Extending a continuous complex valuation to a complex Borel measure

Let $$X$$ be a locally compact Polish space and let $$\mathcal{O}(X)$$ be the set of open subsets of $$X$$. A complex valuation on $$X$$ is a function $$v: \mathcal{O}(X) \to \mathbb{C}$$ such that $$v(\varnothing)=0$$ and $$v(U \cup V)+v(U\cap V) = v(U)+v(V)$$. Further, $$v$$ is called continuous if for a sequence $$U_0 \subseteq U_1 \subseteq U_2 \subseteq \dots$$ of open sets we have that $$v(U_n)$$ converges to $$v(\bigcup_i U_i)$$ for $$n \to \infty$$.

Can a continuous valuation be extended to a complex Borel measure on $$X$$?

For valuations with values in $$[0,+\infty)$$ that are monotonous, I believe this follows from Theorem 4.4 in Extension of valuations on locally compact sober spaces by Alvarez-Manilla.

In the special case $$X = \mathbb{R}$$, I believe the answer is yes. I'm still interested in more general cases as well.

The proof for $$X = \mathbb{R}$$ goes as follows. Comments, shorter proofs or references are welcome.

Suppose that $$v : \mathcal{O}(X) \to \mathbb{C}$$ is a continuous complex valuation. Then we can define its real part $$\mathrm{Re}(v)$$ and imaginary part $$\mathrm{Im}(v)$$ via $$\mathrm{Re}(v)(U) = \mathrm{Re}(v(U))$$ and $$\mathrm{Im}(v)(U) = \mathrm{Im}(v(U))$$. This gives two valuations with values in $$\mathbb{R}$$, and we can write $$v = \mathrm{Re}(v) + i\, \mathrm{Im}(v)$$. So if we want to show that $$v$$ can be extended to a complex Borel measure, we can assume without loss of generality that $$v$$ takes values in $$\mathbb{R}$$.

We now show that there are only countably many points that have a "point mass" associated to it. Let $$v : \mathcal{O}(X) \to \mathbb{R}$$ be a continuous valuation, and define $$w(x) = v(\mathbb{R})-v(\mathbb{R}-\{x\})$$ for every real number $$x$$. Further, we write $$W = \{ x \in \mathbb{R} : w(x) \neq 0\}$$. We show that $$W$$ is countable. If it was not, then there would be an $$N \in \mathbb{N}$$ such that $$W’ = \{ x \in \mathbb{R} : w(x) > 1/N \}$$ is still uncountable. Uncountable subsets of $$\mathbb{R}$$ have a limit point, so we take a limit point $$\tilde{x} \in W'$$ and a sequence $$(x_n)_n$$ in $$W’$$ such that $$x_n$$ converge to $$\tilde{x}$$ and such that $$W’’ = \{ x_1, x_2, x_3, \dots \} \cup \{ \tilde{x} \}$$ is homeomorphic to $$\{0,1,1/2,1/3,\dots\} \cup \{ 0 \}$$. Now set $$U_0 = \mathbb{R} - W’’$$ and $$U_n = U_0 \cup \{x_1, \dots, x_n\}$$ for each $$n>0$$. Then $$U_n \cap (\mathbb{R} - \{x_n\}) = U_{n-1}$$ and $$U_n \cup (\mathbb{R}-\{x_n\}) = \mathbb{R}$$, so we see that $$v(U_{n-1}) + v(\mathbb{R}) = v(U_n) + v(\mathbb{R}-\{x_n\})$$, and as a result $$v(U_n) = v(U_{n-1}) + w(x_n) \geq v(U_{n-1}) + 1/N$$. But because $$v$$ is continuous and $$U_n$$ is an increasing union of open subsets, $$v(U_n)$$ should converge, a contradiction. So $$W$$ is countable.

We now show that $$v$$ extends to a (signed) Borel measure on $$\mathbb{R}$$. Define the cumulative distribution function $$F(t) = v(\mathbb{R})-v((t,+\infty))$$. Continuity of $$v$$ shows that $$F$$ is right continuous. We want to construct a (signed) Borel measure that has $$F$$ as cumulative distribution function. Since $$F$$ is right continuous, a Borel measure like this exists if and only if $$F$$ is of bounded variation (this is explained well here). Suppose that $$F$$ is not of bounded variation. Then we can find $$a_0 < a_1 < \dots < a_n$$ such that $$\sum_{i = 0}^{n-1} | F(a_{i+1}) - F(a_i) | > 1$$. We claim that we can assume $$a_0,\dots,a_n \notin W$$, for $$W$$ as defined above. Then because $$a_i \notin W$$ for each $$a_i$$, we find that $$F(a_{i+1}) - F(a_i) = v((a_i,a_{i+1}))$$ for each $$i$$. To make sure that $$a_i \notin W$$, we can replace $$a_i$$ by $$a_i + \epsilon_i \notin W$$ for $$\epsilon>0$$, with $$\epsilon$$ small enough such that still $$\sum_{i = 0}^{n-1} |v((a_i+\epsilon_i,a_{i+1}+\epsilon_{i+1}))| = \sum_{i = 0}^{n-1} | F(a_{i+1}+\epsilon_{i+1}) - F(a_i+\epsilon_i) | > 1$$. This is possible because $$W$$ is countable and $$F$$ is right continuous. If we now consider the disjoint open intervals $$I_1^{(0)},\dots,I_{n_0}^{(0)}$$ given by $$(-\infty, a_0+\epsilon_0)$$, $$(a_0+\epsilon_0, a_1+\epsilon_1)$$, …, $$(a_{n-1}+\epsilon_{n-1}, a_n + \epsilon_n)$$, $$(a_n+\epsilon_n,+\infty)$$, then we still have $$\sum_{j=1}^{n_0} |v(I_j)| > 1$$.

Now $$F$$ again fails to be of unbounded variation over at least one of the intervals $$I^{(0)}_j$$. We choose precisely one such interval, and we subdivide this particular interval further in the way described above, such that we end up in total with a new set of open intervals $$I_1^{(1)}, \dots, I_{n_1}^{(1)}$$ that again partition $$\mathbb{R}$$ (up to finitely many points) and such that $$\sum_{j =1}^{n_1} |v(I_j)| > 2$$. By repeating this process, we make in step $$k$$ an open partition $$I_1^{(k)}, \dots, I_{n_k}^{(k)}$$ of $$\mathbb{R}$$ (up to finitely many points) with $$A(k) = \sum_{j=1}^{n_k} |v(I_j^{(k)})| > k$$.

After renumbering, we can assume that the interval in step $$k$$ that gets subdivided in step $$k+1$$, is precisely the interval $$I_1^{(k)}$$. Further, by choosing the intervals small enough in each step, we can assume that $$v(I_1^{(k)})$$ converges to a constant $$C \in \mathbb{R}$$ as $$k$$ goes to infinity. Now let $$\Omega$$ be the set of intervals that do not have a further subdivision, i.e. the intervals of the form $$I_j^{(k)}$$ for some $$k$$ and $$j\geq 2$$. Then $$\sum_{I \in \Omega} v(I)$$ converges unconditionally to $$v(U)$$ with $$U = \bigsqcup_{I \in \Omega} I$$. But then the series is also absolutely convergent by Riemann's Rearrangement Theorem, so $$A = \sum_{I \in \Omega} |v(I)| < +\infty$$. We now see that $$k < A(k) = \sum_{j=1}^{n_k} |v(I_j^{(k)})| \leq A + |v(I_1^{(k)})|$$. Because $$|v(I_1^{(k)})|$$ converges to $$C$$ for $$k \to +\infty$$, this gives a contradiction.

Because $$F$$ is right continuous and of bounded variation, there is a continuous Borel measure $$\mu$$ such that $$F(t) = \mu((-\infty, t])$$. It remains to show that $$\mu$$ restricts to $$v$$ on open subsets. First one can show that $$\mu(\mathbb{R}) = v(\mathbb{R}) = \lim_{x\to+\infty} F(x)$$. Then from the definition of $$F$$ it follows that $$\mu((a,+\infty))=v((a,+\infty))$$ for all $$a \in \mathbb{R}$$. Further, \begin{align*}\mu((-\infty,b)) &= \lim_{\epsilon \to 0} \mu((-\infty,b-\epsilon]) = \lim_{\epsilon \to 0} F(b-\epsilon) = \lim_{\epsilon \to 0} v(\mathbb{R}) - v((b-\epsilon,+\infty)) \\ &= \lim_{\epsilon \to 0} v((-\infty,b-\epsilon)) = v((-\infty,b))\end{align*}, where in the second last equality we use that we can take the limit over those $$\epsilon > 0$$ with $$b-\epsilon \notin W$$. Since $$\mu$$ and $$v$$ agree on the intervals $$(a,+\infty)$$ and $$(-\infty,b)$$, they agree on all open sets.