Lipschitz function and lebusgue measure problem Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a Lipschitz function, with Lipschitz constant $C$. Prove that for any subset $E \subseteq \mathbb{R}$, we have $m^{*}(f(E)) \le Cm^{*}(E)$
I am not sure where to start and have no idea how to prove this...I hope someone can explain and help me.
 A: I'm assuming $m^*$ denotes outer measure.
Let's review what $m^*(E)$ means:
$$
m^*(E) = \inf\left\{\left.\sum_{k=1}^\infty l(I_k) \,\, \right| \,\, E \subseteq \bigcup_{k=1}^\infty I_k\right\},
$$
where $\{I_k\}_{k=1}^\infty$ must be a countable collection of nonempty open, bounded intervals.
First suppose $m^*(E) = \infty$. Then the inequality holds trivially, unless $C = 0$; but if $C = 0$, then $f = 0$, and again, the inequality holds trivially (since $m^*(f(E)) = m^*(x)$ for some $x \in \mathbb{R}$, and thus $m^*(f(E)) = 0$).
So, now we suppose $m^*(E) < \infty$, i.e. $m^*(E) \in \mathbb{R}$. Then for any $\varepsilon > 0$, by the definition of $m^*(E)$, there exists a countable collection $\{I_k\}_{k=1}^\infty$ of nonempty open, bounded intervals covering $E$ for which $\sum_{k=1}^\infty l(I_k) < m^*(E) + \varepsilon$. Now, for each $k$, since $f$ is Lipschitz with Lipschitz constant $C$ (if you're not sure what this means, look it up on Wolfram or Wikipedia; it's pretty straightforward), it must be that $f(I_k) \subseteq [x, x+C\cdot l(I_k)]$ for some $x \in \mathbb{R}$ (you'll have to show this, but it shouldn't be too crazy), which, by the monotonicity of outer measure, in turn implies that $m^*(f(I_k)) \leq m^*([x, x+C\cdot l(I_k)]) = C\cdot l(I_k)$. Now, note separately that $f(E) \subseteq f(\bigcup_{k=1}^\infty I_k)$ (since $E \subseteq \bigcup_{k=1}^\infty I_k$), and therefore $m^*(f(E)) \leq m^*(f(\bigcup_{k=1}^\infty I_k)) = m^*(\bigcup_{k=1}^\infty f(I_k))$ (since $f(\bigcup_{k=1}^\infty I_k) = \bigcup_{k=1}^\infty f(I_k)$; you will also have to show this). Finally, we can use what we showed about $m^*(f(I_k))$, together with the countable subadditivity of outer measure and the fact that $\varepsilon$ is any number greater than $0$, to conclude the proof.
I hope I didn't give away too, too much of this problem (though I fear I might've); let me know if you still have any questions after reading through this and working through the problem for a while more.
A: It is quite easy for an open interval E (let $a$ be the center of $E$ and consider the distances of $f(a)$ and $f(x)$ for any other $x\in E$).
Then extend the proposition to any open subset $E$ of $\mathbb{R}$ (which is a countable union of disjoint open intervals then).
Finally, use the definition of an outer measure to extend the proposition to any $E\subseteq\mathbb{R}$.
