Why Discrete Spaces Reduce the Representation to Weights
On a countable discrete space, every singleton is open, and every function is continuous. The compact subsets are exactly the finite subsets, so a function belongs to \(C_0(X)\) precisely when it becomes small outside finite sets. This makes the representation of a positive functional concrete: its values on the singleton indicators determine the whole functional. The essential constraint is that these values must be summable.
Let \(X\) be a finite or countably infinite set with the discrete topology. Its Borel sigma-algebra is \(\mathcal{B}_X=\mathcal{P}(X)\), the power set. A function \(f:X\to\mathbb{R}\) vanishes at infinity if, for every \(\varepsilon>0\), the set \(\{x:|f(x)|\ge\varepsilon\}\) is compact. On this space that means the set is finite. The space \(C_0(X)\), equipped with the supremum norm \(\|f\|_\infty=\sup_{x\in X}|f(x)|\), consists of exactly these functions. For \(x\in X\), the indicator \(\chi_{\{x\}}\) belongs to \(C_0(X)\).
Recovering the Weights and Proving the Formula
For each point, define \(a_x=L(\chi_{\{x\}})\). Positivity gives \(a_x\ge0\). To see why these weights must be summable, take any finite subset \(F\subseteq X\). Its indicator is in \(C_0(X)\), is nonnegative, and has supremum norm at most one. Therefore $$ \sum_{x\in F}a_x=L(\chi_F)\le \|L\|\,\|\chi_F\|_\infty\le\|L\|. $$ For a countably infinite \(X\), enumerate its points and let \(F_n\) be the first \(n\) points. The partial sums \(\sum_{x\in F_n}a_x\) increase and are bounded above by \(\|L\|\), so their limit exists and is at most \(\|L\|\). In the finite case, take \(F=X\). In either case, \(\sum_{x\in X}a_x\le\|L\|\).
Proof. It remains to establish the representation and the reverse norm inequality. For a function \(g\) supported on a finite set \(F\), linearity gives \(g=\sum_{x\in F}g(x)\chi_{\{x\}}\), so $$ L(g)=\sum_{x\in F}g(x)L(\chi_{\{x\}})=\sum_{x\in F}a_xg(x). $$ If \(X\) is infinite, choose an enumeration \(X=\{x_1,x_2,\ldots\}\) and put \(f_n=f\chi_{F_n}\). Since \(f\in C_0(X)\), for every \(\varepsilon>0\) the set \(E=\{x:|f(x)|\ge\varepsilon\}\) is finite. Once \(F_n\) contains \(E\), \(\sup_{x\notin F_n}|f(x)|<\varepsilon\), and hence \(\|f-f_n\|_\infty\le\varepsilon\). Thus \(f_n\to f\) in supremum norm. Boundedness of \(L\) implies \(L(f_n)\to L(f)\). Also, $$ \sum_{x\in X}a_x|f(x)|\le \|f\|_\infty\sum_{x\in X}a_x<\infty, $$ so the finite sums \(L(f_n)=\sum_{x\in F_n}a_xf(x)\) tend to the absolutely convergent series \(\sum_xa_xf(x)\). In the finite case the finite-support formula already applies to every \(f\). This proves the representation.
For any \(f\in C_0(X)\), the representation and the triangle inequality give $$ |L(f)|\le\sum_xa_x|f(x)|\le\|f\|_\infty\sum_xa_x. $$ Taking the supremum over \(\|f\|_\infty\le1\) gives \(\|L\|\le\sum_xa_x\). Conversely, for every finite \(F\), \(\chi_F\in C_0(X)\) and \(\|\chi_F\|_\infty\le1\), while \(L(\chi_F)=\sum_{x\in F}a_x\). Hence \(\|L\|\ge\sum_{x\in F}a_x\). Taking \(F=X\) if it is finite, or \(F=F_n\) and then \(n\to\infty\) if it is infinite, gives \(\|L\|\ge\sum_xa_x\). Finally, any representation must have \(a_x=L(\chi_{\{x\}})\), proving uniqueness. \(\square\)
The weights define a measure on every subset by \(\mu(A)=\sum_{x\in A}a_x\). Nonnegativity and summability make this a finite measure; countable additivity follows by partitioning the nonnegative series over disjoint subsets. Since every subset is measurable, the measure space \((X,\mathcal{P}(X),\mu)\) is complete. A finite measure is sigma-finite. It is regular: every subset is open, and, for any \(A\subseteq X\), finite subsets of \(A\) increase to \(A\), so continuity from below approximates \(\mu(A)\) from within by measures of compact sets. In particular, this is a Radon measure.
For a nonnegative function \(f\) on \(X\), its integral with respect to this measure is the weighted sum \(\int f\,d\mu=\sum_xa_xf(x)\), with \(+\infty\) allowed. This follows by applying the Monotone Convergence Theorem to the increasing finite partial sums of the nonnegative function. For real-valued \(f\in C_0(X)\), the sum is absolutely convergent because \(\sum_xa_x|f(x)|\le\|f\|_\infty\sum_xa_x\); consequently \(L(f)=\int f\,d\mu\).
Worked Example: A Functional on a Three-Point Space
Let \(X=\{a,b,c\}\) with the discrete topology and define \(L(f)=2f(a)+3f(c)\). Since all functions on \(X\) belong to \(C_0(X)\), the singleton indicators determine the weights: $$ a_a=L(\chi_{\{a\}})=2,\qquad a_b=L(\chi_{\{b\}})=0,\qquad a_c=L(\chi_{\{c\}})=3. $$ Thus \(\mu(\{a\})=2\), \(\mu(\{b\})=0\), and \(\mu(\{c\})=3\). For example, \(\mu(\{a,c\})=2+3=5\), and \(\mu(X)=2+0+3=5\).
Take \(f(a)=1\), \(f(b)=-4\), and \(f(c)=2\). The represented integral is $$ \int_X f\,d\mu=2(1)+0(-4)+3(2)=2+0+6=8. $$ The original formula for \(L\) gives \(L(f)=2(1)+3(2)=8\), as required. The norm is \(\|L\|=2+0+3=5\). Indeed, \(|L(f)|\le 2|f(a)|+3|f(c)|\le5\|f\|_\infty\), and \(f=\chi_{\{a,c\}}\) has norm \(1\) and \(L(f)=5\), so this bound is attained.
Worked Example: Geometric Weights on the Natural Numbers
Let \(X=\mathbb{N}=\{1,2,\ldots\}\), and set \(a_n=2^{-n}\). The weights are nonnegative and $$ \sum_{n=1}^\infty 2^{-n}=\frac{1/2}{1-1/2}=1. $$ They therefore define a positive bounded functional \(L(f)=\sum_{n=1}^\infty 2^{-n}f(n)\) on \(C_0(\mathbb{N})\), with \(\|L\|=1\). Its measure satisfies \(\mu(\mathbb{N})=1\).
The function \(f(n)=1/n\) belongs to \(C_0(\mathbb{N})\), since for each \(\varepsilon>0\) only finitely many \(n\) satisfy \(1/n\ge\varepsilon\). Its integral is $$ \int f\,d\mu=\sum_{n=1}^\infty\frac{1}{n2^n}=\ln 2. $$ To evaluate the series, for \(0\le t\le1/2\) expand \(1/(1-t)=\sum_{k=0}^\infty t^k\). The terms are nonnegative, so the Monotone Convergence Theorem permits integration term by term on \([0,1/2]\): $$ \int_0^{1/2}\frac{1}{1-t}\,dt =\sum_{k=0}^\infty\int_0^{1/2}t^k\,dt =\sum_{k=0}^\infty\frac{(1/2)^{k+1}}{k+1} =\sum_{n=1}^\infty\frac{1}{n2^n}. $$ The integral on the left is \([-\ln(1-t)]_0^{1/2}=\ln2\). As a check on the measure, the total weight is the geometric sum \((1/2)/(1-1/2)=1\), matching the functional norm.
Worked Example: Measuring Even Integers with Different Weights
On \(X=\mathbb{N}\), take \(a_n=3/4^n\). These weights sum to $$ \sum_{n=1}^\infty\frac{3}{4^n} =3\frac{1/4}{1-1/4}=1, $$ so the associated measure is a probability measure and the functional norm is \(1\). For the set \(A=\{2,4,6,\ldots\}\), the even indices are \(n=2k\) with \(k\ge1\), and therefore $$ \mu(A)=\sum_{k=1}^\infty\frac{3}{4^{2k}} =3\frac{1/16}{1-1/16}=\frac15. $$ The odd integers have measure \(1-\frac15=\frac45\), which also follows by summing their weights directly.
Now let \(f(n)=2^{-n}\), which vanishes at infinity. Then $$ \int f\,d\mu=\sum_{n=1}^\infty\frac{3}{4^n2^n} =3\sum_{n=1}^\infty\frac{1}{8^n} =3\frac{1/8}{1-1/8}=\frac37. $$ As a check, \(\int f\,d\mu\) is a weighted average of values \(2^{-n}\) between \(0\) and \(1/2\), so it must lie between \(0\) and \(\frac12\); indeed \(0<\frac37<\frac12\). The odd and even masses also check against the total: \(\frac45+\frac15=1\).
What the Assumptions Prevent
A common mistake on an infinite discrete space is to treat the constant function \(1\) as an element of \(C_0(X)\). If \(X=\mathbb{N}\), then \(\{n:|1|\ge1/2\}=\mathbb{N}\), which is not compact in the discrete topology because it is infinite. Thus \(1\notin C_0(\mathbb{N})\), and \(L(1)\) is not a permitted test of a functional defined only on \(C_0(\mathbb{N})\).
For a concrete failure, assign weight \(a_n=1\) to every \(n\). The formal sum \(\sum_na_n\) is infinite, so these weights cannot define a bounded functional on \(C_0(\mathbb{N})\). In fact, the finite-support functions \(\chi_{\{1,\ldots,N\}}\) have supremum norm \(1\), while their formal weighted sums equal \(N\). A bounded functional would have to satisfy \(N\le\|L\|\) for every \(N\), which is impossible. The summability hypothesis is exactly what rules out this failure.
The representation is also specific to a discrete space: each singleton indicator is continuous and belongs to \(C_0(X)\), so it can be used to recover an individual weight. On a general topological space, singleton indicators need not be continuous; the representation theorem then requires the broader theory of Radon measures rather than this direct point-by-point calculation.
Check Your Understanding
Use the singleton weights and the summability condition to organize each calculation.
- On a finite discrete space \(X=\{u,v\}\), let \(L(f)=4f(u)+f(v)\). Find the representing measure of every subset and compute \(\|L\|\).
- For weights \(a_n=5/6^n\) on \(\mathbb{N}\), compute the total mass and the measure of the even integers.
- Explain why \(\chi_{\{1,\ldots,N\}}\) belongs to \(C_0(\mathbb{N})\), while the constant function \(1\) does not.
- Show that if \(L\) is represented by summable nonnegative weights, then \(|L(f)|\le\|f\|_\infty\sum_na_n\) for every \(f\in C_0(\mathbb{N})\).
- For weights \(a_n=2^{-n}\), compute \(\int g\,d\mu\) when \(g(n)=\chi_{\{1,2,3\}}(n)\), and verify the result directly from the measure of \(\{1,2,3\}\).