I would love some help with the following question:
Let $f$, a differentiable function in $[0, 1]$, such that $|f'(x)| > \leq M$ for every $x \in [0,1]$. Prove that for every Lebesgue measurable set, $E \subset [0,1]$, $f(E)$ is Lebesgue measurable, and $\mu(f(E)) \leq M\times\mu(E)$.
I know how to prove that $f(E)$ is measurable by proving for compact sets, $F_{\sigma}$ sets, sets with $\mu(E) = 0$, and then any Lebesgue measurable set. The second part, I'm having trouble with:
First question: It seems to me like the proof that $\mu(f(E)) \leq M\times\mu(E)$ should be follow the same lines above (so that I can prove both requirements together), but I can't think of a way to do it. Ideas?
Second question: I thought about the following proof, but I'm not sure it's correct:
- Choose an open set, $G$, such that $E \subset G$ and $\mu(G) \leq \mu(E) + \epsilon$
- Use Egorov's Theorem to get subsets of $G$: $G_n$, such that $\mu(G\setminus G_n) < \frac{1}{n}$ and $ f'_n := \frac{f(x + 1/n) - f(x)}{1/n} $ converges uniformly to $f'$ on $G_n$.
- Divide $G_n$ to intervals shorter than the $\delta $ we got from uniform convergence, and show $\mu(f(G_n)) \leq M\times\mu(G_n)$
- WLG, $G_n$ is non decreasing and $G = \lim_{ n \to \infty }G_n$, hence $f(G) = \lim_{ n \to \infty }f(G_n)$, and $\mu(f(G)) = \lim_{ n \to \infty }\mu(f(G_n)) \leq \lim_{ n \to \infty }M\times(\mu(G) - 1/n) = M\times\mu(G)$
- $\mu(f(E)) \leq \mu(f(G)) \leq M\times\mu(G) \leq M\times(\mu(E) + \epsilon)$ for every $\epsilon$.
Does this make sense? Is there a simpler way?
Thank you!