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Here is a question about decomposition of measures in singular parts and in positive and negative parts.

$\newcommand{\RR}{\mathbb{R}}$ Let $\Omega_{1/2}$ be compact subsets of $\RR^d$ equipped signed measures $\alpha_{1/2}$. Let $\lambda$ denote the Lebesgue measure on $\Omega_{1/2}$. We decompose $$ \alpha_i = f_i + \eta^+_i-\eta^-_i $$ where $f_i$ is the Radon-Nikodym derivative of $\alpha_i$ with respect to $\lambda$, and $\eta^+_i-\eta^-_i$ is the Hahn-Jordan decomposition of of the respective singular parts. Put differently, we decompose both $\alpha_i$s into an $L^1$-function, a positive singular measure and a negative singular measure.

Further define the outer sum of $\alpha_1$ and $\alpha_2$ by $$ \alpha_1\oplus\alpha_2 = \alpha_1\times \lambda + \lambda\times\alpha_2. $$

Does it hold that $$ (\alpha_1\oplus\alpha_2)^+ = \Big((f_1 + \eta^+_1)\oplus(f_2 + \eta^+_2)\Big)^+, $$ in other words, is the positive part of $\alpha_1\oplus\alpha_2$ unaffected, if we omit the negative part of the singular parts?

Is suspect that the answer is affirmative, but I have trouble calculating the positive part (I tried using the characterization $\mu^+(A) = \sup_{E\subset A}\mu(A)$ but the choice of $E$ get quite messy…)

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We have $$ \alpha_1\oplus\alpha_2 = (f_1+\eta_1^+)\times\lambda + \lambda\times\alpha_2 -\eta_1^-\times\lambda , \quad\quad\quad\quad (1) $$ and if we drop the $\eta_1^-$ from $\alpha_1$, then the last term is missing. This last term, however, is singular with respect to the first two terms: it's supported by a set $S_1\times\Omega_2$, with $$ \lambda(S_1)=|f_1|(S_1)=\eta_1^+(S_1)=0 . $$ This means that the negative part of $\alpha_1\oplus\alpha_2$ is just the sum of the negative part of the first two terms from the RHS of (1) and $\eta_1^-\times\lambda$. In other words, dropping this last term will not affect the positive part.

The full claim then follows by applying this observation twice.

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  • $\begingroup$ Seems about right. That's how to handle the two negative singular parts! Btw, have you seen this outer sum of measures before? It arises naturally in my work and I wonder if it is a structure that has been studied before... $\endgroup$
    – Dirk
    Dec 5, 2018 at 17:59
  • $\begingroup$ @Dirk: No, I haven't seen this before, but then most of my measures are spectral measures, on $\mathbb R$... $\endgroup$
    – Christian Remling
    Dec 5, 2018 at 20:31
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I do not think it is true - take d=1, the measures singular on say [-2,2], one being dirac(0)-dirac(1) and the other one dirac(-1)-dirac(0); then on the product space [-2,2]x[-2,2], the positive and negative parts of your outer product of the measures have support outside sets of the form AxB, where A and B contain 0, but not -1 or 1, and have same one dimensional Lebesgue measure since the outer product just vanishes there. So the outer product's Hahn-Jordan decomposition positive part has support outside of those sets.

On the other hand, the outer product of the positive parts of the measures doesn't vanish on such sets (the first term doesn't, while the second term does) and it is a singular positive measure on the product set with support strictly containing the support of the positive measure above.

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    $\begingroup$ Probably I don't get you counter example, but what do you mean by outer product? Is this what I called outer sum above? As far as I see, you proposed counterexample does not work... $\endgroup$
    – Dirk
    Dec 5, 2018 at 15:28
  • $\begingroup$ I apologize but I thought that you meant the product measure above when defining the outer sum; otherwise, obviously, my example doesn't work since I have no clue what the product there means. $\endgroup$
    – Conrad
    Dec 5, 2018 at 15:58

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