From Compact Support to Vanishing at Infinity
On a locally compact Hausdorff space, a continuous function that vanishes at infinity need not have compact support. Nevertheless, it can be approximated uniformly by compactly supported continuous functions. This makes it possible to extend the representation of positive functionals on \(C_c(X)\) to the larger space \(C_0(X)\). Boundedness is essential: it ensures that the measure produced from the compactly supported part has finite total mass.
Throughout, \(X\) is a locally compact Hausdorff space, and all functions in this tutorial are real-valued. Write \(C_c(X)\) for the continuous functions on \(X\) with compact support, and \(C_0(X)\) for the continuous functions that vanish at infinity: \(f\in C_0(X)\) means that \(\{x\in X: |f(x)|\ge\varepsilon\}\) is compact for every \(\varepsilon>0\). Equip \(C_0(X)\) with the supremum norm \(\|f\|_\infty=\sup_{x\in X}|f(x)|\). A linear functional \(L:C_0(X)\to\mathbb R\) is bounded if there is a finite constant \(C\) such that \(|L(f)|\le C\|f\|_\infty\) for every \(f\in C_0(X)\); it is positive if \(L(f)\ge0\) whenever \(f\ge0\).
We also use two facts about \(X\) established in the earlier material. First, \(C_c(X)\) is uniformly dense in \(C_0(X)\). Second, for every compact \(K\subset X\), there is a cutoff \(u\in C_c(X)\) with \(0\le u\le1\) and \(u=1\) on \(K\). More generally, if \(K\) is compact and \(U\) is open with \(K\subset U\), the cutoff can be chosen to have support in \(U\).
The Representation Theorem for \(C_0(X)\)
Proof. Restrict \(L\) to \(C_c(X)\). This restriction is positive, so the earlier representation result gives a regular Borel measure \(\mu\), finite on compact sets, with \(L(\varphi)=\int_X\varphi\,d\mu\) for \(\varphi\in C_c(X)\). We first show that \(\mu\) is finite. For each compact \(K\subset X\), choose \(u\in C_c(X)\) such that \(0\le u\le1\) and \(u=1\) on \(K\). Then $$ \mu(K)=\int_X\chi_K\,d\mu\le\int_Xu\,d\mu=L(u)\le C\|u\|_\infty\le C, $$ where \(C\) is a bound for \(L\), and the first inequality follows from \(\chi_K\le u\). By inner regularity on the open set \(X\), \(\mu(X)=\sup\{\mu(K):K\subset X\text{ compact}\}\le C\). This also covers \(X=\varnothing\), when \(\mu(X)=0\).
Now fix \(f\in C_0(X)\). By the uniform density of \(C_c(X)\), there is a sequence \((\varphi_n)\) in \(C_c(X)\) with \(\|\varphi_n-f\|_\infty\to0\). Boundedness gives \(|L(\varphi_n)-L(f)|\le C\|\varphi_n-f\|_\infty\to0\). Since \(\mu(X)<\infty\), the integral satisfies \(\left|\int_X\varphi_n\,d\mu-\int_Xf\,d\mu\right|\le\mu(X)\|\varphi_n-f\|_\infty\to0\). The equality \(L(\varphi_n)=\int_X\varphi_n\,d\mu\) holds for every \(n\), so taking limits proves \(L(f)=\int_Xf\,d\mu\).
For uniqueness, suppose that \(\mu\) and \(\nu\) are finite regular Borel measures representing \(L\). If \(U\subset X\) is open and \(\varphi\in C_c(X)\) satisfies \(0\le\varphi\le1\) and \(\operatorname{supp}(\varphi)\subset U\), then \(\int_X\varphi\,d\mu\le\mu(U)\). Conversely, for each compact \(K\subset U\), choose such a cutoff with \(\varphi=1\) on \(K\). Then \(\mu(K)\le\int_X\varphi\,d\mu\). Inner regularity on \(U\) therefore gives $$ \mu(U)=\sup\left\{\int_X\varphi\,d\mu:\varphi\in C_c(X),\ 0\le\varphi\le1,\ \operatorname{supp}(\varphi)\subset U\right\}. $$ The same formula holds for \(\nu\), and the integrals agree because both measures represent \(L\). Thus \(\mu(U)=\nu(U)\) for every open \(U\). For any Borel set \(E\), outer regularity gives \(\mu(E)=\inf_{U\supset E,\ U\text{ open}}\mu(U)\) and the corresponding formula for \(\nu(E)\). Hence \(\mu(E)=\nu(E)\), proving uniqueness. \(\square\)
Compactly Supported Cutoffs Detect Total Mass
Proof. For every \(u\in C_c(X)\) with \(0\le u\le1\), the representation gives \(L(u)=\int_Xu\,d\mu\le\mu(X)\). This proves that the supremum is at most \(\mu(X)\). For the reverse inequality, fix a compact \(K\subset X\) and choose \(u\in C_c(X)\) with \(0\le u\le1\) and \(u=1\) on \(K\). Then \(L(u)=\int_Xu\,d\mu\ge\mu(K)\). Taking the supremum over compact \(K\), and using inner regularity of \(\mu\) on \(X\), gives \(\sup_u L(u)\ge\mu(X)\). The two inequalities prove the formula, including the case \(X=\varnothing\). \(\square\)
Worked Example: A Functional on a Compact Interval
Let \(X=[0,1]\) with its usual topology and define \(L(f)=\int_0^1 f(x)\,dx\) for \(f\in C_0(X)=C(X)\). The measure \(m\) restricted to the Borel subsets of \([0,1]\) is finite and regular, and \(L(f)=\int_Xf\,dm\). For the test function \(f(x)=x^2\), direct integration gives $$ L(f)=\int_0^1x^2\,dx=\left[\frac{x^3}{3}\right]_0^1=\frac13. $$ The representing measure gives the same value: \(\int_Xx^2\,dm=1/3\). Since \(X\) itself is compact, the constant function \(u=1\) belongs to \(C_c(X)\), takes values in \([0,1]\), and \(L(u)=\int_0^1 1\,dx=1\). No admissible cutoff can have integral greater than \(m(X)=1\), so the supremum of cutoff values is exactly the total mass. Check: the mass of \([0,1]\) is its length, \(1-0=1\), agreeing with the cutoff computation.
Worked Example: A Functional Supported at Two Points
Take \(X=\mathbb R\) and define \(L(f)=2f(-1)+3f(2)\) for \(f\in C_0(\mathbb R)\). The estimate $$ |L(f)|\le2|f(-1)|+3|f(2)|\le5\|f\|_\infty $$ shows boundedness, and \(f\ge0\) implies \(L(f)\ge0\). The measure \(\mu=2\delta_{-1}+3\delta_2\) is finite and regular, with \(\mu(\mathbb R)=2+3=5\). For \(h(x)=\max(1-|x+1|,0)\), we have \(h(-1)=1\) and \(h(2)=0\), so \(L(h)=2(1)+3(0)=2\). The integral against the proposed measure is also \(2h(-1)+3h(2)=2\). For \(q(x)=\max(1-|x-2|,0)\), the values are \(q(-1)=0\) and \(q(2)=1\); hence \(L(q)=2(0)+3(1)=3\), again matching the integral.
To detect the total mass with one cutoff, let \(u(x)=1\) on \([-1,2]\), let \(u(x)=x+2\) on \([-2,-1]\), let \(u(x)=3-x\) on \([2,3]\), and let \(u(x)=0\) outside \([-2,3]\). These formulas agree at their endpoints, so \(u\in C_c(\mathbb R)\), and \(0\le u\le1\). As \(u(-1)=u(2)=1\), \(L(u)=2+3=5=\mu(\mathbb R)\). Check: for every admissible cutoff \(v\), \(L(v)=2v(-1)+3v(2)\le2+3=5\), so this value is maximal.
Worked Example: A Finite Measure on a Noncompact Space
On \(X=\mathbb R\), define $$ L(f)=\int_{-\infty}^0 f(x)e^x\,dx \qquad (f\in C_0(\mathbb R)). $$ Since \(\int_{-\infty}^0e^x\,dx=[e^x]_{-\infty}^0=1\), we have \(|L(f)|\le\|f\|_\infty\), and \(L\) is positive. Its representing measure is \(d\mu(x)=\chi_{(-\infty,0]}(x)e^x\,dx\), with total mass \(1\). For \(f(x)=e^{-|x|}\), which is continuous and vanishes at infinity, \(f(x)=e^x\) when \(x\le0\), so $$ L(f)=\int_{-\infty}^0e^{2x}\,dx=\left[\frac{e^{2x}}2\right]_{-\infty}^0=\frac12. $$ The measure integral gives the same result.
Here is an explicit sequence of compactly supported cutoffs. For each integer \(n\ge1\), let \(u_n(x)=0\) for \(x\le-n-1\), \(u_n(x)=x+n+1\) for \(-n-1\le x\le-n\), \(u_n(x)=1\) for \(-n\le x\le0\), \(u_n(x)=1-x\) for \(0\le x\le1\), and \(u_n(x)=0\) for \(x\ge1\). These pieces agree at the endpoints and \(0\le u_n\le1\). The part on \([0,1]\) contributes zero to \(L(u_n)\), since the density is zero for \(x>0\). On the plateau, $$ \int_{-n}^0e^x\,dx=1-e^{-n}. $$ On the left ramp, substituting \(t=x+n+1\) gives $$ \int_{-n-1}^{-n}(x+n+1)e^x\,dx=e^{-n-1}\int_0^1te^t\,dt =e^{-n-1}\big[(t-1)e^t\big]_0^1=e^{-n-1}. $$ Therefore \(L(u_n)=1-e^{-n}+e^{-n-1}=1-(1-e^{-1})e^{-n}\), which tends to \(1\). Each value is at most \(1\), and the limit equals the total mass, as the cutoff formula predicts. Check: at \(n=1\), the formula gives \(L(u_1)=1-e^{-1}+e^{-2}\), also obtained by adding the plateau integral \(1-e^{-1}\) and ramp integral \(e^{-2}\).
A Common Pitfall: The Constant Function
On a noncompact space, it is generally wrong to evaluate \(L(1)\) to find the total mass: the constant function \(1\) does not belong to \(C_0(X)\) when \(X\) is noncompact. For example, in the preceding example \(1\notin C_0(\mathbb R)\), so the expression \(L(1)\) is not defined by the given functional. Writing \(L(1)=\int_{-\infty}^0e^x\,dx=1\) may suggest the right number, but it uses a function outside the domain. The valid calculation uses the functions \(u_n\in C_c(\mathbb R)\) and their values \(L(u_n)=1-(1-e^{-1})e^{-n}\), whose supremum is \(1\).
The distinction matters even though the resulting measure is finite. The representation theorem gives an integral identity only for functions in its stated domain, and compactly supported cutoffs provide a legitimate way to recover total mass without adding a constant function to that domain.
Check Your Understanding
Use the theorem and cutoff formula to check each conclusion, paying attention to the function space on which the functional is defined.
- Why does restricting a bounded positive functional on \(C_0(X)\) to \(C_c(X)\) produce a positive functional to which the earlier representation result applies?
- In the theorem’s proof, where is boundedness used to show that the representing measure is finite?
- For \(L(f)=4f(0)\) on \(C_0(\mathbb R)\), identify the representing measure and compute its total mass using a compactly supported cutoff equal to \(1\) at \(0\).
- Explain why outer regularity and equality of two finite regular measures on every open set imply equality on every Borel set.
- On a noncompact space, why can the expression \(L(1)\) be undefined even when the representing measure has finite total mass?