Complex Measures & Total Variation
Two of the function spaces we have dualized return their own kind. The dual of a Hilbert space is
again a space of vectors through the
Riesz Representation
Theorem,
and the dual of \(L^p\) is the conjugate space \(L^q\) for \(1 \leq p \lt \infty\). The continuous
functions on a compact
space behave differently. A continuous linear functional on \(C(X)\) cannot in general be written
as integration against an \(L^q\) density, because \(C(X)\) carries the supremum norm and its dual
must account for evaluation-like behavior concentrated on small sets. The objects that do the work
are measures. The dual of \(C(X)\) turns out to be a space of measures on \(X\),
with the functional acting by integration \(f \mapsto \int f \, d\mu\). The goal of this page is
to make that identification precise and to prove it.
Establishing this requires measures that may take complex values, since the functionals on \(C(X)\) are
complex-linear. We therefore begin by extending the notion of a signed measure to the complex setting and
by constructing the total variation, the device that measures the size of such an object and ultimately
supplies the dual norm. We work over a compact (or, where noted, locally compact) Hausdorff space \(X\),
and we write \(\Omega\) for a \(\sigma\)-algebra of subsets of \(X\).
Recall the real-valued notion. A
signed
measure
on a measurable space \((X, \Omega)\) is a countably additive set function
\(\nu : \Omega \to [-\infty, \infty]\) with \(\nu(\varnothing) = 0\), assuming at most one of the
values \(\pm\infty\). The complex case removes the infinite values entirely, since the sum of a
convergent complex series must be finite.
Definition: Complex Measure
Let \((X, \Omega)\) be a measurable space. A complex measure on
\((X, \Omega)\) is a function \(\mu : \Omega \to \mathbb{C}\) that is countably additive: for
every sequence \((A_n)_{n \geq 1}\) of pairwise disjoint sets in \(\Omega\),
\[
\mu\!\left(\bigsqcup_{n=1}^\infty A_n\right) = \sum_{n=1}^\infty \mu(A_n),
\]
the series converging in \(\mathbb{C}\). Taking all \(A_n = \varnothing\) for \(n \geq 2\)
forces \(\mu(\varnothing) = 0\). A complex measure takes only finite values by definition.
Writing \(\mu = \operatorname{Re}\mu + i \operatorname{Im}\mu\), the real and imaginary parts
\((\operatorname{Re}\mu)(\Delta) = \operatorname{Re}\bigl(\mu(\Delta)\bigr)\) and
\((\operatorname{Im}\mu)(\Delta) = \operatorname{Im}\bigl(\mu(\Delta)\bigr)\) are each real-valued
and countably additive, hence finite signed measures. Applying the
Jordan
decomposition
to each gives four positive finite measures \(\mu_1, \mu_2, \mu_3, \mu_4\) with
\[
\mu = (\mu_1 - \mu_2) + i(\mu_3 - \mu_4),
\]
where \(\mu_1 \perp \mu_2\) and \(\mu_3 \perp \mu_4\) are mutually singular. Each of the four is
finite. If a real-valued countably additive \(\nu\) has Jordan parts concentrated on disjoint sets
\(A\) and \(B\), then \(\nu^+(X) = \nu^+(A) = \nu(A)\) and \(\nu^-(X) = \nu^-(B) = -\nu(B)\),
and \(\nu\) takes no infinite value, so both are real numbers. A real-valued signed
measure that happens to be finite is the special case \(\operatorname{Im}\mu = 0\). The
development below therefore subsumes the real theory, with the four-measure decomposition
collapsing to the two-measure Jordan form.
The Total Variation
For a finite signed measure the
total
variation is defined through the Jordan decomposition as \(|\nu| = \nu^+ + \nu^-\). A complex
measure has no such order structure, so we take instead the supremum-over-partitions description, which
agrees with \(\nu^+ + \nu^-\) in the real case and extends verbatim to the complex one.
Definition: Total Variation of a Complex Measure
Let \(\mu\) be a complex measure on \((X, \Omega)\). The total variation of
\(\mu\) is the set function \(|\mu| : \Omega \to [0, \infty]\) defined by
\[
|\mu|(\Delta) = \sup\left\{\, \sum_{j=1}^{m} |\mu(E_j)| : \{E_j\}_{j=1}^{m}
\text{ is a measurable partition of } \Delta \,\right\},
\]
the supremum running over all finite partitions of \(\Delta\) into pairwise disjoint
measurable sets.
For a finite signed measure \(\nu\), choosing the partition \(\{P \cap \Delta,\, N \cap \Delta\}\)
given by a Hahn decomposition \(X = P \sqcup N\) yields
\(\sum_j |\nu(E_j)| = \nu^+(\Delta) + \nu^-(\Delta)\). No partition can exceed this. For any
measurable \(E_j\), the values \(\nu^+(E_j)\) and \(\nu^-(E_j)\) are nonnegative, so
\[
|\nu(E_j)| = |\nu^+(E_j) - \nu^-(E_j)| \leq \nu^+(E_j) + \nu^-(E_j),
\]
and summing over any partition \(\{E_j\}\) of \(\Delta\) gives
\(\sum_j |\nu(E_j)| \leq \nu^+(\Delta) +
\nu^-(\Delta)\) by additivity of
\(\nu^+\) and \(\nu^-\). The supremum definition thus recovers \(\nu^+ + \nu^-\), confirming the
two notions coincide where both apply.
The central structural fact is that this supremum is itself a measure, and a finite one. The
finiteness is the substantive part. It is what makes the dual norm well defined.
Theorem: The Total Variation is a Finite Positive Measure
Let \(\mu\) be a complex measure on \((X, \Omega)\). Then \(|\mu|\) is a positive measure on
\((X, \Omega)\), and it is finite: \(|\mu|(X) \lt \infty\).
Proof
Countable additivity.
Let \(\Delta = \bigsqcup_{n=1}^\infty \Delta_n\) be a disjoint union of measurable sets. We show
\(|\mu|(\Delta) = \sum_n |\mu|(\Delta_n)\).
For the inequality \(|\mu|(\Delta) \leq \sum_n |\mu|(\Delta_n)\), let \(\{E_j\}_{j=1}^m\) be
any finite measurable partition of \(\Delta\). Each \(E_j = \bigsqcup_n (E_j \cap \Delta_n)\),
so by countable additivity of \(\mu\) and the triangle inequality,
\[
\sum_{j=1}^m |\mu(E_j)| = \sum_{j=1}^m \left| \sum_{n=1}^\infty \mu(E_j \cap \Delta_n) \right|
\leq \sum_{j=1}^m \sum_{n=1}^\infty |\mu(E_j \cap \Delta_n)|
= \sum_{n=1}^\infty \sum_{j=1}^m |\mu(E_j \cap \Delta_n)|.
\]
For each fixed \(n\), the sets \(\{E_j \cap \Delta_n\}_{j=1}^m\) form a measurable partition
of \(\Delta_n\), so \(\sum_{j=1}^m |\mu(E_j \cap \Delta_n)| \leq |\mu|(\Delta_n)\). Hence
\(\sum_{j=1}^m |\mu(E_j)| \leq \sum_{n=1}^\infty |\mu|(\Delta_n)\). Taking the supremum over
all partitions \(\{E_j\}\) of \(\Delta\) gives
\(|\mu|(\Delta) \leq \sum_{n=1}^\infty |\mu|(\Delta_n)\).
For the reverse inequality \(|\mu|(\Delta) \geq \sum_n |\mu|(\Delta_n)\), fix \(N \geq 1\). For
each \(n \leq N\) pick a real number \(t_n \lt |\mu|(\Delta_n)\) and a measurable partition
\(\{E^{(n)}_j\}_j\) of \(\Delta_n\) with \(\sum_j |\mu(E^{(n)}_j)| \gt t_n\), which the defining
supremum permits whether or not \(|\mu|(\Delta_n)\) is finite. The collection
\(\{E^{(n)}_j : 1 \leq n \leq N,\, j\} \cup \{\Delta \setminus \bigsqcup_{n \leq N} \Delta_n\}\)
is a finite measurable partition of \(\Delta\), so
\[
|\mu|(\Delta) \geq \sum_{n=1}^N \sum_j |\mu(E^{(n)}_j)| \gt \sum_{n=1}^N t_n.
\]
Taking the supremum over admissible \(t_1, \ldots, t_N\) gives
\(|\mu|(\Delta) \geq \sum_{n=1}^N |\mu|(\Delta_n)\), and
then \(N \to \infty\) gives \(|\mu|(\Delta) \geq \sum_{n=1}^\infty |\mu|(\Delta_n)\). Combined
with the first inequality, \(|\mu|\) is countably additive. Since \(|\mu|(\varnothing) = 0\)
and \(|\mu| \geq 0\), it is a positive measure.
Finiteness.
Decompose \(\mu = (\mu_1 - \mu_2) + i(\mu_3 - \mu_4)\) into
four positive finite measures as above. For any measurable \(E\),
\[
|\mu(E)| = \bigl| (\mu_1 - \mu_2)(E) + i(\mu_3 - \mu_4)(E) \bigr|
\leq \mu_1(E) + \mu_2(E) + \mu_3(E) + \mu_4(E).
\]
Hence for any finite partition \(\{E_j\}\) of \(X\),
\[
\sum_j |\mu(E_j)| \leq \sum_j \sum_{k=1}^4 \mu_k(E_j) = \sum_{k=1}^4 \mu_k(X),
\]
the last equality by additivity of each \(\mu_k\). Taking the supremum over partitions,
\(|\mu|(X) \leq \mu_1(X) + \mu_2(X) + \mu_3(X) + \mu_4(X) \lt \infty\), since each \(\mu_k\)
is finite.
The definition itself gives \(|\mu(E)| \leq |\mu|(E)\) for every measurable \(E\), by taking the
trivial partition \(\{E\}\) of \(E\). The total variation therefore dominates \(\mu\)
setwise while being a genuine positive measure, and it assigns to all of \(X\) a finite number.
That number will serve as the norm of \(\mu\).
Integrating Against a Complex Measure
The functionals we are heading for act by integration, so we must say what \(\int f \, d\mu\)
means. Neither ingredient is covered by an integration theory built for real functions and
positive measures. Both extensions are made by splitting into real and imaginary parts,
and the four-measure decomposition supplies the splitting on the measure side.
Definition: The Integral Against a Complex Measure
Let \(\pi\) be a finite positive measure on \((X, \Omega)\). Call \(f : X \to \mathbb{C}\)
measurable when \(\operatorname{Re} f\) and \(\operatorname{Im} f\) are, and
\(\pi\)-integrable when in addition \(\int |f| \, d\pi \lt \infty\). For such an \(f\), set
\[
\int f \, d\pi = \int \operatorname{Re} f \, d\pi + i \int \operatorname{Im} f \, d\pi,
\]
each term on the right being the Lebesgue integral of a real function. Now let \(\mu\) be a
complex measure on \((X, \Omega)\), decomposed as
\(\mu = (\mu_1 - \mu_2) + i(\mu_3 - \mu_4)\) into finite positive measures as above, and let
\(f\) be \(|\mu|\)-integrable. The
integral of \(f\) against \(\mu\) is
\[
\int f \, d\mu = \left( \int f \, d\mu_1 - \int f \, d\mu_2 \right)
+ i \left( \int f \, d\mu_3 - \int f \, d\mu_4 \right).
\]
Two points make this legitimate. First, the four integrals converge. Uniqueness of the Jordan
decomposition gives \(\mu_1 + \mu_2 = |\operatorname{Re}\mu|\), and for any measurable partition
\(\{E_j\}\) of \(\Delta\) the bound \(|\operatorname{Re} z| \leq |z|\) on complex numbers gives
\(\sum_j |(\operatorname{Re}\mu)(E_j)| \leq \sum_j |\mu(E_j)| \leq |\mu|(\Delta)\). Taking the
supremum over partitions and using the partition description of the total variation of a finite
signed measure established above, \(|\operatorname{Re}\mu| \leq |\mu|\) setwise. The same holds
for the imaginary part, so every \(\mu_k \leq |\mu|\) and \(f\) is integrable against each of
them.
Second, the value does not depend on the decomposition. Let
\(\mu = (\tilde\mu_1 - \tilde\mu_2) + i(\tilde\mu_3 - \tilde\mu_4)\) be any other
decomposition into finite positive measures against each of which \(f\) is integrable. Taking
real parts gives \(\mu_1 - \mu_2 = \tilde\mu_1 - \tilde\mu_2\), hence
\(\mu_1 + \tilde\mu_2 = \tilde\mu_1 + \mu_2\) as positive measures. The integral against a
positive measure is additive in the measure, so
\(\int f \, d\mu_1 + \int f \, d\tilde\mu_2 = \int f \, d\tilde\mu_1 + \int f \, d\mu_2\),
and every term is finite, so
\(\int f \, d\mu_1 - \int f \, d\mu_2 = \int f \, d\tilde\mu_1 - \int f \, d\tilde\mu_2\).
The imaginary parts are handled identically.
Three consequences follow at once and will be used below without further comment. Adding the
decompositions of two complex measures produces one for their sum, and expanding a decomposition
after multiplying through by a complex scalar produces one for the scalar multiple, so the
integral is linear in the measure. Linearity in the function is inherited from the real case.
When \(\mu\) is itself a finite positive measure we may
take \(\mu_1 = \mu\) and \(\mu_2 = \mu_3 = \mu_4 = 0\), so the definition returns the ordinary
integral. And for a measurable function taking the value \(c_j\) on each member of a finite
measurable partition \(\{E_j\}\) of \(X\), the definition unwinds through
\(\int f \, d\mu_k = \sum_j c_j \, \mu_k(E_j)\) to
\(\int f \, d\mu = \sum_j c_j \, \mu(E_j)\).
Regular Borel Measures and M(X)
A complex measure as defined above lives on an abstract measurable space. To represent functionals
on \(C(X)\) we need measures that interact correctly with the topology of \(X\). The mass of a set
must be approximable by the mass of compact sets from inside and open sets from outside. These
regularity conditions tie the measure to the topology and are what make the representation unique.
We now place the measures on the natural \(\sigma\)-algebra and impose regularity.
Throughout, \(X\) is a locally compact Hausdorff space. The Borel
\(\sigma\)-algebra \(\Omega\) is the smallest \(\sigma\)-algebra of subsets of \(X\)
containing all
open sets.
Its members are the Borel sets. A compact subset and an open subset of \(X\) are
Borel, so the approximation conditions below are well posed.
Definition: Regular Borel Measure
Let \(X\) be a locally compact Hausdorff space with Borel \(\sigma\)-algebra \(\Omega\). A positive
measure \(\mu\) on \((X, \Omega)\) is a regular Borel measure if
(a) \(\mu(K) \lt \infty\) for every
compact
set \(K \subseteq X\).
(b)
\(\mu(E) = \sup\{\, \mu(K) : K \subseteq E,\ K \text{ compact} \,\}\) for every
\(E \in \Omega\) (inner regularity).
(c)
\(\mu(E) = \inf\{\, \mu(U) : U \supseteq E,\ U \text{ open} \,\}\) for every \(E \in \Omega\)
(outer regularity).
A complex measure \(\mu\) on \((X, \Omega)\) is a regular Borel measure if its total
variation \(|\mu|\) is a regular Borel measure in the sense above.
When \(X\) is compact, \(X\) is itself a compact set, so condition (a) reduces to the single
requirement \(\mu(X) \lt \infty\). For the total variation of a complex measure that requirement is
automatic by the preceding section. Conditions (b) and (c) remain the substance.
They say a Borel set is squeezed between the compact sets it contains and the open sets that
contain it. The case relevant to \(C(X)\) is that of a compact \(X\), while the locally compact
formulation is stated as primary because the representation theorem and its corollary on
\(C_0(X)\) hold at that generality, with the compact case as the specialization \(C_0(X) = C(X)\).
The Space M(X)
The regular Borel measures of finite total variation form a vector space, on which the total variation of
the whole space supplies a norm.
Definition: The Space \(M(X)\)
Let \(X\) be a locally compact Hausdorff space. Denote by \(M(X)\) the set of all
complex-valued regular Borel measures on \(X\). It is a vector space over \(\mathbb{C}\) under
the pointwise operations \((\mu + \nu)(E) = \mu(E) + \nu(E)\) and
\((\lambda\mu)(E) = \lambda\,\mu(E)\). For \(\mu \in M(X)\), set
\[
\|\mu\| = |\mu|(X),
\]
the total variation of \(\mu\) over all of \(X\), which is finite by the preceding section.
That the pointwise operations preserve regularity, and that \(\|\cdot\|\) is a norm, must be checked. We
record both as a single proposition.
Proposition: \(M(X)\) is a Normed Space
The space \(M(X)\) is a complex vector space, and \(\|\mu\| = |\mu|(X)\) defines a norm on it.
Proof
Vector-space structure.
If \(\mu, \nu \in M(X)\) and
\(\lambda \in \mathbb{C}\), then \(\mu + \nu\) and \(\lambda\mu\) are complex measures, being
pointwise combinations of countably additive set functions. We must check they remain regular,
meaning that \(|\mu + \nu|\) and \(|\lambda\mu|\) are regular Borel measures. From the
partition definition, for any \(E\) and any partition \(\{E_j\}\),
\[
\sum_j |(\mu + \nu)(E_j)| \leq \sum_j |\mu(E_j)| + \sum_j |\nu(E_j)|
\leq |\mu|(E) + |\nu|(E),
\]
so taking the supremum gives \(|\mu + \nu|(E) \leq |\mu|(E) + |\nu|(E)\), that is,
\(|\mu + \nu| \leq |\mu| + |\nu|\) setwise. Likewise \(|\lambda\mu| = |\lambda|\,|\mu|\),
since each term \(|\lambda\mu(E_j)| = |\lambda|\,|\mu(E_j)|\) scales by \(|\lambda|\).
It remains to deduce that \(|\mu + \nu|\) is regular. The same argument applies to
\(|\lambda\mu| = |\lambda|\,|\mu|\), which is regular because a nonnegative scalar multiple of
a regular measure is regular. Write \(\rho = |\mu + \nu|\) and \(\sigma = |\mu| + |\nu|\), so
that \(\rho \leq \sigma\) setwise. Both are finite positive measures, and \(\sigma\) is
regular, being a sum of two regular measures (approximate each summand separately and add).
The key is an elementary fact. Because \(\rho\) and \(\sigma\) are finite and
\(\rho \leq \sigma\), for any two Borel sets \(A \subseteq B\) we have
\[
\rho(B) - \rho(A) = \rho(B \setminus A) \leq \sigma(B \setminus A),
\]
so whenever \(\sigma\) approximates a set well, \(\rho\) does too.
Inner regularity. Fix \(E \in \Omega\) and \(\varepsilon \gt 0\). By inner regularity
of \(\sigma\) there is a compact \(K \subseteq E\) with
\(\sigma(E \setminus K) \lt \varepsilon\). Applying the fact above with \(A = K\), \(B = E\),
\[
\rho(E) - \rho(K) \leq \sigma(E \setminus K) \lt \varepsilon,
\]
so \(\rho(K)\) comes within \(\varepsilon\) of \(\rho(E)\). Since \(\varepsilon\) was
arbitrary, \(\rho(E) = \sup\{\rho(K) : K \subseteq E,\ K \text{ compact}\}\).
Outer regularity. Similarly, by outer regularity of \(\sigma\) there is an open
\(U \supseteq E\) with \(\sigma(U \setminus E) \lt \varepsilon\), and the fact above with
\(A = E\), \(B = U\) gives \(\rho(U) - \rho(E) \leq \sigma(U \setminus E) \lt \varepsilon\).
Hence \(\rho(E) = \inf\{\rho(U) : U \supseteq E,\ U \text{ open}\}\). Thus
\(\rho = |\mu + \nu|\) is regular, and \(\mu + \nu, \lambda\mu \in M(X)\).
Positivity and definiteness.
Clearly \(\|\mu\| = |\mu|(X) \geq 0\). If
\(\|\mu\| = 0\), then \(|\mu|(X) = 0\), so \(|\mu|(E) = 0\) for every \(E \in \Omega\) by
monotonicity. Since \(|\mu(E)| \leq |\mu|(E) = 0\), we get \(\mu(E) = 0\) for all \(E\), that
is, \(\mu = 0\). Conversely \(\|0\| = 0\).
Homogeneity.
\(\|\lambda\mu\| = |\lambda\mu|(X) = |\lambda|\,|\mu|(X) = |\lambda|\,\|\mu\|\), using
\(|\lambda\mu| = |\lambda|\,|\mu|\) from above.
Triangle inequality.
\(\|\mu + \nu\| = |\mu + \nu|(X) \leq |\mu|(X) + |\nu|(X) = \|\mu\| + \|\nu\|\), using
\(|\mu + \nu| \leq |\mu| + |\nu|\) from above. Thus \(\|\cdot\|\) is a norm.
Support and Point Masses
Two further notions attached to a measure will be needed when the representation is applied: the set on
which a measure genuinely lives, and the simplest measures of all, concentrated at a single point.
Definition: Support of a Measure
Let \(\mu\) be a complex measure on a locally compact Hausdorff space \(X\) with Borel
\(\sigma\)-algebra \(\Omega\). The support of \(\mu\) is the set
\[
\operatorname{supp}\mu = X \setminus \bigcup\{\, V : V \text{ is open and } |\mu|(V) = 0 \,\}.
\]
It is the set of points every open neighborhood of which carries positive total variation.
The complement of \(\operatorname{supp}\mu\) is a union of \(|\mu|\)-null open sets, hence open,
so \(\operatorname{supp}\mu\) is closed. For \(\mu \in M(X)\) this null complement is itself
\(|\mu|\)-null, and crucially no countability of \(X\) is needed. Writing
\(W = X \setminus \operatorname{supp}\mu\), every compact \(C \subseteq W\) is covered by finitely
many of the null open sets whose union is \(W\), so \(|\mu|(C) = 0\). Inner regularity of
\(|\mu|\) then gives \(|\mu|(W) = \sup\{|\mu|(C) : C \subseteq W \text{ compact}\} = 0\).
Consequently \(\int_X f \, d\mu = \int_{\operatorname{supp}\mu} f \, d\mu\) for every integrable
\(f\), since the integrand contributes nothing off the support. This is exactly the step that lets
one replace \(X\) by the support of a measure without losing any of its mass. A regular measure
therefore has \(\operatorname{supp}\mu\) as the smallest closed set whose complement is
\(|\mu|\)-null. Regularity is what makes that description available. Without it the union of the
null open sets need not itself be null.
Definition: Dirac Measure
For a point \(x \in X\), the Dirac measure (or point mass)
\(\delta_x\) is the positive measure defined by
\[
\delta_x(E) = \begin{cases} 1 & x \in E, \\\\ 0 & x \notin E, \end{cases}
\]
for \(E \in \Omega\). It is a regular Borel measure with \(\|\delta_x\| = 1\) and
\(\operatorname{supp}\delta_x = \{x\}\), and integration against it is evaluation:
\(\int_X f \, d\delta_x = f(x)\) for every \(f \in C(X)\). A scalar multiple
\(\alpha\delta_x\) with \(\alpha \in \mathbb{C}\) is the complex measure
\(E \mapsto \alpha\,\delta_x(E)\), with \(\|\alpha\delta_x\| = |\alpha|\).
Countable additivity of \(\delta_x\) holds because at most one set in a disjoint family contains
\(x\). Regularity holds because \(\{x\}\) is compact and the value on any Borel set is determined
by whether it contains \(x\). Together with their unimodular multiples \(\alpha\delta_x\), where
\(|\alpha| = 1\), these point masses are the extreme points of the unit ball of \(M(X)\), a fact that
surfaces once the representation identifies \(M(X)\) with a dual space.
From Functionals to Measures
We now connect the two sides. One direction is easy and explicit. Every measure produces a
functional by integration. The other direction, that every functional arises this way, is the
substance of the representation theorem and occupies the rest of this section. Throughout,
\(C_0(X)\) is the space of continuous functions \(f : X \to \mathbb{C}\) that vanish at
infinity, meaning that for each \(\varepsilon \gt 0\) the set
\(\{|f| \geq \varepsilon\}\) is compact, equipped with the supremum norm
\(\|f\|_\infty = \sup_x |f(x)|\). When \(X\) is compact this is simply \(C(X)\), since every
continuous function then has compact support automatically.
Each Measure Defines a Functional
Lemma: Integration Against a Measure is a Functional
Let \(\mu \in M(X)\) and define \(F_\mu : C_0(X) \to \mathbb{C}\) by
\(F_\mu(f) = \int f \, d\mu\). Then \(F_\mu\) is a bounded linear functional,
\(F_\mu \in C_0(X)^*\), and
\[
\|F_\mu\| = \|\mu\| = |\mu|(X).
\]
Proof Sketch
Boundedness and the upper estimate (full).
Linearity of \(F_\mu\) is the
linearity of the integral. For a function \(s = \sum_j c_j \chi_{E_j}\) taking finitely many
values over a measurable partition \(\{E_j\}\), the setwise bound \(|\mu(E)| \leq |\mu|(E)\)
recorded above gives \(\left| \int s \, d\mu \right| = \left| \sum_j c_j \mu(E_j) \right|
\leq \sum_j |c_j| \, |\mu|(E_j) = \int |s| \, d|\mu|\). Every \(f \in C_0(X)\) is a uniform
limit of such functions, and both sides survive that limit because \(|\mu|\) and the four
measures \(\mu_k\) dominating \(\mu\) are finite, so
\(\left| \int f \, d\mu \right| \leq \int |f| \, d|\mu|\) for every \(f \in C_0(X)\). Together
with the bound \(|f(x)| \leq \|f\|_\infty\) for \(f \in C_0(X)\), this gives
\[
|F_\mu(f)| = \left| \int f \, d\mu \right| \leq \int |f| \, d|\mu|
\leq \|f\|_\infty \, |\mu|(X) = \|f\|_\infty \, \|\mu\|.
\]
Hence \(F_\mu\) is bounded with \(\|F_\mu\| \leq \|\mu\|\).
The lower estimate (sketch, resting on the approximation principle stated
after the proof).
For the reverse inequality \(\|F_\mu\| \geq \|\mu\|\) we must produce
functions \(f\) with \(\|f\|_\infty \leq 1\) and \(|F_\mu(f)|\) close to \(|\mu|(X)\). The
polar decomposition of \(\mu\) writes \(\mu = h \, |\mu|\) for a Borel function \(h\) with
\(|h| = 1\) almost everywhere with respect to \(|\mu|\). This is the measure-theoretic
statement that \(\mu\) and \(|\mu|\) differ only by a unimodular phase. The natural candidate
is \(f = \bar{h}\), for which
\[
\int \bar{h} \, d\mu = \int \bar{h} \, h \, d|\mu| = \int |h|^2 \, d|\mu| = |\mu|(X),
\]
but \(\bar{h}\) is only Borel, not continuous, and need not lie in \(C_0(X)\). The remedy is
to approximate \(\bar{h}\) by a continuous function. There is a classical approximation
principle. A Borel function that is bounded by \(1\) can, off a set of arbitrarily small
\(|\mu|\)-measure, be matched by a continuous function of supremum norm at most \(1\).
Granting such an approximant \(\phi\) with \(\|\phi\|_\infty \leq 1\) and
\(\int |\phi - \bar{h}| \, d|\mu| \lt \varepsilon\), we get
\[
|F_\mu(\phi)| = \left| \int \phi \, d\mu \right|
\geq \left| \int \bar{h} \, d\mu \right| - \left| \int (\phi - \bar{h}) \, d\mu \right|
\geq |\mu|(X) - \int |\phi - \bar{h}| \, d|\mu| \gt |\mu|(X) - \varepsilon.
\]
Since \(\|\phi\|_\infty \leq 1\), this forces \(\|F_\mu\| \geq |\mu|(X) - \varepsilon\), and
letting \(\varepsilon \to 0\) gives \(\|F_\mu\| \geq \|\mu\| = |\mu|(X)\). With the upper
estimate, \(\|F_\mu\| = \|\mu\|\).
The approximation principle used here expresses the regularity of \(|\mu|\) at the level of
functions. Just as a Borel set is approximable by compact and open sets, a bounded Borel function
is approximable, in the mean, by a continuous one of the same supremum bound. We use it as a
stated principle. Its proof builds the continuous approximant on the compact sets supplied by
inner regularity.
The map \(\mu \mapsto F_\mu\) is therefore a norm-preserving linear injection of \(M(X)\) into
\(C_0(X)^*\): linear because the integral is linear in the measure, injective because
\(\|F_\mu\| = \|\mu\|\) forces \(F_\mu = 0 \Rightarrow \mu = 0\). We approach surjectivity, the
recovery of a measure from an arbitrary functional, in two steps: first reduce to a
positive functional, then realize a positive functional by a positive measure.
Reduction to a Positive Functional
Call a linear functional \(I : C_0(X) \to \mathbb{C}\) positive if \(I(f) \geq 0\)
whenever \(f \geq 0\). Positivity is the order-theoretic analogue of being represented by a positive
measure, and it is the form in which a measure is easiest to extract. The first step shows every bounded
functional dominates a positive one of the same norm.
Lemma: A Bounded Functional Yields a Positive Functional
Let \(F : C_0(X) \to \mathbb{C}\) be a bounded linear functional. For \(f \in C_0(X)\) with
\(f \geq 0\), define
\[
I(f) = \sup\{\, |F(g)| : g \in C_0(X),\ |g| \leq f \,\}.
\]
Then \(I\) extends to a positive linear functional on \(C_0(X)\) with \(\|I\| = \|F\|\) and
\(|F(g)| \leq I(|g|)\) for all \(g\).
Proof Sketch
For \(f \geq 0\), the defining supremum is finite because
\(|F(g)| \leq \|F\| \, \|g\|_\infty \leq
\|F\| \, \|f\|_\infty\)
whenever \(|g| \leq f\), so \(0 \leq I(f) \leq \|F\| \, \|f\|_\infty\) and \(\|I\| \leq \|F\|\).
Taking \(g = f\) gives \(I(f) \geq |F(f)|\). The reverse norm bound comes from the domination
\(|F(g)| \leq I(|g|)\) verified at the end of this proof, which yields
\(|F(g)| \leq \|I\| \, \|g\|_\infty\) for every \(g\) and hence \(\|F\| \leq \|I\|\). Thus
\(\|I\| = \|F\|\). One checks
additivity on nonnegative functions: \(I(f_1 + f_2) = I(f_1) + I(f_2)\) for
\(f_1, f_2 \geq 0\). The inequality \(\geq\) comes from combining near-optimal \(g_1, g_2\)
into \(g_1 + g_2\) after adjusting phases so that \(|F(g_1)| + |F(g_2)| = |F(g_1 + g_2)|\).
The inequality \(\leq\) comes from splitting any \(g\) with \(|g| \leq f_1 + f_2\) as
\(g = g_1 + g_2\) with \(|g_i| \leq f_i\), setting \(g_i = g f_i / (f_1 + f_2)\) where the
denominator is positive and \(0\) elsewhere. Positive homogeneity \(I(t f) = t I(f)\) for
\(t \geq 0\) is immediate from the definition. An additive, positively homogeneous functional
on the nonnegative cone extends uniquely to a linear functional on \(C_0(X)\) by writing a
real function as a difference of its positive and negative parts and a complex function
through real and imaginary parts. The extension is positive by construction and satisfies
\(|F(g)| \leq I(|g|)\) since \(|g| \leq |g|\) places \(g\) among the competitors defining
\(I(|g|)\).
A Positive Functional Comes From a Measure
The second step is the heart of the representation. A positive linear functional on \(C_0(X)\) is
integration against a positive regular Borel measure. This is the classically named Riesz
representation for positive functionals. Its full construction of the measure is long, and we give
the core idea.
Theorem: Positive Functionals are Positive Measures
Let \(X\) be a locally compact Hausdorff space and let \(I : C_0(X) \to \mathbb{C}\) be a
positive bounded linear functional. Then there is a unique positive regular Borel measure
\(\nu\) on \(X\) such that
\[
I(f) = \int f \, d\nu \quad \text{for every } f \in C_0(X),
\]
and \(\|I\| = \nu(X)\).
Proof Sketch
The measure is built outside-in, starting from open sets. Write \(C_c(X)\) for the continuous
functions of
compact
support, a subspace of \(C_0(X)\). For an open \(U \subseteq X\), define
\[
\nu(U) = \sup\{\, I(\phi) : \phi \in C_c(X),\ 0 \leq \phi \leq 1,\ \operatorname{supp}\phi
\subseteq U \,\},
\]
the supremum of the functional over continuous "bump" functions trapped inside \(U\). This is
monotone in \(U\) and, using that a bump subordinate to a union can be split into bumps
subordinate to the pieces, countably subadditive. For an arbitrary set \(E\) one then sets
\[
\nu(E) = \inf\{\, \nu(U) : U \supseteq E,\ U \text{ open} \,\},
\]
an outer measure by construction. The technical core is to show that this outer measure is
countably additive on the Borel sets, which amounts to the Carathéodory measurability
of open sets. That is where the bulk of the work lies, and where the existence of continuous
bumps separating compact sets from closed sets (a separation property of locally compact
Hausdorff spaces) is used repeatedly. Outer regularity holds by definition. Inner regularity
follows because the bumps witnessing \(\nu(U)\) have compact support. Once \(\nu\) is a
measure, the identity \(I(f) = \int f \, d\nu\) is verified first for
\(0 \leq f
\leq 1\) by sandwiching \(f\) between bumps adapted to the
level sets \(\{f \gt t\}\) and integrating in \(t\), then for general \(f\) by linearity.
Uniqueness follows because two regular Borel measures agreeing as functionals agree on open
sets through the bump-supremum formula, hence on all Borel sets by outer regularity. Finally
\(\|I\| = \nu(X)\). The bound \(|I(f)| \leq \nu(X)\|f\|_\infty\) comes from
\(I(f) = \int f \, d\nu\), and equality is approached by bumps increasing to \(1\).
The two steps compose. An arbitrary bounded functional \(F\) yields a positive functional \(I\)
with \(|F(g)| \leq I(|g|)\), and \(I\) is integration against a positive measure \(\nu\). The
remaining task is to recover \(F\) itself, not just its positive envelope, as integration against
a complex measure built from \(\nu\). Assembling this is the content of the representation theorem
in the next section.
The Riesz Representation Theorem
Everything is now in place. The previous section produced, from a bounded functional \(F\), a positive
functional \(I\) dominating it, and realized \(I\) as integration against a positive measure \(\nu\). We
complete the circle by recovering \(F\) itself as integration against a complex measure, and we record the
resulting identification of \(C_0(X)^*\) with \(M(X)\).
Theorem: Riesz Representation for \(C_0(X)\)
Let \(X\) be a locally compact Hausdorff space. For \(\mu \in M(X)\), let
\(F_\mu \in C_0(X)^*\) be the functional \(F_\mu(f) = \int f \, d\mu\). Then the map
\[
M(X) \longrightarrow C_0(X)^*, \quad \mu \longmapsto F_\mu,
\]
is an isometric isomorphism. It is linear, \(\|F_\mu\| = \|\mu\|\) for every \(\mu\), and
every bounded linear functional on \(C_0(X)\) equals \(F_\mu\) for a unique \(\mu \in M(X)\).
In words, the dual of \(C_0(X)\) is the space of measures \(M(X)\).
Proof
Linearity of \(\mu \mapsto F_\mu\) and the isometry \(\|F_\mu\| = \|\mu\|\) were established
earlier, and the isometry makes the map injective. The equation \(F_\mu = 0\) forces
\(\|\mu\| = 0\), hence \(\mu = 0\). Only surjectivity remains. Given \(F \in C_0(X)^*\), we
construct \(\mu \in M(X)\) with \(F = F_\mu\).
Form the positive functional \(I\) dominating \(F\), and let \(\nu\) be the positive regular
Borel measure with \(I(g) = \int g \, d\nu\). The domination
\(|F(g)| \leq I(|g|) = \int |g| \, d\nu\) says that \(F\), viewed on \(C_0(X)\) sitting inside
\(L^1(\nu)\), is bounded for the \(L^1(\nu)\)-norm:
\[
|F(g)| \leq \int |g| \, d\nu = \|g\|_{L^1(\nu)}.
\]
Continuous functions of compact support approximate \(L^1\) functions in mean, a consequence
of the regularity of \(\nu\), so \(C_0(X)\) is dense in \(L^1(\nu)\). The functional \(F\)
therefore extends uniquely to a bounded linear functional on \(L^1(\nu)\) of norm at most
\(1\). The dual of \(L^1\) is \(L^\infty\), the endpoint case \(p = 1\) of the \(L^p\) duality
recalled at the start of this page within the general theory of
dual spaces, and it applies here because \(\nu\) is a finite measure.
The extension is therefore given by a function
\(\phi \in L^\infty(\nu)\) with \(\|\phi\|_\infty \leq 1\):
\[
F(g) = \int g \, \phi \, d\nu \quad (g \in C_0(X)).
\]
Now define \(\mu\) by \(\mu(E) = \int_E \phi \, d\nu\), the measure with density \(\phi\)
against \(\nu\). It is a complex measure, and its total variation is
\(|\mu|(E) = \int_E |\phi| \, d\nu\), so \(|\mu| \leq \nu\). Since \(\nu\) is a finite regular
Borel measure, so is the dominated \(|\mu|\), whence \(\mu \in M(X)\). For every
\(g \in C_0(X)\),
\[
F_\mu(g) = \int g \, d\mu = \int g \, \phi \, d\nu = F(g),
\]
the middle equality being the change of variables for a density. Thus \(F = F_\mu\).
Uniqueness is the injectivity already shown. Therefore \(\mu \mapsto F_\mu\) is an isometric
isomorphism of \(M(X)\) onto \(C_0(X)^*\).
The harder direction was routed through the duality of \(L^p\) at \(p = 1\). The regularity of
\(\nu\) makes \(C_0(X)\) dense in \(L^1(\nu)\), and that denseness is what lets a density \(\phi\)
represent \(F\). Regularity is therefore essential to the construction, not cosmetic.
The Compact Case and Annihilators
When \(X\) is compact, \(C_0(X) = C(X)\) and the theorem reads \(C(X)^* = M(X)\). Every continuous
linear functional on \(C(X)\) is integration against a unique complex regular Borel measure, with
the dual norm equal to the total variation. This is the form used when \(C(X)\) is the ambient
algebra. In particular, if \(\mathcal{A} \subseteq C(X)\) is a closed subspace, its
annihilator
\[
\mathcal{A}^\perp = \{\, \varphi \in C(X)^* : \varphi(f) = 0 \text{ for all } f \in \mathcal{A} \,\}
\]
becomes, under the identification, the space of measures that integrate to zero against every
member of \(\mathcal{A}\):
\[
\mathcal{A}^\perp = \Bigl\{\, \mu \in M(X) : \int f \, d\mu = 0 \text{ for all } f \in \mathcal{A}
\,\Bigr\}.
\]
An element of \(\mathcal{A}^\perp\) is now a concrete object, a measure on \(X\), to which the
full machinery of support, total variation, and integration applies. This is precisely what turns
an abstract functional-analytic hypothesis about \(\mathcal{A}\) into a question about measures
one can localize and manipulate.
The extreme points of the unit ball of \(M(X)\) make the same identification vivid. For a compact
\(X\), those extreme points are exactly the unimodular point masses \(\alpha\delta_x\) with
\(|\alpha| = 1\) and \(x \in X\). Among all measures of total variation \(1\), the indecomposable
ones are concentrated at a single point. A measure that lives on more than one point is an average
of measures supported on smaller sets and so cannot be extreme, while a point mass admits no such
splitting. Reading this through the Riesz isomorphism, the extreme points of the dual ball of
\(C(X)\) are the scaled evaluations \(f \mapsto \alpha f(x)\), the functionals that read off a
single value. The structure of the dual ball is thereby reduced to the points of \(X\) themselves,
the starting point for extracting a distinguished measure from an extremal functional and the
bridge by which compactness arguments on the dual ball return statements about \(C(X)\).