Prove that $A$ is Jordan measurable and $v(A)=0$

Question

Let $A = \{1/n\mid n \in\mathbb N\} \subset\mathbb R.$ Prove that $A$ is Jordan measurable and $v(A) = 0.$

I understand what it means for it to be Jordan Measurable however I don't understand the steps needed to prove it for this particular question.

Answer 1

Let $\varepsilon$ be a small positive number. The set $[0,\varepsilon/2)$ contains all but finitely many members of the set $A.$ Let $m$ be the number of exceptions—the points of $A$ that are not within that interval. Each of those exceptions $x$ is within the interval $[x,x+\varepsilon/(2m)).$ So $A$ is a subset of the union of those $m+1$ intervals, whose Jordan measure is $\varepsilon.$

Thus no matter how small you make $\varepsilon>0,$ the set $A$ has Jordan measure not more than $\varepsilon.$