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.


2 Answers 2


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

  • $\begingroup$ Thanks! That was indeed not too hard. Clever use of Radon Nikodym... $\endgroup$
    – Dirk
    Oct 18, 2018 at 13:26
  • $\begingroup$ 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 $g$. $\endgroup$ Oct 18, 2018 at 13:27
  • $\begingroup$ I think he just takes $\lambda =|\mu| + |\nu|$. $\endgroup$
    – Dirk
    Oct 18, 2018 at 13:40
  • $\begingroup$ OK. I get now. Nice $\endgroup$ 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 $$ \Vert \mu_+-\nu_+\Vert_*\leq \Vert\mu-\nu\Vert_*. $$


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