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…)



Positive part of "outer sums" of measures
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.


