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Section 4.1 of my text, "Elementary Probability Theory with Stochastic Processes" by Chung, states that a discrete random variable is a numerically valued function on a countable sample space.

However, I have also seen definitions that state that a random variable is a measurable function.

Will discrete RVs automatically be measurable?

On one hand, I think that they should be measurable since countable sets have measure zero.

On the other hand, what if we choose our sigma-algebra $\cal{F}$ such that $X^{-1}(B)\not \in \cal{F}$ for some measurable $B$ in the range of $X$?

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    I would define a "discrete random variable" as a random variable that can take a finite or countably infinite number of values in $\mathbb{R}$. – 2017-01-14
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    I am not sure how this definition is consistent with definition that a random variable is a measurable function from the sample space into the reals. – 2017-01-14
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    Yes, so, a discrete random variable is a measurable function from the sample space to the reals that can take an at most countably infinite number of values. By definition of random variable, all sets of the form $X^{-1}((-\infty, t])$ are assigned a probability measure (for all $t \in \mathbb{R}$). For discrete random variables it suffices to assign a probability mass for each discrete value that can be taken. – 2017-01-14
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    The definition I give is not the same as your book: (i) The book wants the sample space $S$ to be finite or countably infinite, while I want $\{r \in \mathbb{R} : r = X(\omega) \mbox{ for some $\omega \in S$}\}$ to be finite or countably infinite, (ii) The book implicitly assumes you use the sigma-algebra of all subsets of $S$, and you have a prob mass for each outcome in $S$. You are correct that if $S=\{a,b,c\}$ and the sigma-algebra is $\mathcal{F}=\{S, \phi\}$, then the map $X:S\rightarrow\mathbb{R}$ given by $X(a)=1, X(b)=2, X(c)=3$ is not measurable and so is not a random variable. – 2017-01-14

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