$E\subset \mathbb R^n$ is Lebesgue measurable. A continuous injection $f:E\to \mathbb R^n$ maps sets of zero measure to sets of zero measure. Prove: If $m(f(E))$ is finite, then for any $\varepsilon＞0$, there exists $\delta＞0$, such that for any measurable subset $Z\subset E$ with $m(Z)＜\delta$, we have $m(f(Z))＜\varepsilon$.
I find this problem is similar to the property of absolute continuous functions, but the method seems not working in this problem. I don't know how to solve this problem.
If not, there exists $\epsilon >0$ and sets $Z_n$ with $m(Z_n)<\frac 1 {2^{n}} $ and $m(f(Z_n)) \geq \epsilon$ for all $n$. Let $Z=\lim \sup Z_n$. Using injectivity of $f$ check that $f(Z)=\lim \sup f(Z_n)$. Now, $\sum_n m(Z_n) <\infty$ and this implies $m(Z)=0$. But $m(f(Z))=m(\lim \sup f(Z_n))\ge \lim \sup m(f(Z_n)) \ge \epsilon$, a contradiction.
[ The fact that $m(f(E)) <\infty$ is required to say that $m(\lim \sup f(Z_n))\ge \lim \sup m(f(Z_n))$. Apply Fatou's Lemma to the complements].