# 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.

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 $$|f_+-g_+|\le |f-g|$$.

• Thanks! That was indeed not too hard. Clever use of Radon Nikodym...
– Dirk
Commented Oct 18, 2018 at 13:26
• Here you tacitly assume that both $|\mu|$ and $|\nu|$ are absolutely continuous with respect to the same positive measure $\lambda$ so you can invoke the densities $f$ and $g$. Commented Oct 18, 2018 at 13:27
• I think he just takes $\lambda =|\mu| + |\nu|$.
– Dirk
Commented Oct 18, 2018 at 13:40
• OK. I get now. Nice Commented Oct 18, 2018 at 14:06

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 $$\Vert \mu_+-\nu_+\Vert_*\leq \Vert\mu-\nu\Vert_*.$$