$U \subset \mathbb R^n$ is an open set, $f \in C^1(U, \mathbb R^m)$. If $E \subset U$ is a null set, then $f(E)$ is also a null set.

Question

I am trying to prove the following statement:

Suppose $U \subset \mathbb R^n$ is an open set, $f \in C^1(U, \mathbb R^m)$. If $E \subset U$ is a null set, then $f(E)$ is also a null set.

The professor gave a hint: "any open subset in $\mathbb R^n$ can be exhuasted from inside by a finite union of closed boxes."

This statement is of course false, as a finite union of closed sets is closed. Could the teacher have meant countable? That would help immensely to prove what I want to prove. It's obviously true for an uncountable union of closed boxes, but what about countable? Is this true? If so, why?

Answer 1

It is true for countable instead. I spent far too long creating this diagram in TikZ some time ago, it may help: Can you see how iterating this process gives the open set as a countable union of boxes? It additionally shows that each of the boxes can have rational (in fact dyadic) coordinates.

Answer 2

My guess is that he meant "Any open subset in $\Bbb{R}^n$ can be exhausted from inside by a sequence of sets, all of which are finite unions of closed boxes." The word "exhausted" in this context implies that one is considering a sequence of sets, not a single set.

Answer 3

If you know the Lindelof property, there is a simple proof: For every $x\in U,$ there exists an open cube $C_x$ such that $x\in C_x\subset \overline {C_x} \subset U.$ By the Lindelof property, there exists a countable subcollection $\{C_{x_1}, C_{x_2}, \dots \} \subset \{C_x|x\in U\}$ such that

$$U=\bigcup_{x\in U}C_x = \bigcup_{k=1}^{\infty}C_{x_k}.$$

We then have

$$U=\bigcup_{x\in U}C_x \subset \bigcup_{k=1}^{\infty}\overline {C_{x_k}} \subset U.$$

The collection $\{\overline {C_{x_1}}, \overline {C_{x_2}}, \dots \}$ thus has the desired property.