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 $L$, [step 1.1, L2] 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, $\mathrm{\mid }\int fd\mu \mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }\mu \mathrm{\mid }(X)$, so integration is bounded with norm at most $\mathrm{\mid }\mu \mathrm{\mid }(X)$. The definition of total variation and regular approximation by compactly supported phase functions gives functions with $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$ and integrals arbitrarily close to $\mathrm{\mid }\mu \mathrm{\mid }(X)$; hence equality of norms.