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 $\nu \perp \mu$.
A null set is a non-zero set $E$ with measure zero. In this case, because we are in the $n$ dimensional case, we will be talking of $E := E_1 \times \ldots \times E_n$, where $E_i$ is a null-set each.
My question:
Is it always the case, that a measure $\nu$ singular with respect to all representing measures $\mu$ for some point $z \in \mathbb{D}^n$ is always concentrated on $E$?
I tried finding some information about this topic, but unfortunately, I wasn't able to find any papers which could be useful for me, even after searching for hours. I really hope someone will be able to help me here.
The only thing I've found is that the singular measure from the Lebesgue decomposition is concentrated on null sets, but here, I am talking about a specific case of a singular measure (not related to the Lebesgue decomposition at all).
