Is taking the positive part of a measure a continuous operation?

If $\mu =f\cdot \lambda$ for a positive measure $\lambda$ (i.e., $\mu(A)=\int_A fd\lambda$), isn't then $\mu_+= f_+ \cdot\lambda$ (where $f_+$ is the positive part $\max\{f,0\}$ of $f$) and $\|\mu\|=\int|f|d\lambda$? Then $\|\mu_+-\nu_+\| \le \|\mu-\nu\|$ just follows from Radon-Nikodym (applied to $\lambda=|\mu|+|\nu|$) and $|f_+-g_+|\le |f-g|$.

The Borel measures on $\Omega$ can be identified with continuous linear functionals $\newcommand{\bR}{\mathbb{R}}$ $\mu:C(\Omega)\to\bR$. Assume that $\Omega$ is compact so $C(\Omega)$ is Banach space. Denote by $\Vert-\Vert$ the sup norm on $C(\Omega)$ and by $\newcommand{\eM}{\mathscr{M}}$ $\eM(\Omega)$ the dual of $C(\Omega)$ equipped with the dual norm $$ \Vert \mu\Vert_*:=\sup_{\Vert f\Vert\leq 1}|\mu(f)|. $$ Set $$ C(\Omega)_+:=\big\{ f\in C(\Omega):\;\; f(x)\geq 0,\;\;\forall x\in\Omega)\big\}. $$Let $\mu\in \eM(\Omega)$ and $f\in C(\Omega)$. Then, for any $f\in C(\Omega)_+$ we have (see Theorem 4.3.2. of R.E. Edwards: Functional Analysis) $$ \mu_+(f)=\sup_{0\leq g\leq f} \mu(g). $$ Let $\mu,\nu\in \eM(\Omega)$, and $f\in C(\Omega)_+$. Then for any $0\leq g\leq f$ we have

$$ \mu(g)-\nu(g)\leq \Vert\mu-\nu\Vert_*\Vert g\Vert \leq \Vert\mu-\nu\Vert_*\Vert f\Vert. $$ Hence, for any $0\leq g\leq f$ $$ \mu(g)\leq \Vert\mu-\nu\Vert_*\Vert f\Vert+\nu(g), $$ and, symmetrically, $$ \nu(g)\leq \Vert\mu-\nu\Vert_*\Vert f\Vert+\mu(g). $$ Taking the sup on both sides of the above inequalities we deduce $$ \mu_+(f)\leq \Vert\mu-\nu\Vert_*\Vert f\Vert+\nu_+(f)\implies \mu_+(f)- \nu_+(f)\leq \Vert\mu-\nu\Vert_*\Vert f\Vert, $$ $$ \nu_+(f)\leq \Vert\mu-\nu\Vert_*\Vert f\Vert+\mu_+(f)\implies \nu_+(f)- \mu_+(f)\leq \Vert\mu-\nu\Vert_*\Vert f\Vert. $$ Hence $$ \big\vert (\mu_+-\nu_+)f\big\vert\leq \Vert\mu-\nu\Vert_*\Vert f\Vert. $$ This implies $$ \Vert \mu_+-\nu_+\Vert_*\leq \Vert\mu-\nu\Vert_*. $$