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Intuitively I expect this to follow from countable additivity, but there are ideas I can't rule out such as:

  1. Select a real number r from the uniform distribution over [0, 1]. If r is exactly 0.5, then let q = 0, else let q = 1
  2. Select r as above. If r is rational let q = r, else let q = 0

EDIT: @GEdgar writes, "In both of your examples, q has discrete distribution. In example 1, q=1 with probability one. In example 2, q=0 with probability one." My thought was that a random variable that is "almost surely" zero might not be the same as one that is identically zero. But the truth is I have no idea what sort of tortured distribution might work here, much like I never would have come up with the Cantor distribution if someone had asked me whether all random real variables have to be discrete or continuous.

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    $\begingroup$ By definition, yes! $\endgroup$ – Frank Mar 28 '14 at 17:39
  • $\begingroup$ The "distribution" of a random variable is defined in terms of probability, $F(x) = P(X < x)$. So the distributions are the same for two random variables that agree almost surely. $\endgroup$ – GEdgar Mar 28 '14 at 21:44
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In both of your examples, $q$ has discrete distribution. In example 1, $q=1$ with probability one. In example 2, $q=0$ with probability one.

But what do your examples have to do with the proviso "over a countable set"?

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  • $\begingroup$ I think kuzzooroo's thought might be that $q$ can only take a value from a finite or countable set. $\endgroup$ – Henry Mar 28 '14 at 20:07
  • $\begingroup$ Yes, @Henry is right; I mean that the random variable in question can only take on a countable number of values. $\endgroup$ – kuzzooroo Mar 28 '14 at 20:24
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Well, if $\Omega = \{\omega_1,\ldots\}$, then as you say, countable additivity requires that $$ \sum_{i=1}^\infty \mathbb{P}(\{\omega_i\}) = \mathbb{P}(\Omega) = 1 \text{,} $$ so not all of the $\mathbb{P}(\{\omega_i)\})$ can be zero.

But that still allows some rather weird distribution functions, because you can pick weird mappings from $\Omega$ to $\mathbb{R}$. For example, let $\Omega$ be a countable probability space, and $\{q_1,\ldots\}$ be an enumeration of the rational numbers. Then define a random variable $X$ as $$ X \,:\, \Omega \to \mathbb{R} ,\:\, \omega_i \to q_i $$ Since every subset of $\Omega$ is measurable, $X$ surely is a valid random variable, but its cumulative distribution function (CDF) $$ F(x) = \mathbb{P}\left(X^{-1}\left((-\infty,x]\cap \mathbb{Q}\right)\right) $$ isn't going to be a particularly "nice" function. In particular $F$ is potentially discontinuous at every rational $x$ (at all of them if all the $\mathbb{P}(\{\omega_i\})$ are non-zero).

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  • $\begingroup$ So this is discrete but with a messy CDF, right? $\endgroup$ – kuzzooroo Mar 28 '14 at 20:23
  • $\begingroup$ @kuzzooroo Yes, exactly. $\endgroup$ – fgp Mar 28 '14 at 20:35

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