For a set having content zero/measure zero, how can the sum of the volumes of the cover be less than any arbitrary number? I'm having quite a bit of trouble understanding this intuitively. 
From what I've learned, a subset $A$ of $R^n$ has ($n$-dimensional) measure zero if for every $\epsilon > 0$, there is a countable cover $\{U_1, U_2,...\}$ of A by closed rectangles such that $\sum_{i=1}^{\infty} v(U_i) < \epsilon$.
And for Jordan measure otherwise known as content zero, there's the definition; a subset $A$ of $R^n$ has ($n$-dimensional) content zero if for every $\epsilon > 0$, there is a finite cover $\{U_1,...U_n\}$ of A by closed rectangles such that $\sum_{i=1}^{n} v(U_i) < \epsilon$.
Now the idea of A having a cover, i.e A being contained in the union of all $U_i$ makes sense to me. Where I'm confused is the idea of the sum of all the volumes being $< \epsilon$ for any $\epsilon > 0$. If it's a cover, no matter how we choose the intervals, isn't the sum the same?
Not too sure where the flaw in my understanding is but consider $I_1=(a_1,b_1), I_2=(a_2,b_2)$ which has a given sum and is a cover of some set. If we refine this further to say $I_1=(a_1,b_1), I_2=(a_2,b_2), I_3=(a_3,b_3)$, isn't the sum still the same? So that we don't have the sum of the volumes less than any given $\epsilon$.
Some clarification so that this is more intuitive would be appreciated!
 A: Let's think the most trivial case, $A=\{0\}$ in $\mathbb R$. It's clear that $m(A)=0$. Why is that?
For example, $[-1,1]$ is a countable cover of $A$, so $m(A)\leq m[-1,1]=2$. But $[\frac{-1}{2},\frac12]$ is also a countable cover of $A$, which lead us to $m(A)\leq 1$. You can pick up a smaller cover for $A$, $[\frac{-1}{4},\frac14]$, so $m(A)\leq \frac12$.
This is (more or less) an intuitive approach. You can pick a countable cover whose measure is as close to zero you want (that is, smaller than any $\varepsilon$ you pick).
A: It's not the same cover for each $\epsilon$!
For simplicity, ignore content and let $n=1$. A set $A$ has measure $0$ if for each $\epsilon>0$ there is some cover $\mathcal{C}_\epsilon=\{U_i^\epsilon: i\in\mathbb{N}\}$ of $A$ with the sum of the lengths of the intervals $U_i^\epsilon$ less than $\epsilon$. But that cover is allowed to change with $\epsilon$!
No single cover "witnesses" that $A$ has measure zero; instead, the existence of "arbitrarily small" covers does.
