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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 $\mathbb{R}^{m\times n}$
  • $\partial A_j = A_j\cap B$.
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    $\begingroup$ If you allow uncountably many sets just to get what you claim you could make the Ah's singletons. So, by itself this cannot possibly tell much. I am afraid without further details there is not much specufuc to answer. $\endgroup$
    – user9072
    Commented Feb 21, 2012 at 1:51
  • $\begingroup$ Thanks for all the comments. I have added some more information to my question. Best, Julie $\endgroup$
    – Julie
    Commented Feb 21, 2012 at 3:11
  • $\begingroup$ Your additional info imply that $A_j$ is an open set which contains its boundary, making it closed. Did you mean to write that $A_j$ is the closure of an open set? $\endgroup$
    – Yoav Kallus
    Commented Feb 21, 2012 at 4:38
  • $\begingroup$ $R^{mxn}$ is $\sigma$−compact so an uncountable open cover has a countable subcover. So you're good, right? $\endgroup$
    – David Feldman
    Commented Feb 21, 2012 at 4:45
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    $\begingroup$ This question still doesn't make sense. You can't have uncountably many disjoint open balls in euclidean space. $\endgroup$
    – Dan Petersen
    Commented Feb 21, 2012 at 7:28

3 Answers 3

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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 $\mathbb{R}^{m\times n}$).

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  • $\begingroup$ Thanks, this illustrates a property of my problem that I had overlooked. It actually turns out that I have exactly one possible $A_j$. $\endgroup$
    – Julie
    Commented Feb 21, 2012 at 9:32
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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 $t>0$
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  • $\begingroup$ Hi Nick, Thanks for the help. I may be mistaken, but it seems that $\mathcal{U}$ does not contain the points on the boundaries of the Aj? $\endgroup$
    – Julie
    Commented Feb 21, 2012 at 6:49
  • $\begingroup$ Sorry, I meant Neil. $\endgroup$
    – Julie
    Commented Feb 21, 2012 at 7:28
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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.

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  • $\begingroup$ Thanks Gerald. I have added some additional information to my question. $\endgroup$
    – Julie
    Commented Feb 21, 2012 at 3:39

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