The Riesz Representation Theorem for Compact Spaces

From Positive Functionals to Measures

On a compact space, continuous functions are bounded, so a positive linear functional on them has a natural candidate for its total size: its value on the constant function $1$. The Riesz representation theorem says much more than that this value is finite. It says that the functional is integration against one uniquely determined finite regular Borel measure. Thus a rule defined algebraically on functions is represented by a measure defined on sets.

Let $K$ be a compact Hausdorff space, and let $C(K)$ denote the real-valued continuous functions on $K$. A linear functional $L:C(K)\to\mathbb{R}$ is positive if $L(f)\geq 0$ whenever $f(x)\geq 0$ for every $x\in K$. The function $1\in C(K)$ denotes the constant function with value $1$. A finite Borel measure $\mu$ on $K$ is regular if, for every $E\in\mathcal{B}_K$, its measure can be approximated from inside by compact sets and from outside by open sets:

\mu(E)=\sup\{\mu(F):F\subseteq E,\ F\text{ compact}\} =\inf\{\mu(U):E\subseteq U,\ U\text{ open}\}.

Here $(K,\mathcal{B}_K,\mu)$ is the measure space in question. “Finite” means $\mu(K)<\infty$; consequently this measure space is also sigma-finite. The measure is defined on the Borel sigma-algebra and is not assumed to be complete.

Theorem (Riesz Representation for Compact Spaces): Let $K$ be a compact Hausdorff space and let $L:C(K)\to\mathbb{R}$ be a positive linear functional. There exists a unique finite regular Borel measure $\mu$ on $(K,\mathcal{B}_K)$ such that, for every $f\in C(K)$, $$ L(f)=\int_K f\,d\mu. $$ The integral is the real-valued Lebesgue integral with respect to $\mu$, and the identity holds for every $f\in C(K)$, not merely almost everywhere.

The measure is finite even when $K$ is not a finite set: applying the identity to $1$ gives $\mu(K)=L(1)<\infty$. The theorem includes $K=\varnothing$: then $C(K)$ consists only of the zero function, $L$ is zero, and the unique representing measure is the zero measure. The construction of the representing measure and the proof of uniqueness are substantial parts of the representation theorem; they are developed in the subsequent measure-construction results. We use the theorem here in its stated form.

Consequences of Positivity

Positivity imposes an order bound on $L$. Equip $C(K)$ with the supremum norm $\|f\|_\infty=\sup_{x\in K}|f(x)|$, when $K$ is nonempty. The following estimate shows directly that $L$ is bounded with respect to this norm.

Proposition: If $K$ is a nonempty compact Hausdorff space and $L:C(K)\to\mathbb{R}$ is positive and linear, then $$ |L(f)|\leq L(1)\,\|f\|_\infty\qquad(f\in C(K)). $$ In particular, $L(1)\geq 0$. If $L(1)=0$, then $L$ is the zero functional.

Proof. Positivity applied to the nonnegative constant function $1$ gives $L(1)\geq0$. For any $f\in C(K)$, the definition of the supremum norm gives $-\|f\|_\infty 1\leq f\leq\|f\|_\infty 1$ at every point of $K$. Therefore $f+\|f\|_\infty 1$ and $\|f\|_\infty 1-f$ are nonnegative. Positivity and linearity imply $$ -\|f\|_\infty L(1)\leq L(f)\leq\|f\|_\infty L(1), $$ which is the asserted absolute-value bound. If $L(1)=0$, the bound gives $|L(f)|\leq0$ for every $f$, hence $L(f)=0$ for every $f$. $\square$

The representation theorem also has a converse: a finite Borel measure on $K$ defines a positive linear functional by integration. Regularity is not needed for this direction, but it is part of the theorem's conclusion and is essential to its precise representation statement.

Proposition: Let $K$ be compact Hausdorff and let $\mu$ be a finite Borel measure on $(K,\mathcal{B}_K)$. Then $L_\mu(f)=\int_K f\,d\mu$ defines a positive linear functional on the real space $C(K)$, and $$ |L_\mu(f)|\leq\mu(K)\|f\|_\infty. $$

Proof. Every $f\in C(K)$ is bounded and Borel measurable, so it is integrable with respect to the finite measure $\mu$: indeed, $|f|\leq\|f\|_\infty 1$, whose integral is $\|f\|_\infty\mu(K)<\infty$. Linearity of the Lebesgue integral gives linearity of $L_\mu$. If $f\geq0$ everywhere, monotonicity of the integral gives $L_\mu(f)\geq0$, so the functional is positive. Finally, $-|f|\leq f\leq|f|$ and $|f|\leq\|f\|_\infty 1$; integrating these inequalities gives $$ \left|\int_K f\,d\mu\right|\leq\int_K|f|\,d\mu \leq\|f\|_\infty\mu(K). $$ This proves the assertion. $\square$

Worked Representations

Worked Example: A Functional on a Finite Compact Space

Let $K=\{a,b,c\}$ with the discrete topology, which is compact and Hausdorff. Define $L:C(K)\to\mathbb{R}$ by $$ L(f)=2f(a)+3f(b)+f(c). $$ Every function on this finite space is continuous. The functional is linear, and it is positive because $f\geq0$ implies $2f(a)+3f(b)+f(c)\geq0$. Define the Borel measure $\mu=2\delta_a+3\delta_b+\delta_c$, where $\delta_x(E)=1$ if $x\in E$ and $0$ otherwise. This measure is finite and regular: every subset of $K$ is both open and compact. Its total mass is $\mu(K)=2+3+1=6$.

For any $f\in C(K)$, integration against this measure is a finite sum: $$ \int_K f\,d\mu=2f(a)+3f(b)+f(c)=L(f). $$ For example, if $f(a)=1,\ f(b)=2,\ f(c)=4$, then $L(f)=2\cdot1+3\cdot2+1\cdot4=12$, and the integral is $2+6+4=12$. As a check on the total mass, the constant function $1$ gives $L(1)=2+3+1=6=\mu(K)$. Uniqueness in the theorem says no other finite regular Borel measure on this $K$ represents this functional.

Worked Example: A Density on a Compact Interval

Take $K=[0,1]$ with its usual topology and define, for real $f\in C([0,1])$, $$ L(f)=\int_0^1 f(x)(1+x)\,dx, $$ where $dx$ denotes Lebesgue measure. The function $1+x$ is nonnegative on $[0,1]$, so $f\geq0$ implies $L(f)\geq0$. The representing measure is $\mu(E)=\int_E(1+x)\,dm(x)$ for Borel $E\subseteq[0,1]$; it is a finite regular Borel measure, by regularity of Lebesgue measure and the bounded continuous density.

Evaluate $L$ at $f(x)=1+x$. Substitution into the definition gives $$ L(f)=\int_0^1(1+x)^2\,dx =\int_0^1(1+2x+x^2)\,dx =1+1+\frac13=\frac73. $$ The same value follows from the representing measure: $\int_K f\,d\mu=\int_0^1 f(x)(1+x)\,dx=\frac73$. Also, $L(1)=\int_0^1(1+x)\,dx=1+\frac12=\frac32$, which agrees with the total mass $\mu(K)=\frac32$. These checks use the actual density defining this functional.

Worked Example: Evaluation at a Single Point

Let $K=[-1,1]$ and $L(f)=f(0)$ for real $f\in C(K)$. This functional is linear and positive, since $f\geq0$ everywhere implies $f(0)\geq0$. Its representing measure is the Dirac measure $\delta_0$: for every Borel set $E\subseteq K$, $\delta_0(E)=1$ when $0\in E$, and $\delta_0(E)=0$ otherwise. This is a finite regular Borel measure with total mass $1$.

For the specific function $f(x)=x^2+2x+3$, evaluation gives $L(f)=f(0)=0+0+3=3$. Integration against the proposed measure gives $\int_K f\,d\delta_0=f(0)=3$, so the identity holds for this test function. Testing $1$ also gives $L(1)=1=\delta_0(K)$. The theorem asserts the identity for every continuous $f$, and uniqueness rules out a different finite regular Borel measure representing this functional.

Positivity Is Not Optional

A linear functional on $C(K)$ need not be represented by a positive measure. The hypothesis is about the order on functions: nonnegative input functions must have nonnegative output. For instance, on $K=[0,1]$, consider $T(f)=f(0)-f(1)$. It is linear, but for the nonnegative continuous function $f(x)=x$, its value is $T(f)=0-1=-1$. Thus $T$ is not positive. If it were integration against a positive measure $\mu$, monotonicity of the integral would imply $\int_K x\,d\mu\geq0$, contradicting $T(x)=-1$. The Riesz theorem therefore does not apply.

A related mistake is to assume that the representing measure must be a probability measure. The theorem guarantees finiteness, not total mass $1$. For the finite-space example, $\mu(K)=6$, and for the density example, $\mu([0,1])=3/2$. Both are valid finite measures. The mass is determined by the functional: the representation identity at $1$ gives $\mu(K)=L(1)$. A probability measure arises only when the functional satisfies the additional normalization $L(1)=1$.

The theorem matters because it lets one replace a positive linear rule on continuous test functions with a concrete measure. The measure then assigns masses to Borel sets, while integration recovers the original values on continuous functions. In applications, positivity and the compact Hausdorff setting are not cosmetic conditions: they are part of what makes this exact representation possible.

Check Your Understanding

Use the theorem's hypotheses and identity to answer each question.

For a positive linear functional $L$ on $C(K)$, why must the mass of its representing measure equal $L(1)$?
On $K=\{a,b\}$ with the discrete topology, find the representing measure for $L(f)=4f(a)+f(b)$, and compute its total mass.
For $K=[0,1]$, verify that $L(f)=\int_0^1 2f(x)\,dx$ is positive and state the total mass of its representing measure.
Give a nonnegative continuous function that shows $T(f)=f(0)-f(1)$ is not positive.
Does the theorem require $L(1)=1$? Explain what $L(1)$ determines instead.

The Riesz Representation Theorem for Compact Spaces

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