Decomposing a space to respect a countable collection of measures Let $X$ be a locally compact Hausdorff space. Let $\{ \mu_i\}_{i \in I}$ be a countable collection of finite regular Borel measures on $X$. 

Claim: There exists a partition $X = \bigsqcup_{j \in J} X_j$ of $X$ into countably many Borel sets such that each $X_j$ has the following property: all the restricted measures $\mu_i|_{X_j}$ which do not vanish are mutually absolutely continuous. 

I feel like this claim should follow from standard facts about decomposing measures. Is this true? Does anyone see a snappy proof?
 A: Actually I think this claim should be false, per examples of the following type.

Example: Let $X=[0,1]$. Consider the countable collection of all intervals of the form $[\frac{k}{2^n}, \frac{k+1}{2^n}]$ for $n=0,1,2,\ldots$ and $k=0,1,\ldots,2^{n-1}$. To each such interval, associate the measure on $[0,1]$ which is just Lebesgue measure with support restricted to the given interval.

With a bit of fooling around, one convinces oneself that $X=[0,1]$ does not admit a countable partition respecting the above countable collection of measures, in the sense of my question.
On the other hand, one can show fairly easily that this can be done with an uncountable partition of $X$, though this is not at all helpful for the purpose I'd had in mind. I'm writing an argument (certainly not the best one) for that below the fold, just so that I can record it somewhere. 

For simplicity, assume each $\mu_i$ is a probability measure. Define a new probability measure $\lambda$ by $\lambda = \sum_{i=1}^\infty 2^{-i} \mu_i$. By construction, $\mu_i \ll \lambda$ for every $i$, so we may choose a Radon-Nikodym derivative $\frac{d \mu_i}{d \lambda}$. Let $Y_i$ denote the set of points $x \in X$ where $\frac{d \mu_i}{d \lambda}$ is nonzero. The main thing to note about $Y_i$ is that $\mu_i$ is zero on $X \setminus Y_i$ and mutually absolutely continuous to $\lambda|_{Y_i}$ on $Y_i$. We can push a bit further by considering, for each subset $S \subseteq I$, the corresponding "Venn diagram region" $X_S = \left(\bigcap_{i \in S} Y_i\right) \cap \left( \bigcap_{i \notin S} X \setminus Y_i \right)$. Observe that, for all $i \in S$, we have that $\mu_i|_{X_S}$ is mutually absolutely continuous with $\lambda|_{X_S}$ whereas, for $i \in I \setminus S$, we have that $\mu_i|_{X_S}=0$. The $\{X_S\}_{S \subseteq I}$ form a partition of $X$ with the desired property, but sadly there are $2^{|I|}$ of them. 
