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" ?