What does it mean to say "almost always" ? I have a set, $A$, of $m \times n$ matrices with certain properties and a subset $B$ of $A$. I would like to say that when randomly selecting such a matrix, I am "almost always" never in $B$. I can show that $A$ is the disjoint union of subsets, $A_j$, such that for each $A_j$, the measure of $A_j\cap B$ is 0. However, this union is uncountable. Is this enough to reasonably conclude that my randomly selected matrix will "almost always" never be in $B$?
Thanks for the help!
Best,
Julie  
Additional info: Each $A_j$ has the following properties : 


*

*the interior of $A_j$ is a nonempty open subset of $\mathbb{R}^{m\times n}$ 

*$\partial A_j = A_j\cap B$.

 A: Let $\mathcal{U}$ denote the set of all open balls with rational centre and rational radius that are contained in $A_j$ for some $j$.  This is a countable family with the same union as the original family $\{A_j : j\in J\}$.  Now for $U\in\mathcal{U}$ we have $U\cap B\subseteq A_j\cap B$ for some $j$ so $U\cap B$ has measure zero.  As $\mathcal{U}$ is countable and covers $B$ we conclude that $B$ has measure zero.
UPDATE:  The answer above is correct if the sets $A_j$ are open, as in an earlier version of the question.  The current version seems to be consistent with the following example in which $B$ does not have measure zero:


*
 
*$n=m=1$
 
*$B=[0,\infty)$
 
*$A_t=(-\infty,0]\cup\{t\}$ for $t>0$ 

A: I see some contradiction in your hypotheses: since the $A_j$'s are disjoint and have non empty interiors, the union should be at most countable (there can't exist more $A_j$'s than the cardinal of the set of rational points in $\mathbb{R}^{m\times n}$).
A: No, an uncountable partition will, in general, not be enough.  
For example, any set whatever can be partitioned into single-element subsets, each of measure zero.  
