Verifying that Lipschitz functions carry measure 0 sets to measure 0 sets Let $f: [a,b] \rightarrow \mathbb{R}$  be a Lipschitz function.
Let $\epsilon > 0$.
Let $E$ be a set of measure zero.  There exists countable, bounded, open intervals with the form $I_n=(x_n, y_n)$ such that $E \subseteq \bigcup\limits_n I_n$ and for some $x_k, y_k \in I_k$, then $|f(x_k)-f(y_k)| \leq c|x_k -y_k|$.  I believe this sufficiently incorporates the definition of Lipshitz.
Now I want to cover $E$ with the intervals such that $\sum\limits_n l(I_k) < \frac{\epsilon}{c}$.
So, for every n we pick $I_{n}^{'} \subset f(I_n)$ which are open, bounded, and countable such that $l(I_{n}^{'}) \leq c l(I_{n})$.  The collection $\{I_{n}^{'}\}$ is a cover for $f(E)$ and $\sum\limits_n l(I_{n}^{'}) \leq \sum\limits cl(I_{n})< c\frac{\epsilon}{c} =\epsilon$.
 A: Here is the proof for absolutely continuous functions, which Lipschitz functions are a special case of.
Let $f$ be an absolutely continuous function on $\mathbb{[a,b]}$ and let $A$ be a set of measure zero.  By outer regularity, for any $\delta > 0$ there exists a countable disjoint set of open intervals $(a_i,b_i)$ so that $A\subseteq \cup_i (a_i,b_i)$ and $\sum_i b_i - a_i < \delta$.  
Now fix $\epsilon > 0$, by absolute continuity there is a $\delta > 0$ so that for any finite collection of disjoint intervals $(x_i, y_i)$ with $\sum y_i - x_i < \delta$ we have $\sum_i |f(y_i) - f(x_i)| \leq \epsilon$.  It actually doesn't change anything by taking the collection to be countable (take limits of the finite sums) so long as the sum of the lengths of the partition remains less than $\delta$.
Here's where Lipschitz is nice.  For general absolutely continuous functions, notice that we can without loss of generality assume that the maximum occurs at the right endpoint and the minimum at the left endpoint, otherwise there is a point $z_i$ inside $[x_i,y_i]$ so that $f(z_i)$ is minimal on that interval and a corresponding point for the maximum, so without loss of generality we can replace $[x_i,y_i]$ with the interval of those two points in the definition of absolute continuity and nothing changes.  It follows then that 
$$ \mu(f(A)) \leq \mu(\cup f([x_i,y_i]))\leq \sum_i |f(y_i) - f(x_i)| \leq \epsilon$$
$\epsilon$ is arbitrary so the result follows.
A: OK, now this almost works. You can't necessarily cover $f(E)$ with open intervals contained in $f(I_n)$: consider the constant functions, which are very much Lipschitz, but whose images contain no interval. But as you know you can approximate $f(I_k)$ from the outside as well as you please by open sets, so do that instead.
That's the substantive point, but here's one on exposition, if you're interested:
the way you define the $I_n$ makes it seem like you've fixed them once and for all, when in fact later on you require the sum of their lengths to be less than $\frac{\epsilon}{c}$. Instead, you might say "Then there exists a countable set of open intervals $\{I_k\}$ with $\sum_{k=1}^\infty l(I_k)<\frac{\epsilon}{c}$," and go on from there.
