$U \subset \mathbb R^n$ 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. 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?
 A: 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.
A: 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.
A: 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.
