Is it true that $$f(x) \cdot 1(|f'(x)| > \lambda) = \int_{-\infty}^x f'(t)\cdot1(|f'(t)| > \lambda)\:\mathrm dt?$$
$\cdot$ is the usual multiplication on $\mathbb R$ and $1$ denotes here the usual indicator function: $$\mathsf 1(x) = \begin{cases} 1 & \text{if $x$ is true} \\ 0 & \text{else} \end{cases}$$