Lebesgue outer measure of image of function with bounded derivative I encountered the following problem in my real analysis course this past semester:
Let $f$ be a real-valued function on $[a,b]$. Suppose that $E \subset [a,b]$ and that $f'$ exists and is bounded on $E$, say, by $M$. Prove that $\lambda^*(f(E)) \leq M\lambda^*(E)$.
(source: Exercise 8.18 in A Course in Real Analysis 2nd Ed., by McDonald and Weiss)
Here is what I have so far:
Consider the special case when $E$ is an open set. Then $E$ is the disjoint union of countably many open intervals $(I_n)_n$. Pick any such open interval $I_n$. Let $x,y \in I_n$. By the Mean Value Theorem, $|f(x)-f(y)| \leq M|x-y|$. But $|x-y| \leq \lambda^*(I_n)$. Therefore $|f(x) - f(y)| \leq M\lambda^*(I_n)$ for any $x,y \in I_n$. It follows that $\lambda^*(f(I_n)) \leq M\lambda^*(I_n)$. Summing over all $n$ and using subadditivity of Lebesgue outer measure yields the desired inequality.
For an arbitrary set $E$, I considered using the formal definition of Lebesgue outer measure. Pick a countable set of open intervals $(I_n)_n$ that covers $E$. I may assume without loss of generality that the sum of lengths is within some $\epsilon > 0$ of $\lambda^*(E)$. Then I wish to show that there is some countable set of open intervals $(J_n)_n$ that covers $f(E)$ so that $$\sum_{n=1}^\infty l(f(J_n)) \leq M\sum_{n=1}^\infty l(f(I_n)).$$ However, this seems to me equally insoluble. The fact that $f'$ may not be bounded and may not even exist outside $E$ seems to make things particularly complicated, since an interval $I_n$ may not be contained in $E$. What key insights am I missing?
 A: I think the starting point is that having a bound on $f'$ at a point $x$ does allow you to control the growth of $f$ near $x$, even if $f$ is not differentiable at these nearby points.
The assumption allows us to get local control of $f$ in the following way: for each $x \in E$ we have 
$$ \lim_{y \to x} \frac{\lvert f(x) - f(y) \rvert}{\lvert x - y \rvert} \leq M,$$
so we can choose $\delta_x > 0$ such that
$$0 < \lvert x - y \rvert \leq \delta_x \quad\text{implies}\quad \lvert f(x) - f(y) \rvert \leq (M + \varepsilon)\lvert x - y \rvert.$$
So points near $x$, whether or not they are in $E$, can't be mapped too far from $f(x)$.
Now we have the problem of using this to control the measure of the image of $E$. First we give ourselves a little bit of room around $E$: by definition of outer measure, we can take an open set $U \supseteq E$ such that $\lambda^*(U) \leq \lambda^*(E) + \varepsilon$.
We can also require that $\delta_x$ is small enough that $[x - \delta_x, x+\delta_x] \subset U$. Denote $$B_x = [x-\delta_x, x+\delta_x].$$
We now use these intervals to approximate the measure of $E$ using some covering lemmas: Vitali's covering lemma gives a countable subset $E' \subset E$ and new radii $\delta'_x \leq \delta_x$ for $x \in E'$ such that the intervals $B_x' = [x - \delta_x', x+\delta_x']$ are disjoint and still cover $E$, minus some set $N \subset E$ of outer measure zero. By choice of $\delta_x$, we also have $$f(B_x') \subseteq [f(x) - (M + \varepsilon) \delta_x',\ f(x) + (M + \varepsilon) \delta_x']$$ so that in particular $\lambda^*(f(B_x')) \leq (M + \varepsilon) \lvert B_x' \rvert.$ Furthermore,
$$f(E\setminus N) \subseteq f\left(\bigcup_{x \in E'} B_x'\right) = \bigcup_{x \in E'} f(B_x')$$
so, using that the intervals $B_x'$ are disjoint and are contained in $U$,
$$ \lambda^* (f(E\setminus N)) \leq \sum_{x \in E'} \lambda^*(f(B_x')) \leq \sum_{x \in E'} (M + \varepsilon) \lvert B_x' \rvert = (M + \varepsilon) \lambda\left(\bigcup_{x \in E'} B_x' \right)\leq (M + \varepsilon) \lambda(U) \leq (M + \varepsilon)[\lambda^*(E) + \varepsilon]. \tag{1}$$
Now we need to control what happens on the null set $N$. To do this we can use a different covering lemma: First take an open set $V \supset N$ with $\lambda(V) \leq \varepsilon$, and for each $x \in N$ choose $\eta_x \leq \delta_x$ so that $[x - \eta_x, x+\eta_x] \subset V$. By Besicovitch, there is some collection of points $N' \subset N$ such that the corresponding intervals $I_x = [x - \eta_x, x+\eta_x]$ have bounded overlap; more precisely, there is some constant $C$ (independent of the original cover) such that each point of $\mathbb{R}$ is contained in at most $C$ of the intervals. Similarly to above, but using bounded overlap instead of disjointness, we get the bound
$$ \lambda^*(f(N)) \leq \sum_{x \in N'} \lambda^*( f(I_x)) \leq \sum_{x \in N'} (M + \varepsilon) \lvert I_x \rvert \leq C (M + \varepsilon) \lambda \left( \bigcup_{x \in N'} I_x \right) \leq C(M + \varepsilon) \lambda(V) \leq C(M + \varepsilon) \varepsilon. \tag{2}$$
Combining (1) and (2) via subadditivity, we get
$$ \lambda^*(f(E)) \leq \lambda^*(f(E \setminus N)) + \lambda^*(f(N)) \leq (M + \varepsilon)[\lambda^*(E) + \varepsilon] + C(M + \varepsilon) \varepsilon. $$
Letting $\varepsilon \to 0$ gives the result.
