"Hahn decomposition" for countable or arbitrarily many mutually singular probability measure Let $(\mu_i)_{i\in I}$ be a collection of mutually singular probability measure on $(\Omega, \mathcal F)$.(i.e. any two of them are singular to each other). I am not sure if the following statement is true:

There exists a partition $\Omega=\sum_{i \in I}\Omega_i$ (by $\sum$ I mean the disjoint union) such that $\mu(\Omega_i)=1, \forall i\in I$.

I can only prove this statements for the case when $I$ is finite(by induction). But how to prove (or disprove) that it holds for an arbitrary $I$? Should I use the transfinite induction(and how)?
If it is false for arbitrarily many mutually singular measures, is it still true for countably many singular measures?
Thanks in advance! 
 A: For a countable case you can use intersections. Namely, we say that $\mu_i\perp \mu_j$ if there exists a set $A_{ij}$ such that $\mu_i(A_{ij}) = 1$ and $\mu_j(A_{ij}) = 0$. Let's also assume that take $A_{ji} = \Omega\setminus  A_{ij}$. Now define $A_i :=\bigcap_{j\neq i} A_{ij}$, then by continuity of measure $\mu_i(A_i) = 1$. Also, these sets are mutually disjoint:
$$
A_i\cap A_{i'} = A_i := \bigcap_{j\neq i} A_{ij}\cap \bigcap_{j\neq i'} A_{i'j} \subseteq A_{ii'}\cap A_{i'i} = \emptyset.
$$
If there is some space left after you subtract all $A_i$, you can add it to $A_1$ - this won't affect anything. Hence, this decomposition holds in the countable case. 
For the uncountable case, there are two ways in which the intersection method can fail. First, the uncountable intersection may fail to be measurable - this is more or less doable to overcome if measures are inner regular. However, even in this case there's a second issue: how to assure that measure of the uncountable intersection is still full. This made me thought of the following counterexample. Let $\Omega$ be a unit interval with its Borel $\sigma$-algebra, and $\mu_i = \delta(i)$ for $i\in \Omega$, $\mu_{l}$ is the Lebesgue measure and $I = \Omega\cup \{l\}$. Then if you need to decompose $\mu_i$ for $i\neq l$ you must have $A_i = \{i\}$. However, it leaves no space for the support of the Lebesgue measure. 
So: yes for countable, no for uncountable, with counterexample available in a very nice probability space already.
