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! 🙂
Best Answer
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.