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Let $\mu$ and $\nu$ be two measures on the $\sigma$-Borel set $\mathcal{B}(\mathbb{D}^n)$.

We say that $\mu$ is a representing measure for some point $z \in \mathbb{D}^n$, if

$$\forall_{u \in A(\mathbb{D}^n)} \quad u(z) = \int u d\mu$$

where $A(\mathbb{D}^n)$ is the polydisc algebra.

We say that $\nu$ is singular with respect to $\mu$, if there are such two disjoint sets $A, B \subset \mathbb{D}^n$ that $A \cup B = \mathbb{D}^n$ and $\nu(A) = 0$, $\mu(B) = 0$. We denote $\nu \perp \mu$.

Have there ever been some papers about the relationship between representing measures and measures singular with respect to them? The case which is the most interesting for my advisor and me is the case for the only measure singular with respect to $\mu$ being the zero measure i.e. $\nu = 0$. If that is the case, is there anything more which we can tell about the measure $\mu$? Does it satisfy some unusual or extra properties in this case?

I tried finding something by Google Scholar but couldn't find anything related to these topics.

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  • $\begingroup$ I'm probably misunderstanding something, but for every ($\sigma$-finite) measure $\mu$, there is a $\nu\not= 0$, $\nu\perp\mu$. For example $\nu=\delta_z$ for suitable $z$ works. $\endgroup$
    – Christian Remling
    Commented Oct 19 at 15:07
  • $\begingroup$ @ChristianRemling I think I should've asked it from a different perspective: What properties does $\mu$ has to satisfy so that $\nu=0$ is the only measure singular with respect to $\mu$? I've changed the title of this question $\endgroup$
    – S-F
    Commented Oct 19 at 16:07
  • $\begingroup$ @S-F doesn't Christian's comment explain why what you want to happen never happens? $\endgroup$
    – Sam Hopkins
    Commented Oct 20 at 13:17
  • $\begingroup$ @SamHopkins Yes, unfortunately I agree. A pity, because if it were possible, the problem we're working on would've been solved. I understand that $\delta_z$ is a point measure for $z$ ($\delta_z (x) = 1$ for $x = z$, $\delta_z (x) = 0$ for $x \neq z$)? $\endgroup$
    – S-F
    Commented Oct 21 at 13:09

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