Prove $s<\infty\to S<\infty \to I<\infty$ 
Let $p\geq 1$ and $f\geq 0$ Setting
$I:=\int_X|f|^pd\mu$
$s=\sum_{n\in\mathbb{Z}}2^n\mu(\{x:2^n<f(x)\leq2^{n+1}\})^{\frac{1}{p}}$
$S=\sum_{n\in\mathbb{Z}}2^n\mu(\{x:2^n<f(x)\})^{\frac{1}{p}}$
Prove $s<\infty\to S<\infty \to I<\infty$

I've tried to set $A_n:=\{x:2^n<f(x)\leq2^{n+1}\}$,$B_n:=\{x:2^n<f(x)\}$ to prove that $S<\infty$ and from chebyshev inequality we know $2^n\mu(B_n)=2^n\mu(\{x\in B_N:|f|>2^n\})^{1/p}\leq(\int_{B_n}|f|^p)^{\frac{1}{p}}$
And got stuck here:
$\sum_{n\in\mathbb{Z}}2^n\mu(B_n)^{\frac{1}{p}}\leq\sum_{n\in\mathbb{Z}}(\int_{B_n}|f|^p)^{\frac{1}{p}}=\sum_{n\in\mathbb{Z}}(\int_{\uplus_{k=n}^{\infty}A_k}|f|^p)^{\frac{1}{p}}=\sum_{n\in\mathbb{Z}}\sum_{k=n}^{\infty}(\int_{A_k}|f|^p)^{\frac{1}{p}}\leq\sum_{n\in\mathbb{Z}}\sum_{k=n}^{\infty}(\int_{A_k}(2^{k+1})^p)^{\frac{1}{p}}\leq\sum_{n\in\mathbb{Z}}\sum_{k=n}^{\infty}2^{k+1}\mu(A_k)^{\frac{1}{p}}$
For if we asumme that $S<\infty$ and trying to prove that  $I<\infty$ I've tried :
$\int_X|f|^{\frac{1}{p}}=\int_{\uplus A_n}|f|^{\frac{1}{p}}\underset{(1)}{=}\sum_{n\in\mathbb{Z}}\int_{A_n}|f|^{\frac{1}{p}}\leq\sum_{n\in\mathbb{Z}}(2^{n+1})^{\frac{1}{p}}\mu(A_n)^{\frac{1}{p}}\leq\sum_{n\in\mathbb{Z}}(2^{n+1})^{\frac{1}{p}}\mu(B_n)^{\frac{1}{p}}$
And I'm not sure about the if (1) is correct aswell.
 A: To show that $s<\infty$ implies $S<\infty$ for $p>1$ set
$a_n:=\mu\big(\{x:2^n<f(x)\leq2^{n+1}\}\big)$,
$b_n:=\mu\big(\{x:2^n<f(x)\}\big)$, $q=p/(1-p)$ and
$M_n=\sup\{a_n\bigm| m\geq n\}$. Then,
$$b_n=\sum_{m\geq n}^\infty a_n
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=\sum_{m\geq n}^\infty a_n^{1/p}  a_n^{1/q}
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\leq M_n^{1/q}\sum_{m\geq n}^\infty a_n^{1/p} 
$$
so that
\begin{align*}
  \sum_{n=-k}^k2^nb_n^{1/p}& =
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\sum_{n=-k}^k2^nM_n^{1/pq}\Big\{\sum_{m\geq n}^\infty a_m^{1/p}\Big\}^{1/p}
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  \leq\sum_{n=-k}^k\Big\{\sum_{m\geq n}^\infty 2^n a_m^{1/p}\Big\}^{1/q}
  \Big\{\sum_{m\geq n}^\infty 2^n a_m^{1/p}\Big\}^{1/p} \\
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& =\sum_{n=-k}^k\sum_{m\geq n}^\infty 2^n a_m^{1/p}
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 <\sum_{m=-\infty}^\infty 2^{m+1} a_m^{1/p}=2s.
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\end{align*}
Thus $s<\infty$ implies $S<\infty$ for $p>1$. If $p=1$ a similar
calculations shows the implication.
To see that $S<\infty$ implies $\int_X |f|^p\mathrm{d}\mu<\infty$, let
$B_n:=\{x:2^n<f(x)\}$ and let $\mathcal{X}_{B_n}$ denote the
characteristic function of $B_n$. Note that
\begin{align*}
  \int_{B_{-k}\setminus B_k }|f|^p\mathrm{d}\mu & =
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\sum_{m=-k}^{k-1}\int_{B_{m}\setminus B_{m+1} }|f|^p\mathrm{d}\mu =
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\sum_{m=-k}^{k-1}\int_{X}\big(\mathcal{X}_{B_{m}}-
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\mathcal{X}_{B_{m+1}}\big)|f|^p\mathrm{d}\mu\\
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  & \leq \sum_{m=-k}^{k-1}2^{p(m+1)}\int_{X}\big(\mathcal{X}_{B_{m}}-
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\mathcal{X}_{B_{m+1}}\big)\mathrm{d}\mu
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   \leq \sum_{m=-k}^{k-1}2^{p(m+1)}\int_{X}\mathcal{X}_{B_{m}}
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\mathrm{d}\mu\\ & =2^p\sum_{m=-k}^{k-1}\Big[2^{m}b_m^{1/p}\Big]^p,
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\end{align*}
which converges by the comparison test if $S<\infty$.
