The problem is:
Let $f$ be Lipschitz with constant $M$. Show that if $E \subseteq [0,1]$ is of measure zero, then $f(E)$ is of measure zero.
My attempt:
Let $\epsilon > 0$, and choose a countable covering of $E$ by open intervals $\{U_i\}_{i=1}^\infty$ such that $\sum_{i=1}^\infty |U_i| < \epsilon/M$.
By the intermediate value theorem, $V_i := f((a_i,b_i))$ is also an interval, and for any two points $f(x), f(y) \in V_i$, we have $|f(x) – f(y)| \leq M|x-y| < M|U_i|$. Hence $|V_i| < M|U_i|$, and
$$
\sum_{i=1}^\infty |V_i| < M \sum_{i=1}^\infty |U_i| = M \cdot \frac{\epsilon}{M} = \epsilon.
$$
Since the $V_i$ cover $E$, we conclude that $f(E)$ is of measure zero.
Comments: I have made two large assumptions here: 1) that the $V_i$ cover $f(E)$, but I think this is true since the $U_i$ cover $E$, and 2) the estimate on the measure of $V_i$. I actually don't think that is correct.
Best Answer
As $f$ is $M$-Lipschitz, we know that $m(f(E)) \leq M m(E)$, where $m$ denotes the Lebesgue measure on $\mathbb{R}$. If $E$ has measure $0$, then $0 \leq m(f(E)) \leq M \cdot 0 = 0$.
The proof of $m(f(E)) \leq M m(E)$ goes exactly as your proof, which is correct (note you can prove something more general in the same way).