Why the Function Space Matters
A positive functional represented by a measure need not be bounded on every natural space of continuous functions. The key distinction is what the space allows as test functions: compact support only, vanishing at infinity, or arbitrary continuous functions on a compact space. These choices determine whether the representing measure may have infinite total mass and whether integration is defined and bounded for every function in the space.
Throughout, \(X\) is a locally compact Hausdorff space when it is not explicitly assumed compact. Functions are complex-valued unless stated otherwise; positivity of a linear functional means that it takes nonnegative values on real-valued nonnegative functions. Measures in the representation statements are positive regular Borel measures. A measure is locally finite here if every point has an open neighborhood of finite measure; it is finite if \(\mu(X)<\infty\). These are different conditions on a noncompact space.
For locally compact Hausdorff \(X\), \(C_c(X)\subseteq C_0(X)\). The earlier result on approximating \(C_0(X)\) functions by compactly supported ones says that \(C_c(X)\) is dense in \(C_0(X)\) in the supremum norm. If \(X\) is compact, every continuous function has compact support and vanishes at infinity, so \(C_c(X)=C_0(X)=C(X)\). On noncompact \(X\), \(C(X)\) is not the same space as \(C_0(X)\); arbitrary continuous functions need not be bounded, so there is generally no supremum norm on all of \(C(X)\).
The Three Choices and Their Measure Conditions
The relevant versions of the Riesz representation theorem give the following comparison. In each case, the functional is linear and positive. The representation identities hold for every test function in the stated space.
| Test-function space | Representation and measure condition | Boundedness in the supremum norm |
|---|---|---|
| \(C_c(X)\), \(X\) locally compact Hausdorff | A positive functional is represented by a unique locally finite regular Borel measure; that measure need not be finite. | It is bounded on \(C_c(X)\) exactly when its representing measure is finite. |
| \(C_0(X)\), \(X\) locally compact Hausdorff | A positive bounded functional is represented by a unique finite regular Borel measure. | Boundedness is part of the hypothesis; the functional norm equals the total mass. |
| \(C(X)\), \(X\) compact Hausdorff | A positive functional is represented by a unique finite regular Borel measure. | Every positive linear functional is bounded; its norm equals the total mass. |
The representation results for \(C_c(X)\) and \(C_0(X)\), and the norm-equals-total-mass result for \(C_0(X)\), are the standard Riesz representation results established earlier in the course. The compact-space case can also be checked directly for boundedness: compactness supplies the constant function \(1\), which controls every function in the supremum norm.
Proof. If \(\mu(X)<\infty\), every \(f\in C_c(X)\) is integrable and \[ |L(f)|\le\int |f|\,d\mu\le\|f\|_\infty\mu(X). \] Thus \(L\) is bounded and \(\|L\|\le\mu(X)\). For the reverse inequality, inner regularity gives \(\mu(X)=\sup\{\mu(K):K\subseteq X\text{ compact}\}\). For every compact \(K\), the compactly supported cutoff property for locally compact Hausdorff spaces gives a function \(g\in C_c(X)\) such that \(0\le g\le1\) and \(g=1\) on \(K\). Therefore \(\|g\|_\infty\le1\), and \[ \|L\|\ge L(g)=\int g\,d\mu\ge\mu(K). \] Taking the supremum over compact \(K\) gives \(\|L\|\ge\mu(X)\), hence equality. Conversely, suppose \(L\) is bounded, with \(\|L\|=C<\infty\). For each compact \(K\), choose \(g\in C_c(X)\) with \(0\le g\le1\) and \(g=1\) on \(K\). Then \(\mu(K)\le\int g\,d\mu=L(g)\le C\). Inner regularity gives \(\mu(X)=\sup_K\mu(K)\le C\), so \(\mu\) is finite. This proves both directions. \(\square\)
For the compact-space case, boundedness follows without first constructing a measure.
Proof. For any \(f\in C(X)\), choose a complex number \(\lambda\) with \(|\lambda|=1\) such that \(\lambda L(f)=|L(f)|\) when \(L(f)\ne0\); if \(L(f)=0\), the desired bound holds immediately. Pointwise, \(\operatorname{Re}(\lambda f)\le |f|\le\|f\|_\infty 1\). Positivity and linearity on real-valued functions imply \[ |L(f)|=L(\operatorname{Re}(\lambda f))\le L(|f|)\le\|f\|_\infty L(1). \] Thus \(\|L\|\le L(1)\). Since \(1\) has supremum norm \(1\), \(\|L\|\ge|L(1)|=L(1)\), including the case \(L(1)=0\). Hence \(\|L\|=L(1)\). Under the Riesz representation on compact Hausdorff \(X\), \(L(1)=\int_X1\,d\mu=\mu(X)\), so the norm is the total mass. \(\square\)
Worked Examples: What the Space Allows
Worked Example: A Finite Atomic Measure on a Compact Space
Let \(X=[0,1]\), which is compact Hausdorff, and define \(\mu=2\delta_0+3\delta_{1/2}+\delta_1\). Here \(\delta_x\) is the measure assigning mass \(1\) to a set containing \(x\) and mass \(0\) otherwise. For \(f\in C(X)\), integration gives \[ L(f)=2f(0)+3f(1/2)+f(1). \] The total mass is \(2+3+1=6\). For every \(f\), the triangle inequality gives \[ |L(f)|\le2|f(0)|+3|f(1/2)|+|f(1)|\le6\|f\|_\infty. \] Thus \(\|L\|\le6\). Taking \(f=1\), we have \(\|f\|_\infty=1\) and \(L(f)=2+3+1=6\), so \(\|L\|\ge6\) and \(\|L\|=6\). The calculation agrees with \(L(1)=\mu(X)=6\). Since \(X\) is compact, \(C_c(X)=C_0(X)=C(X)\), and this same functional is bounded on all three spaces.
Worked Example: Lebesgue Measure on the Real Line
Take \(X=\mathbb{R}\) and Lebesgue measure \(m\). The functional \(L(f)=\int f\,dm\) is well-defined and finite for every \(f\in C_c(\mathbb{R})\), since a continuous function is bounded on its compact support. To test boundedness, for \(R>0\) set \(f_R(x)=\max(1-|x|/R,0)\). This function is continuous, supported on \([-R,R]\), and has \(\|f_R\|_\infty=1\). Direct integration gives \[ L(f_R)=2\int_0^R(1-x/R)\,dx =2\left(R-\frac{R}{2}\right)=R. \] Consequently the ratios \(|L(f_R)|/\|f_R\|_\infty=R\) have no finite upper bound as \(R\) increases. For example, \(R=1,2,3\) gives integral values \(1,2,3\), respectively. The measure is locally finite but \(m(\mathbb{R})=\infty\); the theorem predicts that \(L\) is not bounded on \(C_c(\mathbb{R})\).
This does not define a finite-valued integration functional on all of \(C_0(\mathbb{R})\). Indeed, \(h(x)=1/(1+|x|)\) is continuous and vanishes at infinity, but \[ \int_{-T}^{T}h(x)\,dx=2\int_0^T\frac{dx}{1+x}=2\ln(1+T), \] which grows without bound as \(T\to\infty\). Thus \(h\) is not integrable with respect to \(m\). The issue is not whether \(h\) is measurable; it is whether its integral is finite.
Worked Example: Counting Measure on a Discrete Space
Let \(X=\mathbb{N}=\{1,2,\ldots\}\) with the discrete topology and counting measure \(\#\). This is locally compact Hausdorff, and its compact subsets are exactly the finite subsets. Therefore \(C_c(\mathbb{N})\) consists of the finitely supported functions. For each such \(f\), \[ L(f)=\int f\,d\#=\sum_{k\in\mathbb{N}}f(k), \] where only finitely many terms are nonzero, so the sum is finite.
For \(n\ge1\), let \(u_n=\chi_{\{1,\ldots,n\}}\). Its support has \(n\) points, \(\|u_n\|_\infty=1\), and \(L(u_n)=\sum_{k=1}^n1=n\). In particular, the values for \(n=1,2,3\) are \(1,2,3\); as \(n\) grows, the ratio \(L(u_n)/\|u_n\|_\infty=n\) is unbounded. Counting measure is locally finite because each singleton has measure \(1\), but it is not finite because \(\#(\mathbb{N})=\infty\). This is another positive, unbounded functional on \(C_c\).
In this topology, \(C_0(\mathbb{N})\) consists of sequences tending to zero. The sequence \(a(k)=1/k\) belongs to \(C_0(\mathbb{N})\): for every \(\varepsilon>0\), only finitely many \(k\) satisfy \(1/k\ge\varepsilon\). But its counting-measure integral is \(\sum_{k=1}^{\infty}1/k=\infty\). Hence the counting-measure integral is not a finite-valued functional on all of \(C_0(\mathbb{N})\), just as the Lebesgue integral above was not finite on all of \(C_0(\mathbb{R})\).
A Reliable Way to Choose
When a representation question is stated, first identify the function space and the assumptions on the functional. Then check whether the desired measure may have infinite total mass. The following distinctions prevent common errors:
- For positive functionals on \(C_c(X)\), local finiteness is the natural measure condition. Do not infer finite total mass or boundedness in the supremum norm.
- For positive bounded functionals on \(C_0(X)\), the representing measure is finite. Conversely, a finite regular Borel measure defines a bounded integration functional on \(C_0(X)\), with norm equal to its total mass.
- For compact Hausdorff \(X\), \(C_c(X)=C_0(X)=C(X)\); a positive functional is bounded and its representing measure is finite.
A common mistake is to see that \(\int f\,d\mu\) is finite for every \(f\in C_c(X)\) and conclude that this integration functional is bounded. Finiteness for each individual compactly supported test function is not a uniform estimate of the form \(|L(f)|\le C\|f\|_\infty\). The Lebesgue and counting-measure examples disprove that inference: each test function has a finite integral, while the displayed families have supremum norm \(1\) and integrals growing without bound.
Another mistake is to treat \(C_0(X)\) as though every member were integrable against every locally finite measure. The function \(h(x)=1/(1+|x|)\) and the sequence \(a(k)=1/k\) show why that fails for infinite measures. The Riesz theorem for \(C_0(X)\) concerns bounded functionals and finite representing measures, not every possible integral against a locally finite measure.
Check Your Understanding
For each question, distinguish finiteness of an individual integral from boundedness of a functional in the supremum norm.
- Why is integration against counting measure finite for every function in \(C_c(\mathbb{N})\), but not bounded there?
- On a noncompact locally compact Hausdorff space, what measure condition corresponds to a positive functional represented on \(C_c(X)\)?
- Why does the function \(h(x)=1/(1+|x|)\) show that Lebesgue integration is not finite-valued on all of \(C_0(\mathbb{R})\)?
- If \(X\) is compact Hausdorff, explain why \(C_c(X)\), \(C_0(X)\), and \(C(X)\) coincide.
- For a positive functional on compact-space \(C(X)\), how does the value \(L(1)\) determine both its norm and the total mass of its representing measure?