Chapter 13
Radon Measures
Recall that if X is a compact metric space, C(X), the space of continuous
(real-valued) functions on X, is a Banach space with the norm
kf k = sup |f (x)|.
(13.1)
x∈X
We want to identify the dual of C(X) with the space of (finite) signed Borel
measures on X, also known as the space of Radon measures on X.
Before identifying the dual of C(X), we will first identify the set of
positive linear functionals on C(X). By definition, a linear functional
α : C(X) −→ R
(13.2)
is positive provided
(13.3)
f ∈ C(X), f ≥ 0 =⇒ α(f ) ≥ 0.
Clearly, if µ is a (positive) finite Borel measure on X, then
Z
(13.4)
α(f ) = f dµ
is a positive linear functional. We will establish the converse, that every
positive linear functional on C(X) is of the form (13.4).
It is easy to see that every positive linear functional α on C(X) is
bounded. In fact, applying α to f (x) − a and to b − f (x), we see that,
when a and b are real numbers
(13.5)
f ∈ C(X), a ≤ f ≤ b =⇒ aα(1) ≤ α(f ) ≤ bα(1),
179
180
13. Radon Measures
so
|α(f )| ≤ Akf k,
(13.6)
A = α(1).
To begin the construction of µ, we construct a set function µ 0 on the
collection O of open subsets of X by
µ0 (U ) = sup {α(f ) : f ≺ U },
(13.7)
where we say
(13.8)
f ≺ U ⇐⇒ f ∈ C(X), 0 ≤ f ≤ 1, and supp f ⊂ U.
Here, supp f is the closure of {x : f (x) 6= 0}. Clearly µ 0 is monotone. Then
we set, for any E ⊂ X,
(13.9)
µ∗ (E) = inf {µ0 (U ) : E ⊂ U ∈ O}.
Of course, µ∗ (U ) = µ0 (U ) when U is open.
Lemma 13.1. The set function µ∗ is an outer measure.
Proof. By Proposition 5.1, it suffices to show that
[
X
(13.10)
Uj ∈ O, U =
Uj =⇒ µ0 (U ) ≤
µ0 (Uj ),
j≥1
j≥1
so that we have
P an analogue of (5.5). Suppose f ≺ U. We need to show
that α(f ) ≤
µ0 (Uj ). Now, since supp f = K is compact, we have K ⊂
U1 ∪ P
· · · ∪ U` for some finite `. We claim there are g j ≺ Uj , 1 ≤ j ≤ `, such
that
gj = 1 on K. Granted this, we can set fj = f gj . Then fj ≺ Uj , so
α(fj ) ≤ µ0 (Uj ). Hence
X
X
(13.11)
α(f ) =
α(fj ) ≤
µ0 (Uj ),
as desired. Thus Lemma 13.1 will be proved once we have
Lemma 13.2. If K ⊂ X is compact,
P Uj are open, and K ⊂ U1 ∪· · ·∪U` = V,
then there exist gj ≺ Uj such that
gj = 1 on K.
Proof. Set U`+1 = X \ K. Then {Uj : 1 ≤ j ≤ ` + 1} is an open cover of X.
Let {gj : 1 ≤ j ≤ ` + 1} be a partition of unity subordinate to this cover.
(See Exercise 9 at the end of this chapter.) Then {g j : 1 ≤ j ≤ `} has the
desired properties.
Now that we know µ∗ is an outer measure, we prepare to apply Caratheodory’s Theorem.
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13. Radon Measures
Lemma 13.3. The outer measure µ∗ is a metric outer measure.
Proof. Let Sj ⊂ X, and assume
dist(S1 , S2 ) ≥ 4ε > 0.
(13.12)
Take δ > 0. Given U ⊃ S = S1 ∪ S2 , U open, such that µ0 (U ) ≤ µ∗ (S) + δ,
set
Uj = U ∩ {x ∈ X : dist(x, Sj ) < ε}.
It follows that
(13.13)
S j ⊂ Uj ,
U1 ∪ U2 ⊂ U,
U1 ∩ U2 = ∅.
Now, whenever f ≺ U1 ∪ U2 , we have f = f1 + f2 with fj = f U ≺ Uj ,
j
which implies
µ0 (U1 ) + µ0 (U2 ) = µ0 (U1 ∪ U2 ).
Hence
(13.14)
µ∗ (S1 ) + µ∗ (S2 ) ≤ µ0 (U1 ) + µ0 (U2 ) = µ0 (U1 ∪ U2 ) ≤ µ0 (U ).
Thus µ∗ (S) ≥ µ∗ (S1 ) + µ∗ (S2 ) − δ, for all δ > 0, which, together with
subadditivity of µ∗ , yields the desired identity µ∗ (S) = µ∗ (S1 ) + µ∗ (S2 ).
It follows from Proposition 5.8 that every closed set in X is µ ∗ -measurable.
Hence, by Theorem 5.2, every Borel set in X is µ ∗ -measurable, and the restriction of µ∗ to B(X), which we denote µ, is a measure.
We make a few useful comments about µ. First, in addition to (13.7),
we have
µ(U ) = sup {α(f ) : f 4 U },
(13.15)
for U open, where
(13.16)
f 4 U ⇐⇒ f ∈ C(X), 0 ≤ f ≤ 1, f = 0 on X \ U.
To see that (13.7) and (13.15) coincide, take f 4 U. Then set f j = ξj (f ),
where ξj (s) is defined by
0
ξj (s) = 2s −
s
2
j
1
for 0 ≤ s ≤ ,
j
1
2
for
≤s≤ ,
j
j
2
for s ≥ .
j
It follows that fj ≺ U and fj % f (uniformly); hence α(fj ) % α(f ).
Using the identity (13.15), we can establish the following.
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13. Radon Measures
Lemma 13.4. If K ⊂ X is compact, then
µ(K) = inf {α(f ) : f ∈ C(X), f ≥ χ K }.
(13.17)
Proof. Denote the right side of (13.17) by µ 1 (K). It suffices to take the inf
of α(f ) over f < K, where
f < K ⇐⇒ f ∈ C(X), 0 ≤ f ≤ 1, f = 1 on K.
(13.18)
Comparing this with (13.16), we see that f < K ⇔ 1 − f 4 X \ K, so
µ1 (K) = inf {α(1) − α(g) : g 4 X \ K}
(13.19)
= µ(X) − µ(X \ K).
On the other hand, since K is µ∗ -measurable, µ(X \ K) + µ(K) = µ(X), so
the identity (13.17) is proved.
We are now ready to prove
Theorem 13.5. If X is a compact metric space and α is a positive linear
functional on C(X), then there exists a unique finite, positive Borel measure
µ such that
Z
(13.20)
α(f ) = f dµ
for all f ∈ C(X).
Proof. We have constructed a positive Borel measure µ, which is finite since
(13.7) implies µ(X) = α(1). We next show that (13.20) holds. It suffices to
check this when f : X → [0, 1].
In such a case, take N ∈ Z+ and define
(13.21)
We have
(13.22)
P
ϕj (x) = min (f (x), jN −1 ),
0 ≤ j ≤ N,
fj (x) = ϕj (x) − ϕj−1 (x),
1 ≤ j ≤ N.
fj = f and
1
1
χKj ≤ fj ≤ χKj−1 ,
N
N
n
jo
.
Kj = x ∈ X : f (x) ≥
N
Hence
(13.23)
1
µ(Kj ) ≤
N
Z
fj dµ ≤
1
µ(Kj−1 ).
N
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13. Radon Measures
We claim that also
1
1
µ(Kj ) ≤ α(fj ) ≤ µ(Kj−1 ).
N
N
(13.24)
To see this, first note that if Kj−1 ⊂ U is open, then N fj ≺ U, so N α(fj ) ≤
µ0 (U ). This implies the second inequality of (13.24). On the other hand,
N fj < Kj , so Lemma 13.4 gives the first inequality of (13.24).
Summing (13.23) and (13.24), we have
(13.25)
Z
N
N −1
1 X
1 X
µ(Kj ) ≤ f dµ ≤
µ(Kj ),
N
N
j=1
1
N
N
X
j=0
µ(Kj ) ≤ α(f ) ≤
j=1
1
N
N
−1
X
µ(Kj ).
j=0
Hence
(13.26)
Z
1
1
1
α(f ) − f dµ ≤ µ(K0 ) − µ(KN ) ≤ µ(X).
N
N
N
Letting N → ∞, we have (13.20).
Only the uniqueness of µ remains to be proved. To see this, let λ be a
positive Borel measure on X such that
Z
(13.27)
α(f ) = f dλ
for all f ∈ C(X). Let K ⊂ X be compact, and apply this to
−1
fν & χK .
R
R
By
Convergence Theorem, we have fν dµ & χK dµ and
R the Monotone
R
fν dλ & χK dλ, so α(fν ) & µ(K) and α(fν ) & λ(K). Hence µ(K) =
λ(K) for all compact K. Now, by (5.60), for every positive Borel measure
λ on a compact metric space X, we have
(13.28)
(13.29)
fν (x) = 1 + ν dist(x, K)
;
E ∈ B(X) =⇒ λ(E) = sup {λ(K) : K ⊂ E, K compact}.
This proves uniqueness.
Generally, if X is a compact Hausdorff space and λ is a finite (positive)
measure on B(X), λ is said to be regular if and only if (13.29) holds. The
implication of Exercises 10–13 of Chapter 5 is that every finite Borel measure is regular when X is a compact metric space. If X is compact but not
184
13. Radon Measures
metrizable, a finite measure on B(X) need not be regular. The generalization of Theorem 13.5 to this case is that, given a positive linear functional α
on C(X), there is a unique finite regular Borel measure µ such that (13.20)
holds. (Note that if λ is any finite measure on B(X), then (13.27) defines a
positive linear functional on C(X), which then gives rise to a regular Borel
measure.) For this more general case, the construction of µ ∗ is the same
as was done above in (13.7)–(13.9), but the proof that µ ∗ yields a regular
measure on B(X) is a little more elaborate than the proof given above for
compact metric spaces. Treatments can be found in [Fol] and [Ru].
We want to extend Theorem 13.5 to the case of a general bounded linear
functional
ω : C(X) → R.
(13.30)
We start with an analogue of the Hahn decomposition.
Lemma 13.6. If ω is a bounded (real) linear functional on C(X), then
there are positive linear functionals α ± on C(X) such that
ω = α + − α− .
(13.31)
Proof. We first define α+ on
C + (X) = {f ∈ C(X) : f ≥ 0}.
(13.32)
For f ∈ C + (X), set
(13.33)
α+ (f ) = sup {ω(g) : g ∈ C + (X), 0 ≤ g ≤ f }.
The hypothesis that ω is bounded implies |ω(g)| ≤ Kkgk ≤ Kkf k when
0 ≤ g ≤ f, so
(13.34)
0 ≤ α+ (f ) ≤ Kkf k, for f ∈ C + (X),
where K = kωk. Clearly, for c ∈ R,
(13.35)
f ∈ C + (X), c > 0 =⇒ α+ (cf ) = cα+ (f ).
Now, suppose f1 , f2 ∈ C + (X). If g ∈ C + (X) and 0 ≤ g ≤ f1 + f2 , we can
write g = g1 +g2 with g ∈ C + (X) and 0 ≤ gj ≤ fj . Just take g1 = min g, f1 .
Hence
(13.36)
f1 , f2 ∈ C + (X) =⇒ α+ (f1 + f2 ) = α+ (f1 ) + α+ (f2 ).
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13. Radon Measures
We claim that α+ has an extension to a linear functional on C(X), which
would necessarily be positive. In fact, given f ∈ C(X), write
(13.37)
f = f 1 − f2 ,
fj ∈ C + (X),
α+ (f ) = α+ (f1 ) − α+ (f2 ).
A given f ∈ C(X) has many such representations as f = f 1 − f2 ; showing
that α+ (f ) is independent of such a representation and defines a linear
functional on C(X) is a simple application of (13.36); compare the proof of
Proposition 3.7. Note that, if we take f 1 = f + = max(f, 0) and f2 = f − =
max(−f, 0), we see that
(13.38)
|α+ (f )| ≤ K max kf + k, kf − k ≤ Kkf k.
Finally we set α− = α+ − ω. It remains only to show that f ∈ C + (X) ⇒
α− (f ) ≥ 0, i.e., α+ (f ) ≥ ω(f ). But that is immediate from the definition
(13.33), so the lemma is proved.
We can combine Lemma 13.6 and Theorem 13.5 to prove the following,
known as the Riesz Representation Theorem.
Theorem 13.7. If X is a compact metric space and ω is a bounded (real)
linear functional on C(X), then there is a unique finite signed measure ρ on
B(X) such that
(13.39)
ω(f ) =
Z
f dρ,
X
for all f ∈ C(X). Furthermore,
(13.40)
kρk = |ρ|(X) = kωk,
so there is an isometric isomorphism
(13.41)
C(X)0 ≈ M(X).
Here, M(X) denotes the linear space of finite signed measures on B(X),
with norm given by the first identity in (13.40). This is also known as the
space of finite Radon measures on X.
For the proof, write ω = α+ − α−R, as in (13.31), take finite positive
measures µ± on B(X) so that α± (f ) = f dµ± , and set ρ = µ+ − µ− . Thus
we have the identity (13.39).
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13. Radon Measures
We need to prove (13.40). Let ρ = ρ+ − ρ− be the Hahn decomposition
of ρ, so ρ+ ⊥ ρ− and
Z
Z
+
(13.42)
ω(f ) = f dρ − f dρ− ,
for all f ∈ C(X). Consequently
(13.43)
|ω(f )| ≤ kf kρ+ (X) + kf kρ− (X) = kρk · kf k,
so we have kωk ≤ kρk. To prove the reverse inequality, let δ > 0. Suppose ρ ±
are supported on disjoint Borel sets X ± . Let K± be compact sets in X± such
that ρ± (K± ) ≥ ρ± (X) − δ. We have K+ ∩ K− = ∅, so say dist(K+ , K− ) =
ε > 0. Let U± = {x : dist(x, K± ) < ε/4}, so U+ ∩ U− = ∅. Using a simple
±
variant of (13.28), we can construct ϕ ±
ν ≺ U± such that ϕν & χK± . Hence,
as ν → ∞,
−
+
−
(13.44)
ω ϕ+
ν − ϕν → ρ (K+ ) + ρ (K− ) ≥ kρk − 2δ,
−
while kϕ+
ν − ϕν k = 1. This proves the reverse inequality, kρk ≤ kωk, and
establishes (13.40). This inequality also establishes uniqueness, so Theorem
13.7 is proved.
Though it is not needed to prove Theorem 13.7, we mention that µ + −µ−
is the Hahn decomposition of ρ (provided α + is given by (13.33)). The proof
is an exercise.
Applying Alaoglu’s Theorem (Proposition 9.2 and Corollary 9.3) to V =
C(X), we have the following compactness result:
Proposition 13.8. If X is a compact metric space, the closed unit ball in
M(X) is a compact metrizable space, with the weak ∗ topology.
By the definition of weak∗ topology given in Chapter 9, we see that a
sequence µν in M(X) converges weak∗ to µ if and only if
Z
Z
(13.45)
f dµν −→ f dµ,
for each f ∈ C(X). This topology on M(X) is also called the vague topology;
one says µν → µ vaguely provided (13.45) holds for all f ∈ C(X).
There is an important subset of M(X), the set of probability measures.
An element µ of M(X) is called a probability measure if and only if µ is
positive and µ(X) = 1. We use the notation
(13.46)
Prob(X) = {µ ∈ M(X) : µ ≥ 0, µ(X) = 1}.
It is easy to see that
(13.47)
Prob(X) = {µ ∈ M(X) : kµk ≤ 1, µ(X) = 1}.
Hence Prob(X) is a subset of the unit ball of M(X) which is closed in the
weak∗ topology. This has the following useful consequence.
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13. Radon Measures
Corollary 13.9. If X is a compact metric space, the set Prob(X) of probability measures on X is compact (and metrizable) in the weak ∗ topology.
It is often useful to consider Borel measures on a locally compact space
Y, i.e., a Hausdorff space with the property that any y ∈ Y has a compact
neighborhood. In such a case, there is a Banach space C ∗ (Y ), the space of
continuous functions on Y that “vanish at infinity.” We say a continuous
function u : Y → R belongs to C∗ (Y ) if, for any δ > 0, there exists a compact
K ⊂ Y such that |u(y)| < δ for y ∈ Y \ K. We use the sup norm on C ∗ (Y ).
We can construct the “one point compactification” Yb = Y ∪ {∞}, declaring
a set U ⊂ Yb to be open if either U ⊂ Y is open or ∞ ∈ U and Yb \ U = K
is a compact subset of Y. It readily follows that Yb is a compact Hausdorff
space. Also, C∗ (Y ) is naturally isomorphic to a closed linear subspace of
C(Yb ) :
(13.48)
C∗ (Y ) ≈ {u ∈ C(Yb ) : u(∞) = 0}.
Furthermore, given any f ∈ C(Yb ), if we set a = f (∞), then f = g +
a, g(∞) = 0, so
(13.49)
and, for the duals, we have
(13.50)
C(Yb ) ≈ C∗ (Y ) ⊕ R,
C(Yb )0 ≈ C∗ (Y )0 ⊕ R.
In case Y has the additional properties of being metrizable and σcompact, the one-point compactification Yb is also metrizable; cf. Exercise
15. Hence we can appeal to Theorem 13.7 to identify C( Yb )0 with M(Yb ),
the space of finite signed measures on B( Yb ). (Even without these additional
hypotheses, C(Yb )0 can be identified with the space of signed finite regular
Borel measures on Yb , though one must go to another source, such as [Fol]
or [Ru], for a proof.) In the decomposition (13.50), we see that the last
factor on the right consists of multiples of δ ∞ , the measure defined by
δ∞ (S) = 1 if ∞ ∈ S,
0 otherwise.
Consequently we have the identification
(13.51)
C∗ (Y )0 ≈ {µ ∈ M(Yb ) : µ({∞}) = 0} ≈ M(Y ),
where M(Y ) is the space of finite signed measures on B(Y ).
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13. Radon Measures
Exercises
In Exercises 1–3, X is a compact metric space and F : X → X is a continuous map. As in (7.32), we set
(13.52)
F∗ µ(S) = µF −1 (S) = µ F −1 (S) ,
for a Borel measure µ on X. We set F n = F ◦· · · ◦F (n factors) and similarly
define F∗n µ.
1. Define T : C(X) → C(X) by T u(x) = u(F (x)). Show that, under
the identification (13.41), the adjoint T 0 : C(X)0 → C(X)0 is given by
(13.52).
2. Assume µ is a probability measure on X. Set
n
(13.53)
µj = F∗j µ,
νn =
1 X
µj .
n+1
j=0
Show that νn ∈ Prob(X) and
F∗ νn = ν n +
1
µn+1 − µ0 .
n+1
3. Suppose νnj → ν in the weak∗ topology on Prob(X), as j → ∞. Show
that
(13.54)
F∗ ν = ν.
We say ν is an invariant measure for F .
Hint. Given u ∈ C(X), one has
Z
Z
Z
Z
Z
1 u ◦ F dνnj = u dF∗ νnj = u dνnj +
u dµnj +1 − u dµ0 .
nj + 1
4. Let f ∈ L∞ (R). Show that the following are equivalent:
(a) h−1 (τh f − f ) is bounded in L1 (R), for h ∈ (0, 1].
(b) ∂1 f = µR for some finite signed measure µ on R.
x
(c) f (x) = −∞ dµ a.e. on R.
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13. Radon Measures
5. Let f0 (x) denote the right side of (c) above, defined as in Exercise 1 of
Chapter 11. Show that f0 is the unique right continuous function on R
equal to f a.e.
6. Let f : R → R be bounded and right-continuous. Show that f has the
properties (a)–(c) of Exercise 4 if and only if there exists C < ∞ such
that, for any finite set of real numbers x 0 < x1 < · · · < x` ,
X̀
(13.55)
|f (xj ) − f (xj−1 )| ≤ C.
j=1
Hint. To prove one implication, given ν ∈ Z + , set fν (x) = f (ν) if
x ≥ ν, fν (x) = f (−ν) if x ≤ −ν, and
fν (x) = f (−ν + 2−ν j) if − ν + 2−ν j ≤ x < −ν + 2−ν (j + 1),
0 ≤ j < 2ν2ν .
Show that fν → f and {∂1 fν : ν ∈ Z+ } is a bounded set of measures on
R+ . Consider weak∗ limits. One says that f has bounded variation on
R if this property holds, and one writes f ∈ BV(R).
7. Let f ∈ L∞ (R). Show that f is equal a.e. to an element of BV(R) if and
only if you can write f = g1 − g2 a.e., with gj ∈ L∞ (R) monotone %. In
particular, if g ∈ L∞ (R) is monotone %, then ∂1 g is a positive measure.
Reconsider Exercise 1 of Chapter 5, on Lebesgue-Stieltjes measures, in
this light.
Hint. Given f ∈ BV(R), apply the Hahn decomposition to the signed
measure µ arising in Exerise 4.
8. Let f : R → R be bounded and continuous. One says f is absolutely continuous provided that, for every ε > 0, there is a δ > 0 with the property
that, for any finite collection of disjoint intervals, (a 1 , b1 ), . . . , (aN , bN ),
(13.56)
X
(bj − aj ) < δ =⇒
X
|f (bj ) − f (aj )| < ε.
Show that the following are equivalent:
(a) f is absolutely continuous.
(b) ∂1 f = gR ∈ L1 (R).
x
(c) f (x) = −∞ g(y) dy, g ∈ L1 (R).
Hint. If (a) holds, first show that f has bounded variation and use
Exercises 5–6 to get ∂1 f = µ. Then show µ is absolutely continuous
with respect to Lebesgue measure. Compare Exercises 7–9 in Chapter
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13. Radon Measures
10 and also Exercise 1 in Chapter 11.
9. Let {U1 , . . . , U`+1 } be an open cover of a compact metric space X. Show
that there exist open sets Vj , j = 1, . . . , ` + 1, covering X, such that
V j ⊂ Uj . Let
hj (x) = dist(x, X \ Vj ).
P
Show that hj ∈ C(X), supp Hj ⊂ V j ⊂ Uj , and that h = `+1
j=1 hj ∈
C(X) is > 0 on X. Deduce that
gj (x) = hj (x)/h(x)
form a partition of unity of X, subordinate to the cover {U
P1 , . . . , U`+1 }.
Hint. Set Hj (x) = dist(x, X \ Uj ). Show that H =
Hj > 0 on
X, H ∈ C(X), hence H ≥ a > 0. Set Vj = {x ∈ Uj : Hj (x) > a/(`+2)}.
Q
10. The countable infinite product Z = j≥1 {0, 1} is compact, with the
product topology, and metrizable (cf. Appendix A). Let A ⊂ C(Z)
consist of continuous functions depending on only finitely many variables, so an element f ∈ A has the form f (x) = f (x 1 , . . . , xk ), for some
k ∈ Z+ . For such an f , set
X
ϕ(f ) = 2−k
f (x1 , . . . , xk ).
xj ∈{0,1},1≤j≤k
Show that ϕ has a unique extension to a Rpositive linear functional α on
C(Z). Show that, for f ∈ C(Z), α(f ) = Z f dµ, where µ is the product
measure on Z, discussed in Exercises 7–10 of Chapter 6.
11. Generalize Exercise 10 to other countable products of compact metric
spaces, carrying positive Radon measures of total mass one.
12. Let R = I1 ×· · ·×In be a compact cell in Rn , with Iν = [aν , bν ]. Consider
the positive linear functional on C(R) given by f 7→ I(f ), where I(f )
is the Riemann integral of f , discussed in Chapter 1 (for n = 1) and in
the second exercise set at the end of Chapter 7 (for n > 1). Show that
the measure on R produced by Theorem 13.5 coincides with Lebesgue
measure (on the Borel subsets of R), as constructed in Chapter 2 (for
n = 1) and in Chapter 7 (for n > 1).
13. Let X be a compact Hausdorff space. Show that if C(X) is separable,
then X is metrizable.
Hint. Define ϕ : X → C(X)0 by ϕ(y)(f ) = f (y) and use Corollary 9.3.
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13. Radon Measures
14. Let Y be a locally compact Hausdorff space. Assume Y is σ-compact,
S
i.e., there is a countable family of compact K j ⊂ Y such that Y = j Kj .
Show that there is a sequence fk ∈ C∗ (Y ) such that each fk has compact
support and for each y ∈ Y , some fk (y) 6= 0. (Also arrange 0 ≤ fk ≤ 1.)
Note that then
X
f=
2−k fk ∈ C∗ (Y )
k≥1
is > 0 at each point of Y and that Uk = {y ∈ Y : f (y) > 1/k} satisfies
Uk open,
U k compact,
Uk % Y.
15. Let Y be a locally compact metrizable space. Assume Y is σ-compact.
Show that C∗ (Y ) is separable. Deduce that the one-point compactification Yb is metrizable.
Hint. With Uk ⊂ Y as in Exercise
14, show that Vk = {f ∈ C∗ (Y ) :
S
supp f ⊂ U k } is separable and k Vk is dense in C∗ (Y ).
In Exercises 16–17, take C(X) to be the space of complex-valued continuous functions on the compact metric space X. Let ρ be a complex
Borel measure on X, of the form ρ = ν1 + iν2 (as in (8.19)), where νj
are finite signed measures, with associated positive measures |ν j | as in
(8.10). Set |ν| = |ν1 | + |ν2 | and apply the Radon-Nikodym theorem to
νj << |ν| to obtain
ρ = f |ν|.
16. Show that
nX
o
kρk = sup
|ρ(Sk )| : Sk ∈ B(X) disjoint
k≥0
defines a norm on the set MC (X) of complex Borel measures on X.
Show that
Z
kρk = |f | d|ν|.
17. Show that the dual C(X)0 of C(X) is isomorphic to MC (X), with norm
given in Exercise 16.