If $E_\alpha:= \{x\in [a, b]: f '(x) \text{ exists and $|f '(x)| \le \alpha$} \}$, then $\lambda^* (f (E_{\alpha}))\le \alpha \lambda^* (E_{\alpha})$.

Question

The following theorem is to be proven:

Let $f : [a, b]\to \mathbb R$ and for any real number $\alpha > 0$,let $E_\alpha:= \{x\in [a, b]: f '(x) \text{ exists and $|f '(x)| \le \alpha$} \}$. Then $\lambda^* (f (E_{\alpha}))\le \alpha \lambda^* (E_{\alpha})$.

Here, $\lambda^*$ denoted outer measure induced by Lebesgue measure $\lambda$.

One possible way I thought of is to consider $g(t):= \lambda^* (f (E_{t}))- t \lambda^* (E_{t}), t\in [0,\infty)$ and then showing that $g$ is differentiable with non positive derivative. So if $g(0)$ is shown to be $0$ then the result follows by mean value theorem.

The problem is that it seems that $g(0)$ may not be $0$. $g(0)=\lambda^* (f (E_{0})).$ I'm stuck here.

Please help. Thanks.