Question from Folland, criteron for a function to belong to $L^p$ This question is from Folland 6.38,
Show that $f \in L^p $ iff $\sum_{k=-\infty}^ {\infty} 2^{pk} \mu \{{x: |f(x)|>2^{k}}\} \lt \infty$


*

*If $f \in L^p $, I applied the Chebyshev's inequality 

*But for the other direction, I don't know how to begin.


Any advice would be appreciated.
Thanks
 A: 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.
A: Note that 
$$\int|f|^pd\mu=p\int_0^\infty t^{p-1}\mu\{x:|f(x)|>t\}dt$$ 
and 
$$\int_{2^k}^{2^{k+1}}t^{p-1}\mu\{x:|f(x)|>t\}dt\le 2^{(k+1)p}\mu\{x:|f(x)|>2^k\}.$$
