Riesz–Markov–Kakutani (RMK) representation theorem relates a (not necessarily continuous) positive linear map $\Lambda:\mathcal C_c (X) \to \mathbb C$ with a unique non-negative Radon measure on $X$. On the other hand, every finite signed Borel measure $\mu$ on $X$ can be decomposed uniquely as $\mu = \mu^+ – \mu^-$ with $\mu^+,\mu^-$ being non-negative finite Borel measures.
Let $\Lambda:\mathcal C_c (X) \to \mathbb C$ be a linear map.
Can we decompose $\Lambda$ into $\Lambda = \Lambda^+ – \Lambda^-$ with $\Lambda^+, \Lambda^-:\mathcal C_c (X) \to \mathbb C$ being positive?
If yes, we can apply RMK theorem to get $\mu^+$ representing $\Lambda^+$, and $\mu^-$ representing $\Lambda^-$. Then $\mu^+ – \mu^-$ will represent $\Lambda$.
(Rudin) Let
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$X$ be a locally compact Hausdorff space,
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$E:= \mathcal C_c (X)$ the complex vector space of all complex-valued continuous maps on $X$ with compact support, and
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$\Lambda:E \to \mathbb C$ (not necessarily continuous) positive linear. Here $\Lambda$ is positive means $f \in E \text{ s.t. } f(X) \subset \mathbb R_{\ge 0}$ implies $\Lambda (f) \in \mathbb R_{\ge 0}$.
Then there exists a $\sigma$-algebra $\mathfrak{M}$ on $X$ which contains all Borel sets of $X$, and there exists a unique non-negative measure $\mu$ on $\mathfrak{M}$ such that
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$$\Lambda(f) = \int_X f \mathrm d \mu \quad \forall f \in E.$$
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$\mu(K) < \infty$ for every compact set $K \subseteq X$.
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$\mu(E) = \inf \{\mu(V) \mid E \subset V, V \text{ open}\}$ for all $E \in \mathfrak{M}$.
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$\mu(E) = \sup \{\mu(K) \mid K \subset E, K \text{ compact}\}$ for all open set $E \subseteq X$ and for all $E \in \mathfrak{M}$ with $\mu(E) <\infty$.
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If $E \in \mathfrak{M}$ such that $A \subset E$, and $\mu(E)=0$, then $A \in \mathfrak{M}$.
Best Answer
The answer is no, for in the case of a compact space $X$, an affirmative answer would imply that every linear functional on $C(X)$ is continuous, and this is clearly false (under the axiom of choice).