Diffeomorphisms and Lipschitz Condition

Let $A$ be open in $\mathbb R^n$. and let g: $A \to \Bbb R^n$ be locally Lipschitz.
Show that if the subset $E$ of $A$ has measure zero in $\mathbb R^n$, then $g(E)$ has measure zero in $\mathbb R^n$

Many Thanks.

Solutions Collecting From Web of "Diffeomorphisms and Lipschitz Condition"

Let $E\subset A$ have measure zero and put $C_k\subset \mathbb R^n$ be an increasing sequence of compacts such that
$$
A\subset \bigcup\limits_{k=1}^\infty C_k.
$$
Let us denote $E_k = E\cap C_k$ and a measure on $\mathbb R^n$ through $\mu$. If we show that $\mu (g(E_k)) = 0$ for all $k$ it clearly means that $\mu(g(E)) = 0$.

Since $E_k$ has a measure zero as a subset of $E$, for any $\delta>0$ there is a cover of $E_k$ with open balls $\{B(x_i,r_i)\}_{i=1}^m$ such that their total measure does not exceed $\delta$. So,
$$
E_k\subset \bigcup\limits_{i=1}^m B(x_i,r_i).
$$
Note that if $x\in E_k$ then there is $i = 1,\dots,m$ such that $|x-x_i|\leq r_i$. As a result, for any $y\in g(E_k)$ there is $i = 1,\dots,m$ such that $|y-g(x_i)|\leq \lambda_k r_i$ where $\lambda_k$ is a Lipschitz constant for $g$ restricted to $E_k$. As a result,
$$
\mu(g(E_k))\leq\mu\left(\bigcup\limits_{i=1}^m B(g(x_i),\lambda_k r_i)\right)\leq \lambda^n_k \delta.
$$
Since $\delta$ is arbitrary, we obtain that $\mu(g(E_k)) = 0$ and hence $\mu(g(E)) = 0$.