Let $(X,\mathcal{M},\mu)$ be a measure space and $f:X\rightarrow [0,\infty)$ a non-negative measurable function. For $k\in \mathbb{Z}$ consider the level sets$$E_k=\left\{ x\in X : 2^k < f(x) \right\} $$ Prove that $$f\in L^1\implies \sum_{k=-\infty}^{\infty}2^k \mu(E_k)<\infty $$
My trial:
$$\sum_{k=-\infty}^{\infty}2^k\mu(E_k) = \sum_{k=-\infty}^{\infty}\int_{E_k}2^kd\mu\leq \sum_{k=-\infty}^{\infty}\int_{E_k}fd\mu$$
I could not get advanced from here. I am thinking of using the property of $$E_{k+1}\subset E_{k}.$$ Using this property seems reasonable but I cannot come up with good idea. I will very thank for any hint and answer!
Consider $A_k = E_k \cap E_{k+1}^{c} = \{ x \in \mathbb{X} : 2^k < f(x) \leq 2^{k+1} \}$. Observe that the sets $A_k$ are disjoint. Now $$\sum\limits_{-\infty}^{+\infty} 2^k \mu(E_k) = \sum\limits_{k=-\infty}^{+\infty} 2^k \sum\limits_{l \geq k} \mu(A_l) = \sum\limits_{l=-\infty}^{+\infty} \mu(A_l) \cdot \sum\limits_{k \leq l} 2^k = \sum\limits_{l=-\infty}^{+\infty} \mu(A_l) \cdot 2^{l+1} \leq 2 \sum\limits_{l=-\infty}^{+\infty} \int_{A_l} f(x) d \mu \leq 2 \int_{\mathbb{X}} f(x) d \mu < + \infty$$
I know the details are not explained. If something is unclear just point it out!