Let \((E,\mathscr{O})\) be a compact Hausdorff space and \(\mathscr{B}\) its Borel \(\sigma\)-algebra. The set \(C = C(E)\) consists of all continuous functions on \(E\), and is equipped with a complete norm \(\|f\| = \sup_{x\in E}|f(x)|\).
Theorem 1
Every positive continuous linear functional \(\ell\) on the Banach space \(C\) can be represented as
\[ \ell(f) = \int_{E} f(x)\mu(dx), \tag{*} \]
where \(\mu\) is a Borel measure on \(E\) with finite total variation.
Lemma 1
Suppose \(\{E_1,E_2,\dots,E_n\}\) is an open covering of \(E\). There exist continuous functions \(h_1,h_2,\dots,h_n\) satisfying
- \(\forall 1\leq i \leq n, \operatorname{supp}(h_i) \subset E_i\);
- \(\forall 1\leq i \leq n, x\in E, 0\leq h_i(x) \leq 1\);
- \(\forall x \in E, \sum_{i=1}^n h_i(x) =1\).
Proof of Lemma 1
Define \(F_1 = E \backslash \cup_{j=2}^n E_j\) which is closed and \(F_1 \subset E_1\). Since \(E\) is compact Hausdorff, it is also Tychonoff, where open sets separate two disjoint closed sets. Thus, there exists an open set \(V_1\) such that \(F_1 \subset V_1 \subset \bar{V_1} \subset E_1\). By construction, \(\{V_1, E_2, \dots, E_n\}\) is an open cover of \(E\). Inductively, we can prove that there exists an open cover \(\{V_1,\dots,V_n\}\) that is also a refinement of \(\{E_1,E_2,\dots,E_n\}\) and \(\bar{V_i} \subset E_i\) for each \(1\leq i \leq n\). For each \(i\), the sets \(\bar{V_i}\) and \(E_i^c\) are disjoint closed sets. By the Urysohn lemma, there exists a continuous function \(g_i:E \to [0,1]\) such that \(g_i(x)=1\) for \(x\in \bar{V_i}\) and \(g_i(x)=0\) for \(x\in E_i^c\). Thus, \(\operatorname{supp}(g_i) \subset E_i\). Define \(G(x) = \sum_{i=1}^n g_i(x)\). Since \(\{V_1,\dots,V_n\}\) covers \(E\), for any \(x\in E\), there is at least one index \(j\) such that \(x\in V_j \subset \bar{V_j}\), meaning \(g_j(x)=1\). Therefore, \(G(x) \geq 1\) for all \(x\in E\). Define \(h_i(x) = {g_i(x)}/{G(x)}\), which fulfills (1)-(3).
Proof of Theorem 1
Step 1
Define for any open set \(O \in \mathscr{O}\),
\[ \mu(O) = \sup \left\{\ell(f):f\in C, 0 \le f \le 1, \operatorname{supp}(f) \subset O\right\}. \]
By the Urysohn lemma, this set on the RHS is non-empty. We need to show that
1.1. \(\mu(\cup_n O_n) = \lim_n \mu(\sum_{m=1}^nO_m)\) for \(O_n \in \mathscr{O}\).
For any \(f \in C\) with \(0 \le f \le 1\) and \(K = \operatorname{supp}(f) \subset O = \cup_{n=1}^\infty O_n\). Since \(\{O_n\}\) is an open cover of \(K\), there exists a finite subcover such that \(K \subset \cup_{m=1}^{n_0} O_m\) for some integer \({n_0}\). Thus, \(\ell(f) \le \mu(\cup_{m=1}^{n_0} O_m) \le \lim_{n\to\infty} \mu(\cup_{m=1}^n O_m)\). Taking the supremum over all such \(f\) yields \(\mu(O) \le \lim_{n\to\infty} \mu(\cup_{m=1}^n O_m)\). The reverse inequality holds trivially by the monotonicity of the supremum.
Step 2
Define for any subset \(A \in 2^E\),
\[ \mu^*(A) = \inf \left\{\mu(O): O \in \mathscr{O}, A \subset O \right\}. \]
We need to show that
Claim 2.1
\(\mu^*(\cup_n A_n) \le \sum_{n} \mu^*(A_n)\) for \(A_n \in 2^E\).
If \(\sum_{n=1}^\infty \mu^*(A_n) = \infty\), the inequality is trivial. Assume the sum is finite. For any \(\varepsilon > 0\) and each \(n\), there exists \(O_n \in \mathscr{O}\) with \(A_n \subset O_n\) such that \(\mu(O_n) < \mu^*(A_n) + {\varepsilon}/{2^n}\). Denote \(O = \cup_{n=1}^\infty O_n \in \mathscr{O}\). Then \(A = \cup_{n=1}^\infty A_n \subset O\), and by Step 1.1 and finite subadditivity on open sets, \(\mu^*(A) \le \mu(O) \le \sum_{n=1}^\infty \mu(O_n) < \sum_{n=1}^\infty \mu^*(A_n) + \varepsilon\). Since \(\varepsilon\) is arbitrary, the countable subadditivity follows.
Step 3
A subset \(B \in 2^E\) is said to satisfy the Carathéodory condition if \(\mu^*(A) = \mu^*(A\cap B) + \mu^*(A\cap B^c) \text{ for all } A \in 2^E\). We need to show
Claim 3,1
All subsets satisfying Carathéodory condition are indeed a \(\sigma\)-algebra \(\bar{\mathscr{B}}\) that contains \(\mathscr{O}\).
Claim 3.2
\(\mu^*|_{\bar{\mathscr{B}}}\) is a measure.
To show \(\mathscr{O} \subset \bar{\mathscr{B}}\), it suffices to show that any closed set \(F \subset E\) is \(\mu^*\)-measurable. Since subadditivity already provides \(\mu^*(A) \le \mu^*(A \cap F) + \mu^*(A \backslash F)\), we only need the reverse inequality. Fix \(A \in 2^E\) and \(\varepsilon > 0\). By the definition of \(\mu^*\), there exists an open set \(O \supset A\) such that \(\mu(O) < \mu^*(A) + \varepsilon.\) Since \(A \cap F \subset O \cap F\) and \(A \backslash F \subset O \backslash F\), it is sufficient to show \(\mu(O) \ge \mu^*(O \cap F) + \mu(O \backslash F)\). Since \(O \backslash F\) is open, there exists \(g_2 \in C\) with \(0 \le g_2 \le 1\), \(\operatorname{supp}(g_2) \subset O \backslash F\), and \(\ell(g_2) > \mu(O \backslash F) - \varepsilon\). Let \(U = O \backslash \operatorname{supp}(g_2)\). Then \(U\) is an open set containing \(O \cap F\). By the definition of \(\mu(U)\), there exists \(g_1 \in C\) with \(0 \le g_1 \le 1\), \(\operatorname{supp}(g_1) \subset U\), and \(\ell(g_1) > \mu(U) - \varepsilon \ge \mu^*(O \cap F) - \varepsilon\). Since \(\operatorname{supp}(g_1) \cap \operatorname{supp}(g_2) = \emptyset\), we have \(0 \le g_1 + g_2 \le 1\) and \(\operatorname{supp}(g_1 + g_2) \subset O\). By the linearity of \(\ell\) and the definition of \(\mu(O)\),
\[ \mu(O) \ge \ell(g_1 + g_2) = \ell(g_1) + \ell(g_2) > \mu^*(O \cap F) + \mu(O \backslash F) - 2\varepsilon. \]
As \(\varepsilon \to 0\), we obtain \(\mu^*(A) \ge \mu^*(A \cap F) + \mu^*(A \backslash F)\). Thus \(F \in \bar{\mathscr{B}}\), which implies \(\mathscr{O} \subset \bar{\mathscr{B}}\). By Carathéodory's Extension Theorem, \(\bar{\mathscr{B}}\) is a \(\sigma\)-algebra containing \(\mathscr{B}\), and \(\mu^*|_{\bar{\mathscr{B}}}\) is a complete measure.
Step 4
By Lemma 1, we will show (*) is valid.
Assume without loss of generality that \(f \ge 0\) with its range in \([0, M]\). For any \(\delta > 0\), partition \([0, M]\) with \(0 = y_0 < y_1 < \dots < y_m = M\) such that \(y_i - y_{i-1} < \delta\). Define \(E_i = \{x \in E : y_{i-1} - \delta < f(x) < y_i + \delta\} \in \mathscr{O}\). By Lemma 1, there exists a partition of unity \(\{h_1, \dots, h_m\}\) subordinate to \(\{E_i\}\). Since \(f = \sum_{i=1}^m f h_i\) and \((y_{i-1} - \delta )h_i(x) \le f(x) h_i(x) \le (y_i + \delta)h_i(x)\), applying \(\ell\) yields
\[ \sum_{i=1}^m (y_{i-1} - \delta)\ell(h_i) \le \ell(f) \le \sum_{i=1}^m (y_i + \delta)\ell(h_i). \]
Integrating with respect to \(\mu\) yields identical bounds
\[ \sum_{i=1}^m (y_{i-1} - \delta)\int_E h_i(x)\mu(dx) \le \int_E f(x)\mu(dx) \le \sum_{i=1}^m (y_i + \delta)\int_E h_i(x)\mu(dx). \]
By the construction of \(\mu\) on open sets, \(\ell(h_i) = \int_E h_i(x)\mu(dx)\). Because \(y_i - y_{i-1} < \delta\), the upper and lower bounds are within \(2\delta \cdot \mu(E)\) of each other. Since \(\delta > 0\) is arbitrary, we conclude that \(\ell(f) = \int_E f(x)\mu(dx)\).
Theorem 1 can be extended to any continuous linear functional on \(C\), possibly non-positive due to the Banach lattice property of \(C\). Let \(C^+ = \{f \in C: f\ge 0\}\), a positive cone satisfying \(c_1f_1+c_2f_2 \in C^+\) for all \(c_1,c_2 \ge 0\) and \(f_1,f_2\in C^+\). It's easy to observe that \(C = C^+ - C^+\).
Theorem 2
Every continuous linear functional on \(C\) has a representation (*), where \(\mu\) is a signed Borel measure on \(E\) with finite total variation.
Proof
Step 1
For each \(f \in C^+\), define \(\ell^+(f)=\sup{\ell(g):0\leq g\leq f}.\) Clearly, \(\ell^+(f)\geq \ell(0)=0.\) Moreover, since \(\ell\) is continuous, for every \(g\) satisfying \(0\leq g\leq f\), we have \(|\ell(g)|\leq|\ell|,|g|\leq|\ell|,|f|.\) Thus, \(\ell^+(f)\) is finite. We will prove that
Claim 1,1
\(\ell^+\) is additive and positively homogeneous on \(C^+\).
Claim 1.2
\(\ell^+\) extends uniquely to a positive linear functional on \(C\).
To prove 1.1, let \(f_1,f_2\in C^+\). For any \(g_1,g_2\in C^+\) satisfying \(0\leq g_1\leq f_1,0\leq g_2\leq f_2,\) we have \(0\leq g_1+g_2\leq f_1+f_2.\) Therefore, \(\ell(g_1)+\ell(g_2) = \ell(g_1+g_2)\leq\ell^+(f_1+f_2).\) Taking the supremum over all admissible \(g_1\) and \(g_2\) gives \(\ell^+(f_1)+\ell^+(f_2)\leq\ell^+(f_1+f_2).\)
Conversely, let \(g\in C^+\) satisfy \(0\leq g\leq f_1+f_2.\) Define \(g_1=\min(g,f_1),g_2=g-g_1.\) Then \(0\leq g_1\leq f_1.\) Moreover, \(g_2 = g-\min(g,f_1) = \max(g-f_1,0),\) and since \(g\leq f_1+f_2\), it follows that \(0\leq g_2\leq f_2.\) Hence, \(\ell(g)=\ell(g_1)+\ell(g_2)\leq\ell^+(f_1)+\ell^+(f_2).\) Taking the supremum over all \(g\) satisfying \(0\leq g\leq f_1+f_2\), we obtain \(\ell^+(f_1+f_2)\leq\ell^+(f_1)+\ell^+(f_2).\) Therefore, \(\ell^+(f_1+f_2)=\ell^+(f_1)+\ell^+(f_2).\)
It remains to verify positive homogeneity. Let \(c\geq 0\). If \(c=0\), the assertion is immediate. If \(c>0\), then every \(g\) satisfying \(0\leq g\leq cf\) can be written as \(g=ch\) for some \(h\) satisfying \(0\leq h\leq f\). Hence,
\[ \begin{aligned} \ell^+(cf) & = \sup\{\ell(g):0\leq g\leq cf\} = \sup\{\ell(ch):0\leq h\leq f\}\\ & = c\sup\{\ell(h):0\leq h\leq f\} = c\ell^+(f). \end{aligned} \]
Thus, \(\ell^+\) is additive and positively homogeneous on \(C^+\).
To prove 1.2, let \(f\in C\). Since \(f=f^+-f^-\) where \(f^+=\max(f,0),f^-=\max(-f,0),\) define \(\ell^+(f) = \ell^+(f^+)-\ell^+(f^-).\) More generally, suppose that \(f\) has two decompositions \(f=f_1-f_2=g_1-g_2,\) where \(f_i,g_i\in C^+\) for \(i=1,2\). Then \(f_1+g_2=g_1+f_2.\) Since \(\ell^+\) is additive on \(C^+\), we have \(\ell^+(f_1)+\ell^+(g_2)=\ell^+(g_1)+\ell^+(f_2).\) Therefore, \(\ell^+(f_1)-\ell^+(f_2)= \ell^+(g_1)-\ell^+(g_2).\) Thus, the definition of \(\ell^+(f)\) is independent of the chosen decomposition. Let \(f,g\in C\). Choose decompositions \(f=f_1-f_2, g=g_1-g_2,\) with \(f_i,g_i\in C^+\). Then \(f+g=(f_1+g_1)-(f_2+g_2),\) and hence
\[ \begin{aligned} \ell^+(f+g) & = \ell^+(f_1+g_1)-\ell^+(f_2+g_2)\\ &= \ell^+(f_1)+\ell^+(g_1) -\ell^+(f_2)-\ell^+(g_2)\\ &= \ell^+(f)+\ell^+(g). \end{aligned} \]
Similarly, if \(c\geq 0\), then \(cf=cf_1-cf_2,\) so \(\ell^+(cf)=c\ell^+(f).\) If \(c<0\), then \(cf=(-c)f_2-(-c)f_1,\) and therefore \(\ell^+(cf)=c\ell^+(f).\) Thus, \(\ell^+\) extends uniquely to a linear functional on \(C\). Since \(\ell^+(f)\geq 0\) whenever \(f\in C^+\), this extension is positive.
Step 2
Define \(\ell^-=\ell^+-\ell.\) For every \(f\in C^+\), the function \(g=f\) is admissible in the definition of \(\ell^+(f)\). Hence, \(\ell^+(f)\geq \ell(f).\) Therefore, \(\ell^-(f)=\ell^+(f)-\ell(f)\geq 0.\) Thus, \(\ell^-\) is also a positive linear functional, and \(\ell=\ell^+-\ell^-.\) Moreover, every positive linear functional on \(C\) is continuous. Indeed, if \(f\in C\), then \(-|f|\mathbf 1\leq f \leq|f|\mathbf 1.\) By positivity, \(-|f|\ell^+(\mathbf 1)\leq\ell^+(f)\leq|f|\ell^+(\mathbf 1),\) and consequently, \(|\ell^+(f)|\leq\ell^+(\mathbf 1)|f|.\) The same argument applies to \(\ell^-\).
Step 3
By Theorem 1, there exist finite positive Borel measures \(\mu^+\) and \(\mu^-\) on \(E\) such that \(\ell^+(f)=\int_E f(x),\mu^+(dx)\) and \(\ell^-(f)=\int_E f(x),\mu^-(dx)\) for every \(f\in C\). Define \(\mu=\mu^+-\mu^-.\) Then \(\mu\) is a signed Borel measure, and for every \(f\in C\),
\[ \begin{aligned} \ell(f) = \ell^+(f)-\ell^-(f) = \int_E f(x)\mu^+(dx)- \int_E f(x)\mu^-(dx)= \int_E f(x)\mu(dx). \end{aligned} \]
Furthermore, \(|\mu|(E)\leq\mu^+(E)+\mu^-(E)<\infty.\) Hence, \(\mu\) has finite total variation. \(\square\)
Also, for a locally compact, second-countable Hausdorff space \(E′\), we know that it is paracompact, which means any open covering has a locally finite refinement that is also an open covering. In such a space, we have a similar proposition. Let’s denote all bounded continuous functions on \(E′\) as \(C′ = C_0(E′)\) where
\[ C_0(E') = \{\forall \varepsilon>0 \exists K \Subset E' \text{ such that } |f(x)|<\varepsilon \text{ for all }x \notin K.\} \]
Theorem 3
Every continuous linear functional on \(C'\) has a representation (*), where \(\mu\) is a locally finite signed Borel measure on \(E\) with finite total variation.
Lemma 2
Suppose \(\{E_\alpha\}_{\alpha\in I}\) is an open covering of some topological space \(E\). \(E\) is paracompact if and only if there exist continuous functions \(\{h_{\alpha}\}_{\alpha\in I}\) satisfying
- \(\forall \alpha \in I, \operatorname{supp}(h_\alpha) \subset E_\alpha\);
- \(\forall \alpha \in I, x\in E, 0\leq h_\alpha (x) \leq 1\);
- \(\forall x \in E,\) there exists finite index set \(I_0(x) = \{\alpha\in I: h_{\alpha}(x) \neq 0\}\) such that \(\sum_{\alpha \in I_0(x)} h_{\alpha}(x) =1\).
Based on Lemma 2, Theorem 3 can be seen as a combination of Theorem 1 and Theorem 2, whose proof is the same in essence.