On $E_k = \{ 2^k < \left| f(x) \right| \leq 2^{k+1} \}$, we have
$ 2^{pk} < \left|f(x)\right|^{p} \leq 2^{p(k+1)}.$
Thus integrating on $E_k$ we have
$ 2^{pk} \mu (E_k) < \int_{E_k} \left|f(x)\right|^{p} \, d\mu \leq 2^{p(k+1)} \mu(E_k).$
Summing through $k$, we obtain
$ \sum_{k=-\infty}^{\infty} 2^{pk} \mu (E_k) < \int \left|f(x)\right|^{p} \, d\mu \leq 2^{p} \sum_{k=-\infty}^{\infty} 2^{pk} \mu(E_k).$
Finally, note that
$ \sum_{j=-\infty}^{k} 2^{pj} = \frac{2^{kp}}{1 - 2^{-p}}. $
Thus we have
$ \begin{align*} \sum_{k=-\infty}^{\infty} 2^{pk} \mu (E_k) &= (1 - 2^{-p}) \sum_{k=-\infty}^{\infty} \sum_{j=-\infty}^{k} 2^{jp} \mu (E_k) \\ &= (1 - 2^{-p}) \sum_{j=-\infty}^{\infty} \sum_{k=j}^{\infty} 2^{jp} \mu (E_k) \\ &= (1 - 2^{-p}) \sum_{j=-\infty}^{\infty} 2^{jp} \mu \{ 2^j < \left| f(x) \right| \}. \end{align*}, $
from which the conclusion follows.