# Lebesgue outer measure of image of set is less than or equal to the Lebesgue outer measure of set

For a differentiable function $$f : \mathbb{R} \to \mathbb{R}$$ with $$|f'(x)|\leq 1$$ and any set $$E \subset \mathbb{R}$$

$$m^*(f(E)) \leq m^*(E)$$

where $$m^*$$ is the outer Lebesgue measure

First, the definition of the outer measure

$$m^*(E) = \inf \bigg\{ \sum_{n=1}^\infty(b_n - a_n): \bigcup_{n=1}^\infty(a_n, b_n) \supset E \bigg\}$$

From the definition of derivative, we have that

$$|f'(x)| = \lim_{x\to y}\bigg|\frac{f(x)-f(y)}{x-y} \bigg| \leq 1$$

Which gives that $$\forall x \in E \Rightarrow |f(x)-f(y)| \leq |x-y|$$

Therefore, when calculating $$m^*(f(E))$$, we know that $$f(E) \subset f(\bigcup_{n=1}^\infty (a_n, b_n)) = \bigcup_{n=1}^\infty (f((a_n, b_n))) = \bigcup_{n=1}^\infty (f(a_n), f(b_n))$$

Therefore,

$$\sum_{n=1}^\infty (f(b_n) - f(a_n)) \leq \sum_{n=1}^\infty (b_n - a_n)$$

Which, by definition means that

$$m^*(f(E)) \leq m^*(E)$$

Is this good enough? I think I might be missing something or not making this rigurous enough.

I don't knoe if this applies for any arbitrary subset $$E \subset \mathbb{R}$$ or just an interval. If it is just an interval I can define $$E$$ as a countable union of disjoint intervals (previous problem) $$E = \bigcup_{n=1}^\infty I_n$$ which gives

$$m^*(f(E)) = m^* \big( f \big( \bigcup_{j=1}^\infty I_j\big)\big) = m^*\big[ \bigcup_{j=1}^\infty f(I_j) \big] \leq \sum_{n=1}^\infty m^*(f(I_j)) \leq \sum_{n=1}^\infty m^*(I_j)$$

But I am lost in this step. Thanks for the help! :)

You have made a mistake in writing $$f(a_n,b_n)$$ as $$(f(a_n),f(b_n))$$. To correct this error you gave to argue that $$f(a_n,b_n) \subset [f(c_n),f(d_n)]$$ where $$c_n$$ is the point in $$[a_n,b_n]$$ where $$f$$ attains its minimum and where $$d_n$$ is the point in $$[a_n,b_n]$$ where $$f$$ attains its maximum. Hence $$m^{*}(f(a_n,b_n)) \leq m^{*} [f(c_n),f(d_n)] =f(d_n)-f(c_n)\leq b_n-a_n$$ by MVT.