The bounded complex dual of C_0(X) is regular complex measures

On the real vector space of real-valued functions put $A(u)=\mathrm{Re}L(u)$ and $B(u)=\mathrm{Im}L(u)$. Apply [L1], then [L2], to the positive decompositions of both $A$ and $B$. This gives finite regular signed measures $\alpha$ and $\beta$ representing $A$ and $B$. Put $\mu =\alpha +i\beta$. If $f=u+iv$, complex linearity gives $L(f)=L(u)+iL(v)$, whose real and imaginary parts agree exactly with those of $\int fd(\alpha +i\beta )$; hence $\mu$ represents $L$.

If two finite regular complex measures $\mu$ and $\nu$ represent $L$, then the real and imaginary signed parts of their difference $\mu -\nu$ integrate every real $C_c$ function to zero. For either signed part, move its negative Jordan component to the other side; the two resulting positive Radon measures have equal integrals on $C_c$. The positive-measure uniqueness in [L2] makes those positive measures equal, so both signed parts of $\mu -\nu$ vanish and $\mu =\nu$.

Conversely, the total-variation integral bound gives $\mathrm{\mid }\int fd\mu \mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }\mu \mathrm{\mid }(X)$, so integration is bounded with norm at most $\mathrm{\mid }\mu \mathrm{\mid }(X)$. If $\mathrm{\mid }\mu \mathrm{\mid }(X)=0$, then $\mu =0$ and equality is immediate. Otherwise fix $0<\epsilon <\mathrm{\mid }\mu \mathrm{\mid }(X)$. By [L3], choose a unit-bounded simple $s={\sum }_{j=1}^m{c}_j{\mathbf{1}}_{E_j}$, with disjoint Borel $E_j$ and $\mathrm{\mid }{c}_j\mathrm{\mid }\le 1$, such that $\mathrm{\mid }\int sd\mu \mathrm{\mid }>\mathrm{\mid }\mu \mathrm{\mid }(X)-\epsilon$. By [L5], choose compact $K_j\subseteq E_j$ and open $O_j\supseteq K_j$ with $\mathrm{\mid }\mu \mathrm{\mid }(E_j\setminus K_j)<\epsilon \mathrm{/}(8m)$ and $\mathrm{\mid }\mu \mathrm{\mid }(O_j\setminus K_j)<\epsilon \mathrm{/}(8m)$. Replace $O_j$ by $U_j:=O_j\setminus {\bigcup }_{i\mathrm{\ne }j}{K}_i$, still an open neighbourhood of $K_j$. By [L4], choose $0\le f_j\le {\mathbf{1}}_{U_j}$ in $C_c(X)$ with $f_j=1$ on $K_j$. Put $f:=\frac{{\sum }_j{c}_j{f}_j}{\max (1,{\sum }_j{f}_j)}\mathrm{.}$ Then $f\in C_c(X)$, $\mathrm{\mid }f\mathrm{\mid }\le 1$, and $f=s$ on ${\bigcup }_j{K}_j$, because on $K_j$ only $f_j$ is nonzero. Outside that union, $\mathrm{\mid }s-f\mathrm{\mid }\le 2$ and its support lies in ${\bigcup }_j(E_j\setminus K_j)\cup {\bigcup }_j(U_j\setminus K_j)$. Thus $\int \mathrm{\mid }s-f\mathrm{\mid }d\mathrm{\mid }\mu \mathrm{\mid }<\epsilon$, and the variation bound gives $\mathrm{\mid }\int fd\mu \mathrm{\mid }>\mathrm{\mid }\mu \mathrm{\mid }(X)-2\epsilon$. Since $\epsilon$ is arbitrary, $\mathrm{\parallel }L\mathrm{\parallel }\ge \mathrm{\mid }\mu \mathrm{\mid }(X)$, proving equality.