Non-singular measure and uncountable family of pairwise disjoint positive measure sets

Question

First of all, by singular measure I mean a measure $\mu$ on $\langle X,\mathcal{S}\rangle$ (measurable space) for which there exists $x \in X$ s.t. $\{x\}\in\mathcal{S}$ and $\mu(\{x\}) > 0$. A measure is non-singular if it is not singular.

There is an exercise in my professor's notes whose solution I can't quite grasp. The exercise is the following

Show that if $\mu:\mathcal{S} \rightarrow [0;+\infty]$ is non-singular and $\{A_\alpha\mid \alpha < \kappa\}\subseteq\mathcal{S}$ are pairwise disjoint, then $\{\alpha < \kappa\mid \mu(A_\alpha) \neq 0\}$ is countable

Where it is understood that $\mathcal{S}$ is a $\sigma$-algebra, $\mu$ a measure and $\kappa$ is an uncountable cardinal.
It should be easy, but I can't relate the existence of such a sequence with the non-sigularity property of the measure.

I proved the thesis assuming that $\mu$ is finite, using the regularity of $\omega_1$ (and without assuming non-singularity). Any hint for the original statement?

Thanks

Answer 1

It’s not true in the general case. Equip $\mathbb{R}\times\mathbb{R}$ with the discrete sigma algebra and let $\mu(A)$ be $0$ if $A$ is countable and $\infty$ otherwise. Then $\mu$ is singular and $\{\{a\}\times \mathbb{R}: a\in\mathbb{R}\}$ is an uncountable collection of pairwise disjoint sets with non-zero measure.

Answer 2

The statement is false.

A simple counterexample is to take the counting measure on $\mathbb{R}$ and replace every number with a countable set to artificially make the measure non-singular. More formally: $X = \mathbb{R} \times \mathbb{N}$, $\mathcal{S} = \left\{ A \times \mathbb{N} : A \subseteq \mathbb{R} \right\}$ and $\mu(A \times \mathbb{N}) = |A| \in \mathbb{N} \cup \{ \infty \}$. Clearly $\{ \{ x \} \times \mathbb{N} : x \in \mathbb{R} \}$ are pairwise disjoint and the measure of each is $1$.

If we additionaly require that every singleton be in $\mathcal{S}$, then a slight modification will work as another counterexample. Take $X = \mathbb{R} \times [0, 1]$, $\mathcal{S} = \{ A \subseteq \mathbb{R} \times [0, 1] : (\forall x \in \mathbb{R}) \, A_x \text{ is a Borel subset of } [0, 1] \}$ and $\displaystyle \mu(A) = \sum_{x \in \mathbb{R}} \lambda(A_x)$, where $A_x = \{ y \in [0, 1] : (x, y) \in A \}$ means the usual section and $\lambda$ is the Lebesgue measure on $[0, 1]$. Then again $\{ \{ x \} \times [0, 1] : x \in \mathbb{R} \}$ is a family of pairwise disjoint measurable subsets, each of measure $1$.