Lipschitz Images and Lebesgue Outer Measure in $\mathbb{R}^n$

We work in the setting of Product Measures and Lebesgue Measure on ℝⁿ: Standing Notation and Background, whose notation is fixed for every dimension and is used here with a natural number $n$ satisfying $1\le n$: the Euclidean norm $\lVert\,\cdot\,\rVert$, distance $d_{E}$ and notion of openness on $\mathbb{R}^{n}$, the Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R}^{n})$, Lebesgue measure $\lambda_{n}$ and its null sets, the constant $\sigma_{n}$ with $\sigma_{n}^{2}=n$, and the conventions on images $T(A)$ and on Lipschitz maps are all as fixed there. Write $\lambda_{n}^{\ast}$ for Lebesgue outer measure. Then the following hold.

1. (Outer measure bound) (i) Let $E\subseteq\mathbb{R}^{n}$ be nonempty, let $L\in\mathbb{R}$ with $0<L$, and let $T:E\to\mathbb{R}^{n}$ be Lipschitz with constant $L$ for the Euclidean metrics, that is

$$\lVert T(x)-T(y)\rVert\le L\,\lVert x-y\rVert\qquad\text{for all }x,y\in E .$$

Then for every $A\subseteq E$,

$$\lambda_{n}^{\ast}\bigl(T(A)\bigr)\le(2\,\sigma_{n}\,L)^{n}\,\lambda_{n}^{\ast}(A),$$

the right-hand side being read as $\infty$ when $\lambda_{n}^{\ast}(A)=\infty$.

2. (Null sets under a Lipschitz map) (ii) In the situation of claim 1, if $A\subseteq E$ is $\lambda_{n}$-null then $T(A)$ is $\lambda_{n}$-null.

3. (Null sets under a locally Lipschitz map) (iii) Let $U\subseteq\mathbb{R}^{n}$ be open, let $T:U\to\mathbb{R}^{n}$ be locally Lipschitz, and let $A\subseteq U$ be $\lambda_{n}$-null. Then $T(A)$ is $\lambda_{n}$-null.