Let $(X,\mu)$ be a measure space and $ f \in L^p$, $ 0 < p < \infty$. Let $ \lambda_f (t) : = \mu ( \{ x \in X : |f(x)| \ge t \} )$. I want to that exists a constant $c_p$ depending only on $p$ such that $ \int |f| ^p d\mu \ge c_p \sum_{ n \in \mathbb Z} \lambda_f (2^n) 2^{np}$.
I found by drawing the graph and compare the area that $ \int |f|^p d\mu \ge \sum_{ n \in Z} \lambda_f (2^n) 2^{(n-1)p} = \frac{1}{2^p} \sum_{ n \in Z} \lambda_f (2^n) 2^{np}$ . So if I let $ c_p = 2^{-p}$ I'm done. But I tried and failed to prove it formally.
Any help is appreciated.
Since $$2^{np}\mu\{2^n\leqslant |f|^p\lt 2^{n+1}\}\leqslant\int |f|^p\chi_{\{2^n\leqslant |f|\lt 2^{n+1}\}}$$ we obtain after a summation that $$\int |f|^p\geqslant \sum_{n\geqslant 1}2^{np}\mu\{2^n\leqslant |f|^p\lt 2^{n+1}\}=\sum_{n\geqslant 1}2^{np}(\lambda_f(2^n)-\lambda_f(2^{n+1})).$$ The result follows from a summation by parts.