How to argue that discrete random variables do not have a Radon–Nikodym density? Suppose I have a discrete random variable $X\in\{1,2,3,4\}$ with a probability mass $\mu(X=k)=1/4$ for $k=1,2,3,4$. How to rigorously argue that it doesn't have a Radon-Nikodym density with respect to Lebesgue measure $\lambda$?
One argument I have in mind is as follows: as having a Radon–Nikodym density is equivalent to being absolutely continuous with respect to the Lebesgue measure $\lambda$, so we look at whether $X$ is absolutely continuous. It's obvious that $X$ is not absolutely continuous, because for measurable set $[1,1]=1$, the Lebesgue measure is $\lambda([1,1])=\lambda(1)=0$, which doesn't imply $\mu(1)=0$. Is this argument correct? If not, how to correct it (and make it fully rigorous)? Thanks!
(another related question: does there exist any measure $c$, such that a discrete random variable have a density with respect to $c$? Thanks!)
 A: Your argument is okay, but talking about "absolutely continuous" may obscure the fact that the statement really has a trivial proof.
Suppose $\mu$ did have a Radon-Nikodym density; call it $f$.  Then it would be the case, for instance, that $$\frac{1}{4} = \mu(\{1\}) = \int_{\{1\}} f\,d\lambda.$$  But $\lambda(\{1\}) = 0$ so $\int_{\{1\}} g \,d\lambda = 0$ for any measurable function $g$ whatsoever, and in particular for $g=f$.  Thus we have a contradiction.
As a more general comment, the Radon-Nikodym theorem says that one measure is absolutely continuous with respect to another if and only if it has a Radon-Nikodym density.  But keep in mind that the backward direction is trivial; it's the much harder forward direction that makes the theorem worthy of being "named".  This is what some authors (e.g. Douglas West in his Graph Theory text) would call a "TONCAS" theorem: The Obvious Necessary Condition is Also Sufficient. So you sound a little bit silly if you call upon a big-name theorem to use only the direction which is obvious.
