From Open-Set Content to an Integral
Let \(X\) be a locally compact Hausdorff space, and let \(C_c(X)\) denote the real-valued continuous functions on \(X\) with compact support. A positive linear functional \(L:C_c(X)\to\mathbb{R}\) assigns a nonnegative number to every nonnegative function. The representation construction associates a regular Borel measure to \(L\). The recovery step is to show that integrating a nonnegative \(f\in C_c(X)\) against this measure gives exactly \(L(f)\).
Use the open-set content \(\lambda(U)=\sup\{L(g):g\in C_c(X),\ 0\le g\le1,\ \operatorname{supp}g\subset U\}\) for open \(U\subset X\). Here \(\operatorname{supp}g\) is the closure of the set where \(g\ne0\). The measure \(\mu\) is the Borel measure constructed from this content. We use the established properties that \(\mu\) is locally finite and regular, and that \(\mu(U)=\lambda(U)\) for every open \(U\). In particular, \(\mu\) is finite on compact sets. No assumption that \(\mu\) is finite, complete, or sigma-finite on all of \(X\) is needed below.
The proof uses a finite collection of level sets at a chosen height increment \(\delta>0\). Strict superlevel sets make the lower estimate compatible with the definition of \(\lambda\); continuous slices give the upper estimate. Both estimates will use the same simple-function bounds for the integral.
Level-Set Bounds
Fix \(f\ge0\) in \(C_c(X)\), and let \(K=\operatorname{supp}f\). If \(f=0\), positivity and linearity give \(L(f)=0=\int f\,d\mu\), so suppose \(f\) may be nonzero. Choose a relatively compact open set \(V\) containing \(K\); such a set exists by local compactness and compactness of \(K\). Local finiteness gives \(\mu(V)<\infty\). Fix \(\delta>0\), choose an integer \(N\ge1\) with \(f(x)\le N\delta\) for all \(x\), and define the open sets \(U_j=\{x\in X:f(x)>j\delta\}\) for integers \(j\ge0\). Each \(U_j\) is contained in \(V\): outside \(K\), \(f=0\), whereas \(j\delta\ge0\). Also \(U_N=\varnothing\).
The lower simple function counts how many strict thresholds lie below \(f(x)\); the upper one includes a first increment on \(V\). Define \(s_\delta=\delta\sum_{j=1}^{N}\chi_{U_j}\) and \(u_\delta=\delta\chi_V+\delta\sum_{k=0}^{N-2}\chi_{U_k}\), where a sum with no indices is zero. At each \(x\), the number of integers \(j\ge1\) with \(j\delta<f(x)\), multiplied by \(\delta\), is at most \(f(x)\). For the upper bound, if \(x\in V\) and \(f(x)>0\), the term \(\delta\chi_V(x)\) covers the first increment, and each \(k\delta<f(x)\) contributes another \(\delta\); since \(f(x)\le N\delta\), these contributions total at least \(f(x)\). If \(f(x)=0\), the upper bound is nonnegative, and outside \(V\) both \(f\) and all the indicators vanish. Thus \(s_\delta\le f\le u_\delta\) everywhere. These are nonnegative measurable simple functions, so monotonicity of the Lebesgue integral gives \(\int s_\delta\,d\mu\le\int f\,d\mu\le\int u_\delta\,d\mu\).
For the functional’s lower bound, fix \(\varepsilon>0\). For each \(j=1,\ldots,N\), the equality \(\mu(U_j)=\lambda(U_j)\) provides a function \(\varphi_j\in C_c(X)\) with \(0\le\varphi_j\le1\), \(\operatorname{supp}\varphi_j\subset U_j\), and \(L(\varphi_j)\ge\mu(U_j)-\varepsilon\). When \(U_j\) is empty, take \(\varphi_j=0\). At any \(x\), \(\varphi_j(x)\ne0\) implies \(f(x)>j\delta\). Consequently \(\delta\sum_{j=1}^N\varphi_j(x)\le f(x)\). Positivity and linearity therefore give \(L(f)\ge\delta\sum_{j=1}^N L(\varphi_j)\ge\int s_\delta\,d\mu-\delta N\varepsilon\). Because this holds for every \(\varepsilon>0\), \(L(f)\ge\int s_\delta\,d\mu\).
For the upper bound, split \(f\) into continuous slices \(h_j=\min\{\delta,(f-(j-1)\delta)^+\}\), \(j=1,\ldots,N\), where \(r^+=\max\{r,0\}\). Each \(h_j\) is continuous, nonnegative, and compactly supported, and the slices sum to \(f\): at a point with value \(t\in[0,N\delta]\), they contribute successive increments of length at most \(\delta\), totaling \(t\). For \(j=1\), \(h_1/\delta\) is admissible in the definition of \(\lambda(V)\), so \(L(h_1)\le\delta\mu(V)\). For \(j\ge2\), the support of \(h_j\) is contained in \(\{f\ge(j-1)\delta\}\), which in turn is contained in the open set \(U_{j-2}=\{f>(j-2)\delta\}\). Thus \(h_j/\delta\) is admissible for \(\lambda(U_{j-2})\), and \(L(h_j)\le\delta\mu(U_{j-2})\). Adding these inequalities proves \(L(f)\le\int u_\delta\,d\mu\).
The two simple-function bounds become close as \(\delta\) decreases. If \(N\ge2\), cancellation of the common terms in their integrals gives \(\int u_\delta\,d\mu-\int s_\delta\,d\mu =\delta\bigl(\mu(V)+\mu(U_0)-\mu(U_{N-1})-\mu(U_N)\bigr) \le2\delta\mu(V)\). Here \(U_0\subset V\), and all measures are nonnegative. If \(N=1\), then \(U_1=\varnothing\), and the difference is \(\delta\mu(V)\), which also is at most \(2\delta\mu(V)\). We have proved \(0\le \int u_\delta\,d\mu-\int s_\delta\,d\mu\le2\delta\mu(V)\). Both \(L(f)\) and \(\int f\,d\mu\) lie between these same bounds. Since \(\mu(V)<\infty\) is fixed, choosing \(\delta\) arbitrarily small forces their difference to be zero. This proves the theorem. \(\square\)
Worked Level-Set Calculations
Worked Example: A Functional on a Finite Discrete Space
Let \(X=\{a,b,c\}\) have the discrete topology, and let \(\mu(\{a\})=2\), \(\mu(\{b\})=1\), and \(\mu(\{c\})=3\). Define \(L(g)=2g(a)+g(b)+3g(c)\). Every function on this finite space is continuous and compactly supported. For \(f(a)=2\), \(f(b)=1\), and \(f(c)=\tfrac12\), compute the integral and compare the level-set bounds using \(\delta=\tfrac12\).
The direct functional value is \(L(f)=2\cdot2+1\cdot1+3\cdot\tfrac12=4+1+\tfrac32=\tfrac{13}{2}\). Take \(V=X\), so \(\mu(V)=2+1+3=6\), and \(N=4\), since \(f\le4\delta=2\). With strict thresholds, \(U_0=X\), \(U_1=\{a,b\}\), \(U_2=\{a\}\), \(U_3=\{a\}\), and \(U_4=\varnothing\). Their measures are respectively \(6,3,2,2,0\). Hence \(\int s_\delta\,d\mu=\tfrac12(3+2+2+0)=\tfrac72\), while \(\int u_\delta\,d\mu=\tfrac12(6+6+3+2)=\tfrac{17}{2}\). The theorem’s bounds give \(\tfrac72\le\tfrac{13}{2}\le\tfrac{17}{2}\). Check: integrating directly gives \(2\cdot2+1\cdot1+3\cdot\tfrac12=\tfrac{13}{2}\), as does the defining formula for \(L\).
Worked Example: A Dirac Measure Detects One Function Value
Take \(X=[0,1]\), with its usual topology, \(\mu=\delta_{1/3}\), and \(L(g)=g(1/3)\). For \(f(x)=2-x\), the functional gives \(L(f)=2-\tfrac13=\tfrac53\), and the integral against the Dirac measure is \(\int f\,d\mu=f(1/3)=\tfrac53\). The level sets also give the theorem’s bounds at \(\delta=\tfrac13\).
Here \(V=X\), \(\mu(V)=1\), and \(N=6\) works because \(f(x)\le2=6\delta\). The point \(1/3\) belongs to \(U_j=\{x:2-x>j/3\}\) exactly when \(\tfrac53>\tfrac j3\), or \(j<5\). Thus \(\mu(U_j)=1\) for \(j=0,1,2,3,4\), and \(\mu(U_5)=\mu(U_6)=0\). It follows that \(\int s_\delta\,d\mu=\tfrac13(1+1+1+1+0+0)=\tfrac43\). For the upper sum, the indices are \(k=0,1,2,3,4\), so \(\int u_\delta\,d\mu=\tfrac13(1+1+1+1+1+0)=\tfrac53\); the first \(1\) is \(\mu(V)\). Therefore \(\tfrac43\le\tfrac53\le\tfrac53\). Check: the upper bound is exact because the atom lies at a point where \(f=5/3\), and direct evaluation gives \(L(f)=5/3\).
Worked Example: Lebesgue Measure and the Function f(x) = x
Let \(X=[0,1]\), let \(\mu\) be Lebesgue measure on this compact space, and take \(L(g)=\int_0^1g(x)\,dx\). For \(f(x)=x\) and \(\delta=\tfrac14\), set \(V=X\) and \(N=4\). The strict superlevel sets are \(U_0=(0,1]\), \(U_1=(\tfrac14,1]\), \(U_2=(\tfrac12,1]\), \(U_3=(\tfrac34,1]\), and \(U_4=\varnothing\). Their measures are \(1,\tfrac34,\tfrac12,\tfrac14,0\).
The lower and upper bounds are \(\int s_\delta\,d\mu=\tfrac14(\tfrac34+\tfrac12+\tfrac14+0)=\tfrac38\) and \(\int u_\delta\,d\mu=\tfrac14(1+1+\tfrac34+\tfrac12)=\tfrac{13}{16}\). The exact integral is \(\int_0^1x\,dx=[x^2/2]_0^1=\tfrac12\), so \(\tfrac38\le\tfrac12\le\tfrac{13}{16}\), as required. Check: the level-set calculation gives \(\tfrac38\) below and \(\tfrac{13}{16}\) above \(\tfrac12\); the direct antiderivative independently verifies the value of \(L(f)\).
Why the Cutoffs and Strict Inequalities Matter
The content \(\lambda(U)\) is defined using continuous functions whose supports lie inside an open set \(U\). It is not defined by simply inserting the indicator of a level set. For example, on \([0,1]\) with Lebesgue measure, the tempting choice \(\chi_{[1/2,1]}\) for \(f(x)=x\) is not continuous: its value jumps at \(x=1/2\). Its integral is \(\tfrac12\), but that computation does not make it an admissible cutoff in the definition of the open-set content. The valid lower argument instead chooses continuous \(\varphi\) with \(0\le\varphi\le1\) and compact support inside the open set \(\{x:x>1/2\}\), and then uses the supremum over all such cutoffs.
Strictness also keeps the pointwise lower estimate correct at a threshold. If \(f(x)=j\delta\), that point contributes nothing to the count of indices satisfying \(j\delta<f(x)\); the sum of the increments counted below \(f(x)\) therefore cannot exceed \(f(x)\). The upper estimate has a different job: its first term \(\delta\chi_V\) covers the initial increment, and the remaining open sets account for later increments. Omitting that first term, or using a set that is not open when invoking \(\lambda\), breaks the corresponding justification.
The proof shows the mechanism behind representation: open-set content records the largest functional values available to compactly supported cutoffs, and finitely many level sets convert those values into upper and lower estimates for an arbitrary nonnegative \(f\in C_c(X)\). Local finiteness is essential to the limiting comparison in this argument: the error is bounded by \(2\delta\mu(V)\), and it tends to zero because the chosen neighborhood has finite measure. No global finiteness assumption is being smuggled into that step.
Check Your Understanding
Use the definitions and estimates above to answer the following questions.
- Why does local finiteness provide a finite quantity in the bound \(2\delta\mu(V)\), even when \(\mu(X)=\infty\)?
- For a fixed \(j\), explain why \(\varphi_j\) supported in \(U_j=\{f>j\delta\}\) contributes at most \(\delta\varphi_j(x)\le f(x)\) at a point \(x\).
- For \(f(x)=x\) on \([0,1]\) and \(\delta=\tfrac12\), list \(U_0,U_1,U_2\), using strict inequalities, and compute their Lebesgue measures.
- In the upper estimate, why is \(\operatorname{supp}h_j\subset U_{j-2}\) for \(j\ge2\), rather than merely \(h_j(x)=0\) outside \(U_{j-2}\)?
- What fails if an indicator of a closed level set is used as though it were a continuous cutoff in the definition of \(\lambda(U)\)?