A proposition to prove the real interpolation of positive exponent Let $p,A\in(0,\infty)$
$\|f\|_{L^{p,\infty}(X,\mu)} := \sup \left\{\lambda\mu(\{|f|\geq\lambda\})^{\frac{1}{p}}\right\}$ .
Show that the following are equivalent
(1) $\|f\|_{L^{p,\infty}(X,\mu)}\leq C_pA$ for some constant $C_p$ 
(2) For any set $E$ of finite measure, there exist a subset $E'\subset E$ with $\mu(E')\geq\frac{\mu(E)}{2}$ s.t.$$\int_{E'} fd\mu =O\left(\mu(E)^{1-\frac{1}{p}}\right)$$ 
I follow the hint and try to prove the proposition for $f(x)=|x|^\frac{-d}{p}$ and $X=\mathbb{R}^d$ but I can't.
Actually, this is a question from a lecture notes of Fourier analysis.
I have no any idea even for $f(x)=|x|^\frac{-d}{p}$. (2)=>(1) is done. However, for (1)=>(2), I don't have any idea about how to find $E'$ s.t. $\mu(E')\geq\frac{\mu(E)}{2}$ by using (1).
 A: $(1)\Rightarrow (2)$
Suppose that $\|f\|_{L^{p,\infty}(X,\mu)}\leq C_pA$. First, for a measurable set $E$, with $0 <\mu(E) <\infty$, define
\begin{align*}
D&=\{x \in E \mid |f(x)|>2^{1/p}A\mu(E)^{-1/p}\},\\
E'&=\{x \in E \mid |f(x)|\leq 2^{1/p}A\mu(E)^{-1/p}\}
\end{align*}
(this means that $E'=E\setminus D$).
Claim: $2\mu(E')\geq\mu(E)$.
In fact, since $D,E'\subset E$ and $\mu(E)<\infty$, we obtain $\mu(E')=\mu(E)-\mu(D)$. Furthermore, as $\sup_{\lambda>0}\lambda \mu(\{x \in X \mid |f(x)|>\lambda\})^{1/p}\leq A<\infty$ , the inequality
$$
\mu(\{x \in X \mid |f(x)|>\lambda\})\leq A^p\lambda^{-p}
$$
holds any positive $\lambda$. Consequently
\begin{align*}
\mu(D)&=\mu(\{x \in E \mid |f(x)|>2^{1/p}A\mu(E)^{-1/p}\})\\ &\leq\mu(\{x \in X \mid |f(x)|>2^{1/p}A\mu(E)^{-1/p}\})\\&\leq A ^p(2^{1/p}A\mu(E)^{-1/p})^{-p}=\frac{\mu(E)}{2}
\end{align*}
and so $\mu(E')=\mu(E)-\mu(D)\geq \mu(E)-\frac{1}{2}\mu(E)= \frac{1}{2 }\mu(E)$.
Finally, by the Hölder's inequality and the $\mu(E')\leq \mu(E)$ (remember that $E'\subset E$)
\begin{align*}
\left|\int_{E'}f(x)\,d\mu(x)\right|\leq \int_{E'}|f(x)|\,d\mu(x)\leq \| \chi_{E'}\|_1\|f\chi_{E'}\|_{\infty}&\leq \mu(E')\left(2^{1/p}A\mu(E) ^{-1/p}\right)\\&\leq \mu(E)\left(2^{1/p}A\mu(E)^{-1/p}\right)\\&= 2^{1/p}A\left(\mu(E)^{1-1/p}\right)\\&=O\left(\mu(E)^{1-1/p}\right ).
\end{align*}
$(2)\Rightarrow (1)$
Conversely, suppose that there are $c_p, A<\infty$ such that, for any $E$ with finite measure, there is $E'\subset E$ with $\mu(E)\leq 2\mu(E')$ satisfying
\begin{align*}
\left|\int_{E'}f(x)\,d\mu(x)\right|\leq c_p A \mu(E)^{1-1/p}.
\end{align*}
For each $\lambda>0$, define
\begin{align*}
T_\lambda^+&=\{x \in X \mid f(x)>\lambda\},\\ T_\lambda^- &=\{x \in X \mid f(x)<-\lambda \}
\end{align*}
so that $\{x \in X \mid |f(x)|>\lambda\}\subset T_\lambda^+\cup T_\lambda^-$ (this information is essential for the proof ).
Claim: $\mu(T_\lambda^+),\mu(T_\lambda^-)<\infty$, for all $\lambda >0$.
Without loss of generality, suppose by contradiction that there is a $\lambda_0>0$ such that the set $T_{\lambda_0}^+$ has infinite measure. Since $\mu$ is $\sigma$-finite, there is a sequence of sets $\{E_j\}_{j\in \mathbb{N}}$, with $\mu(E_j)<\infty$ for all $j$ natural, in such a way that $T_{\lambda_0}^+\subset \bigcup_{j\in \mathbb{N}}E_j$. Thus, setting $F_j = E_j\cap T_{\lambda_0}^+$,
$$
\mu(F_j) = \mu(E_j\cap T_{\lambda_0}^+)\leq \mu(E_j)<\infty
$$
and $T_{\lambda_0}^+ = \bigcup_{j\in \mathbb{N}}F_j$. In particular, like $\mu(T_{\lambda_0}^+)=\infty$, there is a natural number $K$ such that if $E=\bigcup_{j=1}^{K}F_j$, then $\left(\frac{2A c_p}{\lambda_0}\right)^p<\mu\left(E\right)$ and by subadditivity of measure
\begin{align*}
0<\left(\frac{2A c_p}{\lambda_0}\right)^p<\mu\left(E\right)\leq \sum_{j=1} ^K \mu(F_j)<\infty.
\end{align*}
So, it follows from the hypothesis the existence of $E'\subset E$, with $\mu(E)\leq 2\mu(E' )$, in such a way that
\begin{align*}
\lambda_0 \frac{\mu(E)}{2} \leq \lambda_0 \mu(E')\leq\int_{E'}f(x)\,d\mu(x)=\left|\int_ {E'}f(x)\,d\mu(x)\right|\leq A c_p\mu(E)^{1-1/p}
\end{align*}
(remember: the second inequality in the above equation is valid because $f(x)>\lambda_0$ for every $x \in E'$, since $E'\subset E\subset T_{\lambda_0}^+$), i.e.,
\begin{align*}
\mu(E)\leq \left(\frac{2A c_p}{\lambda_0}\right)^p,
\end{align*}
a contradiction (due to the choice of $E$).
Consequently, both $T_\lambda^+$ and $T_\lambda^-$ have finite measure. Thus, if $E=T_\lambda^+$ or if $E=T_\lambda^-$, there is $E'\subset E$ with $\mu(E)\leq 2\mu(E')$ such that
\begin{align*}
 A c_p \mu(E)^{1-1/p}\geq \left|\int_{E'}f(x)\,d\mu(x)\right|\geq \lambda \mu(E' )\geq \lambda \frac{\mu(E)}{2}
\end{align*}
and therefore
\begin{align*}
 \lambda \mu(E)^{1/p}\leq 2c_p A.
\end{align*}
Finally, by the previous inequality
\begin{align*}
\lambda\mu(\{x \in X \mid |f(x)|>\lambda\})^{1/p}&\leq \lambda\left(\mu(T_\lambda^+\cup T_\lambda^-)\right)^{1/p} \\&\leq \lambda\left(\mu(T_\lambda^+)+\mu(T_\lambda^-)\right)^{1/ p}&\\&\leq \lambda\max\left\{2\mu(T_\lambda^+),2\mu(T_\lambda^-)\right\}^{1/p}\\& = 2^{1/p}\left(\lambda\max\left\{\mu(T_\lambda^+)^{1/p},\mu(T_\lambda^-)^{1/p} \right\}\right)\\&\leq 2^{1/p}(2c_p A)= 2^{1+1/p}c_p A
\end{align*}
and the result follows taking the supreme over all $\lambda>0$ (define $C_p=2^{1+1/p}c_p$).
