# 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" ?

• Measure zero sets may have the same cardinality of the reals (an example is the Cantor set), or they can be dense (an example is the set of rationals). May 17 '15 at 9:04
• There are uncountable subsets of $\mathbb R$ that has (Lebesgue-)measure zero, see for instance math.stackexchange.com/questions/459849/… so the cardinalility does not have to be small. $\mathbb N$ has measure zero, I wouldn't call that concentrated. So neither is probably a good intuitive way of thinking of measure zero. My guess would be that f(A) is countable though. May 17 '15 at 9:04
• What exactly do you mean when you say you don't understand what this means? Do you at least know the definition of a zero-measure set? May 17 '15 at 9:25
• @JackM The definition is given in the definition of the exercise. My problem is that can't understand it intuitively. If you had to draw a zero-measure set, how would it look ? May 17 '15 at 9:51

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.