THE RIESZ REPRESENTATION THEOREM KLAUS THOMSEN

The note is devoted to the statement and proof of the Riesz representation theorem. The material is stolen from Walter Rudins book: ’Real and complex analysis’. I have only made slight alterations to Rudins presentation - with the intention of making the arguments more accessible to the students in the course, Analyse 2, E04. These students can judge if I have succeeded in my efforts. In the following I will occasionally refer to the book ’Real Analysis’ by H.L. Royden which was used in Analyse 1. 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) 6= 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 ) = α1 f (x1 ) + α2 f (x2 ). Then Γ is a positive linear functional on Cc (X). More general examples arise from the integration theory most of you know from Analyse 1: 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. Z Λµ (f ) =

f dµ,

X

R (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. The following statement of the theorem contains more detailed information which is very important for applications of the theorem, but the essence is covered by the previous sentence. Version: August 24, 2005. 1

2

KLAUS THOMSEN

The Riesz representation theorem for positive functionals. Theorem 0.1. Let X be a locally compact Hausdorff space, and 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) Z Λ(f ) =

f dµ

X

for every f ∈ Cc (X) and which has the following additional properties: (b) µ(K) < ∞ when K ⊆ X is compact. (c) For every E ∈ M, we have µ(E) = inf {µ(V ) : E ⊆ V, V open} . (d) The relation 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. Proof. 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 . Let’s 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. In particular, it follows that Z Z Z Z µ1 (K) = 1K dµ1 ≤ f dµ1 = f dµ2 ≤ 1V dµ2 = µ2 (V ) X

X

X

(0.1)

X

in this situation, i.e. when K is compact, V is open and K ⊆ V . Since µ2 satisfies (c) we deduce from (0.1) 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 . 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 } . (0.2) 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

THE RIESZ REPRESENTATION THEOREM

3

that µ(V ) = ∞! Note also that when V1 and V2 are both open and V1 ⊆ V2 , then it follows from (0.2) that µ(V1 ) ≤ µ(V2 ). Hence, when we set µ(E) = inf {µ(V ) : E ⊆ V open}

(0.3)

for any subset E ⊆ X, we haven’t change the definition on open sets. In other words, with (0.3) we have an extension of µ from open to arbitrary sets. 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} .

(0.4)

Let M be the subsets E of X with the property that E ∩ K ∈ MF

(0.5)

when K ⊆ X is compact. We prove that µ : M → [0, ∞] has the required properties: First of all, observe that µ is monotone, in the sense that A ⊆ B ⇒ µ(A) ≤ µ(B)

(0.6)

since {µ(V ) : B ⊆ V open} ⊆ {µ(V ) : A ⊆ V open}. Next we establish the following Observation 0.2. If E1 , E2 , E3 , . . . , are arbitrary subsets of X, ! ∞ ∞ [ X µ Ei ≤ µ(Ei ). i=1

(0.7)

i=1

Proof. We prove first that

µ (V1 ∪ V2 ) ≤ µ(V1 ) + µ(V2 )

(0.8)

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 (0.2) that µ2 (V1 ∪ V2 ) ≤ µ(V1 ) + µ(V2 ), proving (0.8). - By induction it follows that ! n n X [ µ(Vi ) (0.9) µ Vi ≤ i=1

i=1

for any finite collection V1 , V2 , . . . , Vn of open sets in X. This is then used to prove Observation 0.2 in the following way: If µ (Ei ) = ∞ for some i, the inequality (0.7) is trivial, so we may assume that µ(Ei ) < ∞ for all i. Let ǫ > 0. It follows from the definition, (0.3), that thereSare open S sets Vi , i = 1, 2, . . . , in X such that ∞ µ(Vi) ≤ µ(Ei ) + 2−i ǫ for all i. Since ∞ E ⊆ i=1 i i=1 Vi we conclude from (0.6) that ! ! ∞ ∞ [ [ µ Ei ≤ µ Vi . (0.10) i=1

i=1

4

KLAUS THOMSEN

S . . . , is an open To estimate the right-hand side, let f ≺ ∞ i=1 Vi . Then Vi , i = 1, 2,S cover of the compact set supp f so there is an n ∈ N such that f ≺ ni=1 Vi . Then ! n [ Λ(f ) ≤ µ Vi , i=1

and by use of (0.9) we find that Λ(f ) ≤

n X i=1

S∞

µ(Vi ) ≤

n X i=1

n ∞ X X
µ(Ei ) + 2−i ǫ ≤ ǫ + µ(Ei ) ≤ ǫ + µ(Ei ). i=1

i=1

S∞ Since f ≺ V was arbitrary we conclude (by using (0.2)) that µ ( i i=1 i=1 Vi ) ≤ P∞ ǫ + i=1 µ(Ei ). Since ǫ > 0 was arbitrary, we conclude that ! ∞ ∞ X [ µ(Ei ). Vi ≤ µ i=1

i=1

In combination with (0.10), this proves (0.7).



The next step is to establish Observation 0.3. If K ⊆ X is compact, then K ∈ MF , and µ(K) = inf {Λ(f ) : K ≺ f } .

(0.11)

Proof. 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 , and hence that 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 ). 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 (0.3) 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 (0.2). Since K ≺ f , we conclude that inf {Λ(f ) : K ≺ f } ≤ µ(V ) ≤ µ(K)+ǫ. Since ǫ > 0 was arbitrary, we have established (0.11). As we saw, µ(K) < ∞, so it follows from (0.6) and the compactness of K that K ∈ MF . - Observation 0.3 is proved.  Observation 0.4. Every open set satisfies (0.4). Hence V ∈ MF when V is open and µ(V ) < ∞. Proof. Let V ⊆ X be open, and let α be a real number such that α < µ(V ). By definition of µ(V ), cf. (0.2), 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 (0.3) 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 (0.6), so Observation 0.4 has been established. 

THE RIESZ REPRESENTATION THEOREM

Observation 0.5. Suppose E = members of MF . Then

S∞

i=1

5

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

µ(E) =

∞ X

µ (Ei ) .

(0.12)

i=1

If, in addition, µ(E) < ∞, then also E ∈ MF . Proof. We prove first that µ (K1 ∪ K2 ) = µ(K1 ) + µ(K2 )

(0.13)

when Ki , i = 1, 2, are compact and disjoint (i.e. K1 ∩ K2 = ∅). 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 Observation 0.3 that there is g ∈ Cc (X) such that K1 ∪ K2 ≺ g and Λ(g) ≤ µ(K1 ∪ K2 ) + ǫ. Then K1 ≺ f g and K2 ≺ (1 − f )g. Hence µ(K1 ) ≤ Λ(f g) and µ(K2 ) ≤ Λ((1 − f )g) by Observation 0.3. Since Λ is linear, we find that µ(K1 ) + µ(K2 ) ≤ Λ(f g) + 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 (0.13) follows from Observation 0.2. To prove (0.12) let ǫ > 0 and pick compact subsets Ki ⊆ E Sinsuch that µ(Ki ) ≥ −i µ(Ei ) − 2 ǫ. This is possible since Ei ∈ MF . For each n, i=1 Ki is a compact subset of E, and it follows from (0.13) that ! n n n n [ X X
X −i µ(E) ≥ µ Ki = µ (Ki ) ≥ µ (Ei ) − 2 ǫ ≥ µ (Ei ) − ǫ. (0.14) i=1

i=1

i=1

i=1

Since n and ǫ > 0 are arbitrary here, we conclude that ∞ X µ(E) ≥ µ (Ei ) .

(0.15)

i=1

Combined with Observation 0.2Sthis yields (0.12). Returning to (0.14) we conclude then that if µ(E) < ∞, KnP = ni=1 Ki is a compact subset P∞ of E such that µ(K) ≥ P ∞ n µ (E ) − ǫ. Since µ (E ) − ǫ = µ(E) − j i i=1 µ (Ei ) = µ(E) < ∞ we j=n+1 i=1 P∞ 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 (0.4) when µ(E) < ∞. Thus E ∈ MF in this case.  Observation 0.6. 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. Proof. It follows from (0.3) that there is an open set V ⊇ E such that µ(V ) ≤ µ(E) + 2ǫ and since (0.4) holds and µ(E) < ∞ there is a compact subset K ⊆ E such that µ(K) ≥ µ(E) − 2ǫ . Note that V = K ∪ (V \K), and that V \K is open. It follows from Observation 0.4 that V \K ∈ MF since µ(V \K) ≤ µ(V ) < ∞, and from Observation 0.3 that K ∈ MF , so we conclude from Observation 0.5 that µ(V ) = µ(K) + µ(V \K), which implies that µ(V \K) = µ(V ) − µ(K) ≤ µ(E) + 2ǫ − µ(E) +

ǫ 2

= ǫ.



6

KLAUS THOMSEN

Observation 0.7. If A ∈ MF , B ∈ MF , then A\B, A ∪ B, and A ∩ B belong to MF . Proof. It follows first from Observation 0.2) that µ(A ∪ B) ≤ µ(A) + µ(B) < ∞, and then from (0.6) that µ(A\B) < ∞, µ(A ∪ B) < ∞ and µ(A ∩ B) < ∞. To it remains just to show that all three sets in the statement satisfy (0.4). To this end, let ǫ > 0. By Observation 0.6 there 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 ) . Of the three last sets two are open and one is compact. By Observation 0.4 and Observation 0.3 they are all in MF and hence Observation 0.2 implies that µ(A\B) ≤ µ (V1 \K1 ) + µ (K1 \V2 ) + µ (V2 \K2 ) ≤ ǫ + µ (K1 \V2 ) + ǫ. Thus 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 Observation 0.5 that A ∪ B = (A\B) ∪ B and A ∩ B = A\ (A\B) are both in MF .  Observation 0.8. M is a σ-algebra in X which contains all Borel sets. Proof. In the following K is an arbitrary compact subset of X. Recall that K ∈ MF by Observation 0.3. 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 Observation 0.7. 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) \ (B · ∪ Bn−1 ) , n ≥ 2. Then the Bi ’s are disjoint S1 ∪ B2 ∪ · · S ∞ A ) = subsets of X and K ∩ ( ∞ n=1 Bn . It follows from Observation 0.7 that n=1 n Bi ∈ MF for all i. Since !! ∞ [ µ K∩ An ≤ µ(K) < ∞ n=1

S∞ by (0.6) and Observation 0.3, we deduce from Observation 0.5 that n=1 Bn ∈ MF . S A ∈ M since K was arbitrary. It follows that ∞ n=1 n We have shown that M is a σ-algebra. If C is a closed subset of X, the intersection C ∩ K is compact and hence an element of MF by Observation 0.3. 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. 

Observation 0.9. MF = {A ∈ M : µ(A) < ∞}. Proof. If A ∈ MF , µ(A) < ∞ by definition. Furthermore, it follows from Observation 0.3 and Observation 0.7 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 Observation 0.4, V ∈ MF , and by Observation 0.6 we can find a

THE RIESZ REPRESENTATION THEOREM

7

compact subset K ⊆ V such that µ (V \K) ≤ ǫ. 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.  Observation 0.10. µ is a measure on M. Proof. Let E1 , E2 , . . . be a sequence S of mutually disjoint P∞ elements of M. If µ (Ei ) = ∞ for some i, it is clear that µ ( ∞ E ) = ∞ = µ (Ei ). Otherwise, it foli=1 i i=1S P∞ ∞ lows from Observation 0.9 that Ei ∈ MF and hence µ ( i=1 Ei ) = i=1 µ (Ei ) by Observation 0.5.  The reader is now asked to observe that the properties of µ stipulated in (b)-(e) of the theorem all hold: (b) follows from Observation 0.3, (c) follows from the definition of µ, (d) follows from Observation 0.9 and Observation 0.4, while (e) follows from Observation 0.9 since the set A of (e) is in MF by the definition of MF . It remains to check that condition (a) holds. To this end it suffices to show that Z Λ(f ) ≤ f dµ (0.16) X

Rfor every real-valued element f of Cc (X). Indeed, itRfollows then that also Λ(−f ) ≤ RX −f dµ holds, which implies first that Λ(f ) ≥ X f dµ and then that Λ(f ) = f dµ. By linearity this yields (a). X We prove (0.16): Choose a < b in R such that f (X) ⊆ [a, b], and let 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. It follows from the definition of µ that there are open sets Wi ⊇ Ei such that µ (Wi ) < µ (Ei ) + nǫ . Set Vi = Wi ∩f −1 (] − ∞, yi + ǫ[). Then Vi is an open set such that Vi ⊇ Ei , f (x) < yi +ǫ for x ∈ Vi , and ǫ µ (Vi ) ≤ µ (Wi ) < µ (Ei ) + . (0.17) n S Note that K ⊆ ni=1 Vi . It follows from Theorem 0.18 in ’Kommentarer’ (the partition of unity theorem) that there 1] such that P[0, Pn are continuous functions hi : X → n hi ≺ Vi , i = 1, 2, . . . , n, and i=1 hi (x) = 1, x ∈ K. Then f = i=1 hi f , and it follows from Observation 0.3 that ! n n X X µ(K) ≤ Λ hi = Λ(hi ). (0.18) i=1

i=1

8

KLAUS THOMSEN

Since hi f ≤ (yi + ǫ)hi , and since yi − ǫ < f (x) on Ei , we find that n n X X Λ(f ) = Λ(hi f ) ≤ (yi + ǫ)Λ(hi ) i=1

= ≤

n X

i=1

(|a| + yi + ǫ)Λ(hi ) −

i=1 n X

(|a| + yi + ǫ)µ(Vi ) −

n X

i=1 n X

|a|Λ(hi) |a|Λ(hi )

(using (0.2))

i=1

i=1 n X

n h ǫi X (|a| + yi + ǫ) µ(Ei ) + |a|Λ(hi ) (using (0.17)) − n i=1 i=1 n i h X ǫ − |a|µ(K) (using (0.18)) (|a| + yi + ǫ) µ(Ei ) + ≤ n i=1 n n n h X X ǫi X ǫ = (yi + ǫ) µ(Ei ) + + |a| (since µ(Ei ) = µ(K)) n n i=1 i=1 i=1

≤

=

n X

(yi − ǫ)µ(Ei ) + 2ǫµ(K) +

i=1

i=1

≤ ≤

Z

Z

f dµ + 2ǫµ(K) + X

n X

n X

ǫ (|a| + yi + ǫ) n

(|a| + yi + ǫ)

i=1

f dµ + 2ǫµ(K) + ǫ (|a| + b + ǫ)

ǫ n

(since

(since

n X

µ(Ei ) = µ(K))

i=1

n X

(yi − ǫ)1Ei ≤ f )

i=1

(since yi ≤ b for all i).

X

Since ǫ > 0 was arbitrary here, this yields (0.16).