Fag 

Det Naturvidenskabelige Fakultet 

Aarhus Universitet 

November 2005 

Klaus Thomsen 17. Riesz’ representation theorem

We read the lecture notes on Riesz’ representation theorem, mostly 

stolen from Walter Rudins book: ’Real and complex analysis’. 


The Riesz representation theorem - the setting 

Throughout the note X will be a fixed locally compact Hausdorff 

space. By Cc(X) we denote the vector space of continuous 

compactly supported functions on X, i.e. a continuous function 

f : X → C is in Cc(X) if and only if 

supp f = {x ∈ X : f (x) = 0} 

is a compact subset of X. 

A linear functional Λ : Cc(X) → C is said to be positive when 

f ∈ Cc(X), f (x) ≥ 0 ∀x ∈ X ⇒ Λ(f ) ≥ 0. 

Some simple examples of positive linear functionals on Cc(X) are 

easy to come by: For example, one may choose a couple of points 

x1,x2 ∈ X and two non-negative real numbers α1,α2, and define 

Γ : Cc(X) → C such that 

Γ(f ) = α1f (x1) + α2f (x2). 

Then Γ is a positive linear functional on Cc(X). 


The Riesz representation theorem - the setting 

More generally: Let µ be a measure defined on a σ-algebra of 

subsets of X which (at least) contains the Borel sets in X. If µ is 

finite on every compact subset of X, i.e. if µ(K) < ∞ when K ⊆ X 

is compact, then any function f ∈ Cc(X) is integrable with respect 

to µ. (This follows from Proposition 15 ii) on p. 267 in Roydens 

book since |f | ≤ M1K, where M = sup {|f (x)| : x ∈ X } < ∞ and 

1K is the characteristic function of K = supp f .) We can therefore 

define Λµ : Cc(X) → C by integration with respect to µ, i.e. 

 

Λµ(f ) = f dµ, 

(sometimes written 

X f (x) dµ(x) to emphasize the variable.) It 

follows from Proposition 15 i)+ ii) on p. 267 in Roydens book that 

Λµ is then a positive linear functional on Cc(X). 

The main content of Riesz’s representation theorem is that every 

positive linear functional on Cc(X) arises in this way. 

X 


The Riesz representation theorem for positive functionals 

Theorem 

Let X be a locally compact Hausdorff space. Let Λ : Cc(X) → C 

be a positive linear functional. 

Then there exists a σ-algebra M in X which contains all Borel sets 

in X, and there exists a unique positive measure µ on M which 

represents Λ in the sense that 

(a) 

 

Λ(f ) = f dµ for every f ∈ Cc(X), 

X 

and which has the following additional properties: 


The Riesz representation theorem for positive functionals 

Theorem 

(b) 

(c) For every E ∈ M, we have 

(d) The relation 

µ(K) < ∞ when K ⊆ X is compact. 

µ(E) = inf {µ(V) : E ⊆ V, V open} . 

µ(E) = sup {µ(K) : K ⊆ E, K compact} 

holds for every open set E, and every E ∈ M with µ(E) < ∞. 

(e) If E ∈ M, A ⊆ E, and µ(E) = 0, then A ∈ M. 


The proof - uniqueness 

The uniqueness part of the statement is that if µ1 and µ2 are both 

measures on M with the properties (a)-(d), then µ1 = µ2. 

We start by proving this. For this we introduce the following 

notation: When f ∈ Cc(X) is a function taking values in [0,1], and 

E ⊆ X we shall write 

f ≺ E, 

when supp f ⊂ E, and 

E ≺ f 

when f (x) = 1 for x ∈ E. 

It follows then from Urysohn’s lemma, Lemma 0.17 in 

’Kommentarer’ that whenever K is a compact subset of X, V an 

open subset of X, and K ⊆ V , then there is an f ∈ Cc(X) such 

that 

K ≺ f ≺ V. 


The proof - uniqueness 

In particular, it follows that 

 

µ1(K) = 1K dµ1 ≤ f dµ1 = 

X 

X 

X 

 

f dµ2 ≤ 

X 

1V dµ2 = µ2(V) 

(1) 

in this situation, i.e. when K is compact, V is open and K ⊆ V . 

Since µ2 satisfies (c) we deduce from (1) that µ1(K) ≤ µ2(K) 

when K ⊆ X is compact. 

Since µ1 and µ2 both have property (d) we conclude that 

µ1(V) ≤ µ2(V) when V ⊆ X is open. 

Since they also both have property (c) we conclude that 

µ1(E) ≤ µ2(E) for all E ∈ M. By symmetry we must also have the 

reversed inequality, and we may therefore conclude that µ1 = µ2. 


The proof - the construction 

We turn now to the construction of the measure µ. First we define 

µ(V) when V ⊆ X is open: 

µ(V) = sup {Λ(f ) : f ≺ V }. (2) 

Since a part of the conditions in ’f ≺ V ’ is that f (X) ⊆ [0,1] it 

follows that µ(V) ≥ 0 because Λ is a positive linear functional. 

Note that it may very well be that µ(V) = ∞! Note also that 

when V1 and V2 are both open and V1 ⊆ V2, then it follows from 

(2) that µ(V1) ≤ µ(V2). 

Hence, when we set 

µ(E) = inf {µ(V) : E ⊆ V } (3) 

for any subset E ⊆ X, we haven’t changed the definition on open 

sets. In other words, with (3) we have an extension of µ from open 

to arbitrary sets. 


The proof - the construction 

This extension will not in general give us a measure defined, as it 

is, on all subsets of X, and a major part of the proof is to identify a 

σ-algebra of sets in X which contains the Borel sets, and on which 

µ does give us a measure. (The trouble, of course, is the countable 

additivity of µ!) 

Let MF be the subsets E of X for which µ(E) < ∞ and 

µ(E) = sup {µ(K) : K ⊆ E compact} . (4) 

Let M be the subsets E of X with the property that 

when K ⊆ X is compact. 

E ∩ K ∈ MF 

Klaus Thomsen 17. Riesz’ representation theorem 

(5)

The proof - checking the properties of M and µ 

We prove that µ : M → [0, ∞] has the required properties: 

First of all, observe that µ is monotone, in the sense that 

since {µ(V) : B ⊆ V } ⊆ {µ(V) : A ⊆ V }. 

Next we establish the following 

A ⊆ B ⇒ µ(A) ≤ µ(B) (6) 

If E1,E2,E3,..., are arbitrary subsets of X, 

 

∞ 

 

∞ 

µ ≤ µ(Ei). 

i=1 

Ei 

i=1 


(7)

Proof of (7) 

We prove first that 

µ (V1 ∪ V2) ≤ µ(V1) + µ(V2) (8) 

when V1 and V2 are open. 

To this end, let g ≺ V1 ∪ V2. It follows then from Theorem 0.18 in 

’Kommentarer’ - the theorem on partitions of unity - that there are 

functions hi ∈ Cc(X),i = 1,2, such that h1(x) + h2(x) = 1 when 

x ∈ supp g, and gi ≺ Vi,i = 1,2. 

It follows that g = gh1 + gh2 so the additivity of Λ shows that 

Λ(g) = Λ(gh1) + Λ(gh2). 

Since ghi ≺ Vi we conclude that Λ(g) ≤ µ(V1) + µ(V2). It follows 

then from (2) that µ2(V1 ∪ V2) ≤ µ(V1) + µ(V2), proving (8). 


Proof of (7) 

By induction it follows that 

 

n 

µ 

i=1 

Vi 

 

≤ 

n 

µ(Vi) (9) 

for any finite collection V1,V2,... ,Vn of open sets in X. 

This is then used to prove (7) in the following way: If µ (Ei) = ∞ 

for some i, the inequality (7) is trivial, so we may assume that 

µ(Ei) < ∞ for all i. 

Let ǫ > 0. It follows from the definition, (3), that there are open 

sets Vi,i = 1,2,... , in X such that Ei ⊆ Vi and 

µ(Vi) ≤ µ(Ei) + 2 −i ǫ for all i. 

i=1 

Since ∞ 

i=1 Ei ⊆ ∞ 

i=1 Vi we conclude that 

µ 

∞ 

i=1 

Ei 

 

≤ µ 

∞ 

i=1 

Vi 

 

. (10) 


Proof of (7) 

To estimate the right-hand side, let f ≺ ∞ i=1 Vi. Then 

Vi,i = 1,2,... , is an open cover of the compact set suppf so 

there is an n ∈ N such that f ≺ n i=1 Vi. 

Then 

 

∞ 

 

Λ(f ) ≤ µ , 

and by use of (9) we find that 

Λ(f ) ≤ 

n 

µ(Vi) ≤ 

i=1 

i=1 

i=1 

Vi 

n 

µ(Ei) + 2 −i ǫ n 

∞ 

≤ ǫ+ µ(Ei) ≤ ǫ+ µ(Ei). 

Since f ≺ ∞ i=1 Vi was arbitrary we conclude (by using (2)) that 

µ ( ∞ i=1 Vi) ≤ ǫ + ∞ i=1 µ(Ei). 

Since ǫ > 0 was arbitrary, we conclude that 

µ ( ∞ i=1 Vi) ≤ ∞ i=1 µ(Ei). 

In combination with (10), this proves (7). 

i=1 


i=1

Proof of (11) 

The next step is to establish 

If K ⊆ X is compact, then K ∈ MF, and 

µ(K) = inf {Λ(f ) : K ≺ f }. 

(11) 

If K ≺ f , and 0 < α < 1, set Vα = {x ∈ X : f (x) > α}. 

Then K ⊆ Vα since f (x) = 1 when x ∈ K, and when g ∈ Cc(X) is 

any function such that g ≺ Vα, we have that αg ≤ f . 

Hence Λ(αg) = αΛ(g) ≤ Λ(f ) because Λ is linear and positive. 

It follows that 

µ(K) ≤ µ(Vα) = sup {Λ(g) : g ≺ Vα} ≤ α −1 Λ(f ). 


Proof of (11) 

In particular we see that µ(K) < ∞, and by letting α → 1 we 

obtain the conclusion that µ(K) ≤ Λ(f ). 

Since K ≺ f was arbitrary, we see that µ(K) ≤ inf {Λ(f ) : K ≺ f }. 

On the other hand, if we let ǫ > 0, it follows from (3) that there is 

an open set V ⊇ K such that µ(V) ≤ µ(K) + ǫ. 

By Urysohn’s lemma there is a function K ≺ f ≺ V , and for this 

function Λ(f ) ≤ µ(V) by (2). Since K ≺ f , we conclude that 

inf {Λ(f ) : K ≺ f } ≤ µ(K) + ǫ. 

Since ǫ > 0 was arbitrary, we have established the equality in (11). 

As we saw, µ(K) < ∞, so it follows from (6) and the compactness 

of K that K ∈ MF. - (11) is proved. 


Proof of (12) 

Every open set satisfies (4). 

Hence V ∈ MF when V is open and µ(V) < ∞. 

Let V ⊆ X be open, and let α be a real number such that 

α < µ(V). By definition of µ(V), cf. (2), there is an f ∈ Cc(X) 

such that f ≺ V and Λ(f ) > α. 

Then K = supp f is a compact subset of V , and when W is an 

open subset containing K, viz. K ⊆ W, then f ≺ W and hence 

µ(W) ≥ Λ(f ). 

It follows from (3) that µ(K) ≥ Λ(f ) > α. Since α < µ(V) was 

arbitrary, we conclude that that sup {µ(K) : K ⊆ V } ≥ µ(V). 

The reversed inequality is trivial, thanks to (6), so (12) has been 

established. 


(12)

Proof of (13) 

Suppose E = 

We prove first that 

∞ 

Ei, where E1,E2,E3,... , are 

i=1 

pairwise disjoint members of MF. Then 

∞ 

µ(E) = µ (Ei). 

i=1 

If, in addition, µ(E) < ∞, then also E ∈ MF. 

(13) 

µ (K1 ∪ K2) = µ(K1) + µ(K2) (14) 

when Ki,i = 1,2, are compact and disjoint. 


Proof of (13) 

By Urysohn’s lemma there is an f ∈ Cc(X) such that 0 ≤ f ≤ 1, 

f (x) = 1,x ∈ K1, and f (x) = 0 when x ∈ K2. 

Let ǫ > 0. It follows from (11) that there is g ∈ Cc(X) such that 

K1 ∪ K2 ≺ g and Λ(g) ≤ µ(K1 ∪ K2) + ǫ. 

Then K1 ≺ fg and K2 ≺ (1 − f )g. Hence µ(K1) ≤ Λ(fg) and 

µ(K2) ≤ Λ((1 − f )g) by (11). 

Since Λ is linear, we find that 

µ(K1) + µ(K2) ≤ Λ(fg) + Λ((1 − f )g) = Λ(g) ≤ µ(K1 ∪ K2) + ǫ. 

Since ǫ > 0 was arbitrary we conclude that 

µ(K1) + µ(K2) ≤ µ(K1 ∪ K2), and then (14) follows from (7). 


Proof of (13) 

To prove the equality in (13) we pick compact subsets Ki ⊆ Ei such 

that µ(Ki) ≥ µ(Ei) − 2−iǫ. This is possible since Ei ∈ MF. 

For each n, n i=1 Ki is a compact subset of E, and it follows from 

(14) that 

 

n 

 

n n 

µ(E) ≥ µ = µ (Ki) ≥ µ (Ei)−2 −i n 

ǫ ≥ µ (Ei)−ǫ. 

i=1 

Ki 

i=1 

i=1 

Since n and ǫ > 0 are arbitrary here, we conclude that 

µ(E) ≥ 

i=1 

(15) 

∞ 

µ (Ei) . (16) 

i=1 

Combined with (7) this yields the equality in (13). 


Proof of (13) 

Returning to (15) we conclude then that if µ(E) < ∞, 

Kn = n i=1 Ki is a compact subset of E such that 

µ(K) ≥ n i=1 µ (Ei) − ǫ = µ(E) − ∞ j=n+1 µ (Ej) − ǫ. 

Since ∞ i=1 µ (Ei) = µ(E) < ∞ we have that ∞ j=n+1 µ (Ej) < ǫ if 

n is large enough. So for n large, Kn is a compact subset of E such 

that µ (Kn) ≥ µ(E) − 2ǫ. 

It follows that E satisfies (4) when µ(E) < ∞. Thus E ∈ MF in 

this case. 


Proof of (17) 

If E ∈ MF, and ǫ > 0, there is a compact subset K ⊆ X 

and an open subset V ⊆ X such that K ⊆ E ⊆ V , and µ(V \K) ≤ ǫ. 

(17) 

It follows from (3) that there is an open set V ⊇ E such that 

µ(V) ≤ µ(E) + ǫ 

2 

and since (4) holds and µ(E) < ∞ there is a compact subset 

K ⊆ E such that µ(K) ≥ µ(E) − ǫ 

2 . 

Note that V = K ∪ (V \K), and that V \K is open. It follows from 

(12) that V \K ∈ MF since µ(V \K) ≤ µ(V) < ∞, 

and from (11) that K ∈ MF, so we conclude from (13) that 

µ(V) = µ(K) + µ(V \K), 

which implies that 

µ(V \K) = µ(V) − µ(K) ≤ µ(E) + ǫ ǫ 

2 − µ(E) + 2 = ǫ. 


Proof of (18) 

If A ∈ MF, B ∈ MF, then A\B, A ∪ B, and A ∩ B belong to MF. 

(18) 

Since µ(A ∪ B) ≤ µ(A) + µ(B) < ∞ (using (7)), it follows from 

(6) that µ(A\B) < ∞, µ(A ∪ B) < ∞ and µ(A ∩ B) < ∞. 

So it remains just to show that all three sets in the statement 

satisfy (4). 

To this end, let ǫ > 0. By (17) there are compact sets Ki,i = 1,2, 

and open sets Vi,i = 1,2, such that K1 ⊆ A ⊆ V1, K2 ⊆ B ⊆ V2, 

and µ(Vi\Ki) ≤ ǫ,i = 1,2. 

Note that 

A\B ⊆ V1\K2 ⊆ (V1\K1) ∪ (K1\V2) ∪ (V2\K2) . 


Proof of (18) 

Of the three last sets two are open and one is compact. (12) and 

(11) they are all in MF and hence (7) implies that 

µ(A\B) ≤ µ (V1\K1)+µ (K1\V2)+µ (V2\K2) ≤ ǫ+µ (K1\V2)+ǫ. 

Then K1\V2 is a compact subset of A\B such that 

µ (K1\V2) ≥ µ(A\B) − 2ǫ. 

Since ǫ > 0 was arbitrary, this shows that A\B ∈ MF. Once this is 

established it follows from (13) that A ∪ B = (A\B) ∪ B and 

A ∩ B = A\(A\B) are both in MF. 


Proof of (19) 

M is a σ-algebra in X which contains all Borel sets. (19) 

In the following K is an arbitrary compact subset of X. Recall that 

K ∈ MF by Observation 11. 

Let A ∈ M, and consider the complement Ac = X \A. By definition 

of M, A ∩ K ∈ MF, and Ac ∩ K = K\(A ∩ K) is therefore the 

set-theoretic difference of two sets from MF. Hence Ac ∩ K ∈ MF 

by (18). This shows that Ac ∈ M since K was arbitrary. 

Consider then a sequence Ai,i = 1,2,3,... , of sets from M. Put 

B1 = A1 ∩ K, and Bn = (An ∩ K) \(B1 ∪ B2 ∪ · · · ∪ Bn−1),n ≥ 2. 

Then the Bi’s are disjoint subsets of X and 

K ∩ ( ∞ n=1 An) = ∞ n=1 Bn. It follows from (18) that Bi ∈ MF for 

all i. 

Since (K ∩ ( ∞ n=1 An)) ≤ µ(K) < ∞ by (6) and (11), we deduce 

from (13) that ∞ n=1 Bn ∈ MF. It follows that ∞ n=1 An ∈ M since 

K was arbitrary. 

It follows that M is a σ-algebra. 


End of proof of (19), beginning of proof of (20) 

If C is a closed subset of X, the intersection C ∩ K is compact and 

hence an element of MF by Observation 11. Thus C ∈ M since K 

was arbitrary, and we see that M contains all closed subsets of X. 

Being a σ-algebra it must therefore contain all Borel subsets of X. 

MF = {A ∈ M : µ(A) < ∞} . (20) 

If A ∈ MF, µ(A) < ∞ by definition. Furthermore, it follows from 

(11) and (18) that A ∩ K ∈ MF for every compact subset K. 

Hence A ∈ M. This proves one of the desired inclusions. 

To prove the other, assume that A ∈ M and that µ(A) < ∞. To 

prove that A ∈ MF, let ǫ > 0. 

By definition of µ there is an open set V ⊇ A such that 

µ(V) < ∞. By (12) V ∈ MF, and by (17) we can find a compact 

subset K ⊆ V such that µ (V \K) ≤ ǫ. 


Proof of (20) 

Since A ∩ K ∈ MF, there is a compact subset H ⊆ A ∩ K such that 

µ(H) ≥ µ(A ∩ K) − ǫ. 

Since A ⊆ (A ∩ K) ∪ (V \K), it follows that 

µ(A) ≤ µ(A ∩ K) + µ (V \K) ≤ µ(H) + ǫ + µ (V \K) ≤ µ(H) + 2ǫ. 

Since ǫ > 0 was arbitrary, we see from this that A ∈ MF, as 

desired. 


Proof of (21) 

µ is a measure on M. (21) 

Let E1,E2,... be a sequence of mutually disjoint elements of M. 

If µ (Ei) = ∞ for some i, it is clear that 

µ ( ∞ i=1 Ei) = ∞ = ∞ i=1 µ (Ei). 

Otherwise, it follows from (20) that Ei ∈ MF and hence 

µ ( ∞ 

i=1 Ei) = ∞ 

i=1 µ (Ei) by (13). 

The reader is now asked to observe that the properties of µ 

stipulated in (b)-(e) all hold: (b) follows from (11), (c) follows from 

the definition of µ, (d) follows from (20) and (12), while (e) follows 

from (20) since the set A of (e) is in MF be definition of MF. 


Proof of a) 

It remains to check that condition (a) holds. To this end it suffices 

to check that 

 

Λ(f ) ≤ f dµ (22) 

for every real-valued element of Cc(X). 

Indeed, it follows then that also Λ(−f ) ≤ 

X −f dµ holds, which 

implies first that Λ(f ) ≥ 

 

X f dµ and then that Λ(f ) = X f dµ. 

By linearity this yields (a). 

We prove (22): Choose a 

ǫ > 0. Choose y0 < y1 < y2 < · · · < yn such that yi − yi−1 < ǫ for 

all i, y0 < a and yn = b. Put 

Ei = {x ∈ R : yi−1 < f (x) ≤ yi} ∩ K, 

i = 1,2,... ,n, where K = supp f . 

Since f is continuous and hence Borel measurable, the Ei’s are 

Borel sets. They are mutually disjoint. 

X 


Proof of a) 

It follows from the definition of µ that there are open sets Wi ⊇ Ei 

such that µ (Wi) < µ (Ei) + ǫ 

n . Set Vi = Wi ∩ f −1 (] − ∞,yi + ǫ[. 

Then Vi is a an open set such that Vi ⊇ Ei,f (x) < yi + ǫ,x ∈ Vi, 

and 

µ (Vi) ≤ µ (Wi) < µ (Ei) + ǫ 

. (23) 

n 

Note that K ⊆ n i=1 Vi. It follows from Theorem 0.18 in 

’Kommentarer’ that there are continuous functions hi : X → [0,1] 

such that hi ≺ Vi,i = 1,2,... ,n, and n i=1 hi(x) = 1,x ∈ K. 

Then f = n i=1 hif , and it follows from (19) that 


Proof of a) 

µ(K) ≤ Λ 

n 

i=1 

hi 

 

= 

n 

Λ(hi). (24) 

Since hif ≤ (yi + ǫ)hi, and since yi − ǫ < f (x) on Ei, we find that 

Λ(f ) = 

= 

≤ 

n 

Λ(hif ) ≤ 

i=1 

n 

(yi + ǫ)Λ(hi) 

i=1 

n 

(|a| + yi + ǫ)Λ(hi) − 

i=1 

n 

(|a| + yi + ǫ)µ(Vi) − 

i=1 

i=1 

n 

|a|Λ(hi) 

i=1 

n 

|a|Λ(hi) (using (2)) 

i=1 


(25)

Proof of a) 

n 

≤ (|a| + yi + ǫ) µ(Ei) + 

i=1 

ǫ 

n 

− |a|Λ(hi) 

n 

i=1 

(using (23)) 

n 

≤ (|a| + yi + ǫ) µ(Ei) + 

i=1 

ǫ 

 

− |a|µ(K) 

n 

(using (24)) 

n 

= (yi + ǫ) µ(Ei) + ǫ 

n 

+ |a| 

n 

ǫ 

n 

n 

(since µ(Ei) = µ(K)) 

= 

i=1 

i=1 

n 

(yi − ǫ)µ(Ei) + 2ǫµ(K) + 

i=1 

(since 

n 

i=1 

i=1 

(|a| + yi + ǫ) ǫ 

n 

n 

µ(Ei) = µ(K)) 

i=1 


(26)

Proof of a) 

 

≤ 

 

≤ 

X 

X 

f dµ + 2ǫµ(K) + 

n 

i=1 

(|a| + yi + ǫ) ǫ 

n 

(since 

n 

(yi − ǫ)1Ei ≤ f ) 

i=1 

f dµ + 2ǫµ(K) + ǫ(|a| + b + ǫ) (since yi ≤ b for all i). 


(27)

Proof of a) 

Since ǫ > 0 was arbitrary here, this yields (22).

17. Riesz' representation theorem - Aarhus Universitet

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