Local Representation Without Finite Total Mass
For a compact space, continuous functions can test the whole space, and the associated positive functional is bounded in the uniform norm. On a noncompact space, compactly supported continuous functions test only bounded regions at a time. A positive functional on those functions can therefore be well behaved on every fixed compact region without being bounded by one constant on all of \(C_c(X)\). Its representing measure may be finite on compact sets but have infinite total mass.
Throughout, \(X\) is a locally compact Hausdorff space, and \(C_c(X)\) denotes the real-valued continuous functions on \(X\) with compact support. The support of \(f\) is the closure of \(\{x:f(x)\ne0\}\). A linear functional \(L:C_c(X)\to\mathbb{R}\) is positive if \(L(f)\ge0\) whenever \(f(x)\ge0\) for every \(x\in X\). Measures below are Borel measures on \((X,\mathcal{B}_X)\); they are not assumed complete or to have finite total mass.
The Locally Compact Representation Theorem
Proof. For existence, use the open-set content associated with \(L\), $$ \lambda(O)=\sup\{L(g):g\in C_c(X),\ 0\le g\le1,\ \operatorname{supp}(g)\subset O\},\qquad O\in\mathcal{B}_X\text{ open}. $$ The supremum is allowed to be \(+\infty\). The open-set-content construction, using open covers and the Caratheodory extension theorem, gives a Borel measure \(\mu\) with \(\mu(O)=\lambda(O)\) for open \(O\). The local finiteness and regularity result for this construction gives that \(\mu\) is Radon and finite on compact sets. The recovery result for the same construction gives \(L(f)=\int_X f\,d\mu\) for every \(f\in C_c(X)\). These are the open-set-content, Caratheodory-measurability, regularity, and functional-recovery results established in the earlier material on this theorem; they require local compactness, the Hausdorff property, and positivity of \(L\), but not a bound on \(L\) over all of \(C_c(X)\).
It remains to prove uniqueness. Suppose \(\mu\) and \(\nu\) are Radon measures finite on compact sets and represent the same \(L\). Fix an open set \(O\subset X\). For any \(g\in C_c(X)\) with \(0\le g\le1\) and \(\operatorname{supp}(g)\subset O\), positivity of integration gives $$ \int_X g\,d\mu\le\mu(\operatorname{supp}(g))\le\mu(O). $$ Taking the supremum over such \(g\) shows \(L\)'s common value in this test-function supremum is at most \(\mu(O)\). Conversely, if \(K\subset O\) is compact, local compactness and the cutoff property for locally compact Hausdorff spaces give a function \(g\in C_c(X)\) such that \(0\le g\le1\), \(g=1\) on \(K\), and \(\operatorname{supp}(g)\subset O\). Hence $$ \mu(K)\le\int_X g\,d\mu=L(g). $$ Taking the supremum over the test functions and then over compact \(K\subset O\), inner regularity on open sets gives $$ \mu(O)=\sup_{\substack{g\in C_c(X)\\0\le g\le1\\\operatorname{supp}(g)\subset O}}L(g). $$ The same argument for \(\nu\) shows \(\mu(O)=\nu(O)\). Finally, for any Borel set \(E\), outer regularity gives \(\mu(E)=\inf_{O\supset E}\mu(O)\) and \(\nu(E)=\inf_{O\supset E}\nu(O)\), where the infima are over open sets. Equality on open sets therefore implies equality on every Borel set. This proves uniqueness. \(\square\)
The local nature of the hypotheses can also be seen directly. Given a compact \(K\subset X\), choose a compactly supported cutoff \(u\) with \(u\ge1\) on \(K\). For any \(f\in C_c(X)\) supported in \(K\), positivity implies $$ -\|f\|_\infty u\le f\le\|f\|_\infty u, \qquad |L(f)|\le \|f\|_\infty L(u). $$ Thus \(L\) has a finite bound when its test functions are confined to \(K\). The bound \(L(u)\) can depend on \(K\); the theorem does not assert one bound works on all of \(X\).
Worked Representations
Worked Example: Lebesgue Measure on the Real Line
Let \(X=\mathbb{R}\) and define \(L(f)=\int_{\mathbb{R}}f\,dm\) for real-valued \(f\in C_c(\mathbb{R})\), where \(m\) is Lebesgue measure. The integral is finite because \(f\) is bounded and vanishes outside a compact set. Linearity follows from linearity of the integral, and \(f\ge0\) implies \(L(f)\ge0\). The representing Radon measure is \(m\).
For an explicit calculation, take \(f(x)=\max(1-|x|,0)\). Its support is \([-1,1]\); it equals \(1+x\) on \([-1,0]\) and \(1-x\) on \([0,1]\). Therefore $$ L(f)=\int_{-1}^0(1+x)\,dx+\int_0^1(1-x)\,dx =\left[x+\frac{x^2}{2}\right]_{-1}^0+\left[x-\frac{x^2}{2}\right]_0^1 =\frac12+\frac12=1. $$ The representing-measure calculation agrees: \(\int f\,dm=1\). As a check, symmetry makes the two triangular halves have equal area, and each has base \(1\) and height \(1\), hence area \(1/2\).
Worked Example: A Radon Measure with Infinite Total Mass
Again let \(X=\mathbb{R}\), and set \(L(f)=\int_{\mathbb{R}}f\,dm\) for \(f\in C_c(\mathbb{R})\). To see explicitly why this valid functional need not be globally bounded in the uniform norm, for each \(R>0\) define $$ f_R(x)= \begin{cases} 1,& |x|\le R,\\ R+1-|x|,& R<|x|<R+1,\\ 0,& |x|\ge R+1. \end{cases} $$ This is continuous, compactly supported, and \(0\le f_R\le1\), so \(\|f_R\|_\infty=1\). Its integral is the area of a rectangle of width \(2R\) and height \(1\), plus two triangular ends, each with base \(1\) and height \(1\): $$ L(f_R)=2R+\frac12+\frac12=2R+1. $$ For \(R=1\), this gives \(L(f_1)=3\); for \(R=2\), it gives \(L(f_2)=5\). In general \(L(f_R)/\|f_R\|_\infty=2R+1\), which has no finite upper bound as \(R\) increases. The representing measure is still finite on every compact interval: for \(A>0\), \(m([-A,A])=2A<\infty\). But \(m(\mathbb{R})=\infty\), since \(m([-A,A])=2A\) is unbounded as \(A\to\infty\). The theorem requires local finiteness, not finite total mass.
Worked Example: Point Evaluation
On \(X=\mathbb{R}\), define \(L(f)=f(0)\) for real-valued \(f\in C_c(\mathbb{R})\). This is linear, and if \(f\ge0\) then \(L(f)=f(0)\ge0\). The measure \(\delta_0\), the Dirac measure at \(0\), is defined by \(\delta_0(E)=1\) if \(0\in E\) and \(0\) otherwise, for Borel \(E\). It is Radon and finite on compact sets. By the definition of integration against a point mass, $$ \int_{\mathbb{R}}f\,d\delta_0=f(0). $$ For the tent function \(f(x)=\max(1-|x|,0)\), \(L(f)=f(0)=1\), and \(\int f\,d\delta_0=1\) as well. More generally, for \(f(x)=\max(2-2|x|,0)\), direct substitution gives \(L(f)=f(0)=2\), and integration against \(\delta_0\) also gives \(2\). These checks illustrate that the measure records the functional's values on every compactly supported continuous test function, not just on one chosen test.
What the Theorem Does Not Say
Common mistake: requiring a finite measure. The preceding Lebesgue example represents a positive functional on \(C_c(\mathbb{R})\), even though \(m(\mathbb{R})=\infty\). Local finiteness is the relevant condition: every compact set has finite measure. An argument that assumes \(L\) must be globally bounded because it is positive has silently confused a local bound with a uniform bound over the whole space.
Common mistake: applying the \(C_0(X)\) theorem without its norm hypothesis. A bounded positive functional on \(C_0(X)\), with the uniform norm, is represented by a finite Radon measure. The \(C_c(X)\) theorem has a different hypothesis and conclusion. The functions \(f_R\) above have uniform norm \(1\), but \(L(f_R)=2R+1\), so this particular functional cannot be extended to a bounded functional on \(C_0(\mathbb{R})\) with the same values on \(C_c(\mathbb{R})\). There is no contradiction: the \(C_c(X)\) theorem does not require such an extension.
Uniqueness has a useful practical interpretation. The measure of an open set is determined by the supremum of \(L(g)\) over continuous test functions between \(0\) and \(1\) supported inside that open set. Compactly supported cutoffs detect compact subsets of the open set; inner regularity then recovers its full measure. Outer regularity extends that information from open sets to all Borel sets.
Check Your Understanding
Use the distinction between local finiteness and finite total mass when answering.
- State the hypotheses on \(X\) and \(L\), and the conclusion of the \(C_c(X)\) representation theorem.
- Why does finiteness of \(\mu(K)\) for every compact \(K\) not imply \(\mu(X)<\infty\)? Give the real-line Lebesgue example.
- For the functions \(f_R\) above, compute \(\|f_R\|_\infty\) and \(L(f_R)\). What do these values show about a global uniform-norm bound?
- Explain how inner regularity and compactly supported cutoffs show that the functional determines the measure of an open set.
- Which additional functional hypothesis is required in the \(C_0(X)\) representation theorem, and why does the Lebesgue functional on \(C_c(\mathbb{R})\) fail it?