Trouble understanding what a measure-zero set is. To begin with some context, I haven't had any exposure to measure theory yet.
I solved the following problem.

A set $A\subset \mathbb R$ such that $\forall \epsilon >0$, there exists countably many open intervals $(I_n)$ such that $A\subset \cup_{n\in\mathbb N}I_n$ and $\sum \operatorname{length}(I)<\epsilon$ is called a measure-zero set.
Let $f\in C^1(\mathbb R,\mathbb R)$. Let $A=\{x\in\mathbb R, f'(x)=0\}$ Prove that $f(A)$ is a measure-zero set.

I fail to understand what this means intuitively. Does it mean that the cardinal of $f(A)$ is small ? Or does it mean $f(A)$ is a "concentrated set" ?
 A: The measure of a set is, essentially, its length. Thus a measure-zero set is one with no length. For example, a point, or a finite set of isolated individual points. The integers are also a zero-measure set. On the other hand, the measure of an interval is the length of that interval. Any set that contains an interval is of course not zero-measure, because its measure is at least the length of that interval.
Thus the answer to your question "If you had to draw a zero-measure set, how would it look ?" is basically that it would look like dust. A collection of isolated points.
All definitions of measure that I'm aware of start with intervals. The idea is that we definitely know what the length of an interval is, so we can build up a more general definition from there. For example, the length of a union of disjoint intervals "should be" the sum of their lengths. Along these same lines is the idea that a set "should have" a measure smaller than any set containing it. Thus if for any $\epsilon > 0$ you can find a set $A_\epsilon$ of measure $\epsilon$ which contains $A$, then $A$ must have zero-measure. That's where this exercice's measure-zero criterion is coming from.
