Is taking the positive part of a measure a continuous operation? Here is question I tried to answer for some time - it seems to be straightforward, but I have trouble figuring it out.
Let $\Omega$ be a compact domain in $\mathbb{R}^n$. For any signed Borel measure $\mu$ on $\Omega$ let $\mu_+$ denote its positive part (obtained by Hahn-Jordan decomposition). My question is:

Is taking the positive part a continuous operation, i.e. does $\mu^n\to \mu$ in the space signed Borel measures imply $\mu^n_+\to \mu_+$ in the same space?

Of course, the answer would be positive if
$$\|w^n_+-w_+\|_{\mathfrak{M}}\leq \|w_n-w\|_{\mathfrak{M}}$$
where $\|\cdot\|_{\mathfrak{M}}$ denotes the variation norm, but I could not see this.
 A: 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 $|f_+-g_+|\le |f-g|$.
A: 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
$$
\Vert \mu_+-\nu_+\Vert_*\leq \Vert\mu-\nu\Vert_*.
$$
