Q:

Locally Lipschitz maps maps sets with outer measure $0$ to sets with outer measure $0$.

Problem: Let $f:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ be locally Lipschitz. If $A\subset \mathbb{R}^{n}$ is measurable with $m^{*}(A)=0$, then $m^{*}(f(A))=0$.

I am trying to understand the proof of this problem, but I am a little confused by the hints that I was given.

Hint: Let $K$ be compact with Lipschitz constant $L_{K}$ show that $f$ maps any cube $Q_{r}$ of side length $r$ contained in $K$ to the set $f(Q_{r})$ which is contained in a cube of side length $L_{K}\sqrt{n}r$.

So I let $K\subset \mathbb{R}^{n}$ be compact with Lipschitz constant $L_{K}$. Then for any cube $Q_{r}\subset K$ with side length $r>0$, we have
    $$m^{*}(f(Q_{r}))\leq L_{K}^{n}m^{*}(Q_{r})=L_{K}^{n}r^{n}\leq L_{K}^{n}n^{\frac{n}{2}}r^{n}.$$
Here is the first part where I am confused. If I am trying to show that $m^{*}(f(A))=0$, why is $\sqrt{n}$ relevant to this at all? I understand that $A$ is a countable union of closed cubes, but I am lost on how this plays into the conclusion.

Hint: Explain why this implies: $m^{∗}(f(A \cap K)) \leq n^{n/2}L_{K}^{n}m^{∗}(A \cap K)$
  and conclude the proof.

It follows that the above inequality is true just by construction and definition of Lipschitz continuity, but again, I am not understanding why these extra factors are coming into play.
Lastly, I don't understand why I am approximating the measure by the inside with compact sets, since the outer measure is approximated by looking at the measure of open sets which $A$ is contained in.
I appreciate any suggestions to help me understand this problem.

A:

Let's start with the statement of the first hint. If $x,y\in Q_r$, then
$$|x-y|=\sqrt{\sum_{i=1}^n(x_i-y_i)^2}\le\sqrt{nr^2}=\sqrt{n}r,$$
so $|f(x)-f(y)|\le L_K\sqrt{n}r$ since $Q_r\subseteq K$. In particular, for fixed $x_0\in Q_r$, we have
$$f(Q_r)\subseteq B(x_0,L_K\sqrt{n}r)\subseteq x_0+[-L_K\sqrt{n}r,L_K\sqrt{n}r]^n$$
which is a cube of side length $2L_K\sqrt{n}r$.
[It seems reasonable to expect that a set of diameter at most $R$ be contained in a cube of side length $R$, but the proof is escaping me for now. In any case, the extra factor of $2$ will not matter.]
This implies that
$$m^*(f(Q_r))\le(2L_K\sqrt{n})^nr^n=(2L_K\sqrt{n})^nm^*(Q_r). $$
Next, fix $\epsilon>0$ and let $K_\epsilon$ be the closure of $K+B(0,\epsilon)$. Take a countable collection $\{Q_i\}$ of cubes such that $Q_i\subseteq\tilde K$, $A\cap K\subseteq\bigcup_iQ_i$, and
$$m^*(A\cap K)\ge\sum_im^*(Q_i)-\epsilon.$$
This is possible from the definition of $m^*$ (and by making each cube of side length smaller than $\epsilon$). Then we have
$$f(A\cap K)\subseteq f\left(\bigcup_iQ_i\right)\subseteq\bigcup_if(Q_i),$$
and so by monotonicity and subadditivity,
$$m^*(f(A\cap K))\le\sum_im^*(f(Q_i))\le(2L_{K_\epsilon}\sqrt{n})^n\sum_im^*(Q_i)\le(2L_{K_\epsilon}\sqrt{n})^n(m^*(A\cap K)+\epsilon).$$
Note that $L_{K_\epsilon}$ is an increasing function of $\epsilon$, so since $m^*(A)=0$ we may let $\epsilon\to0$ and obtain $m^*(A\cap K)=0$ for every compact set $K$. Now just take a countable collection of compact sets that cover $\mathbb R^n$, and you're done.

