Definition: The volume of a bounded set $A\subset\mathbb R^n$ whose characteristic function $1_A$ is integrable over $\mathbb R^n$ is $\int_A 1_A$.
I'm looking for an example of a countably infinite bounded set that has a non-zero volume. (I know that such a set necessarily has measure zero.) This will then show that a countable set need not have volume zero.
If a set is Jordan measurable, then it is Lebsgue measurable and their measures agree. Thus, you cannot have a countable set that is Jordan measurable and is not of Jordan, hence Lebesgue, measure zero.
To clarify Let $A\subseteq {\bf R}^n$ be bounded. Without mention to the Lebesgue integral, we can define the Jordan content (or measure) of $A$ to be the value of the Riemann integral $$c(A)=\int_R \chi_A$$ where $R$ is any closed (bounded) rectangle (or "block", if you may) containing $A$ (observe Riemann integrals aren't defined for sets that are not closed blocks). We can then extend the definition of the Riemann integral, declaring a function $f:B\subseteq\Bbb R^n\to\Bbb R$ to be Riemann integrable over a Jordan measurable set $A\subset B$ if $$\int_B f\cdot\chi_A$$ exists, in which case we denote this by $$\int_A f$$ In particular, the Jordan measure of $A$ is $$\int_A 1$$
As a side note, from Lebesgue's criterion of Riemann integrability, it follows that $A$ is Jordan measurable iff it's boundary $\partial A$ has measure zero: this is where $\chi_A$ is discontinuous. Since $\partial A$ is compact (being bounded, and closed), it has Lebesgue measure zero iff it has Jordan content zero, namely, iff given $\varepsilon >0$ it can be covered by finitely many open blocks whose volumes sum to less than $\varepsilon$.
Consequently, if the Riemann integral $$\int_R \chi_A=\int_A 1$$ exists then it exists as a Lebesgue integral, and it equals, sometimes by definition, sometimes as a theorem, the Lebsgue measure of $A$. Thus, if a set is Jordan measurable, it is Lebesgue measurable, and its measures coincide.