Normability of weak $L^p$-spaces Let  $(X, \mathcal A,\mu)$ be a measure space, and $0<p<\infty$

Definition: The weak $L^p-$space on $(X, \mathcal A,\mu)$
  denoted $L^{p,\infty}(X, \mu)$ is defined as the set of all $\mu$-measurable functions $f$ such that: 
  $$\|f\|_{L^{p,\infty}} = \sup\{ t\mu\left(\{x\in X: |f(x)|>t\}\right)^{1/p}: t>0\}<\infty.$$

One can easily check that the map $f\mapsto \|f\|_{L^{p,\infty}}$ is a quasi norm. namely we have the quasi-triangular inequality
$$\|f+g\|_{L^{p,\infty}} \le \max(2,2^{1/p})\left(\|f\|_{L^{p,\infty}}+\|g\|_{L^{p,\infty}}\right)$$
Now we assume is a $\sigma-$finite measure space. Let $0<r<p<\infty $ we define
$$|\|f|\|_{L^{p,\infty}} = \sup_{0<\mu(E)<\infty} \mu(E)^{-\frac1r+\frac1p} \left(\int_E |f|^rd\mu\right)^{1/r}$$
the surpemun is taken over all measurable subsets $E$ of $X$ of finite measure.

Question: prove that $|\|\cdot|\|_{L^{p,\infty}}$ and $\|\cdot\|_{L^{p,\infty}}$ are equivalent. Then conclude that $L^{p,\infty}$ is normable for $p>1$ and metrisable for $0<p\le 1.$

I was able to prove that, $$\|f\|_{L^{p,\infty}}\le |\|f|\|_{L^{p,\infty}}$$ which is easy from the definition. Now How can I prove that
$$|\|f|\|_{L^{p,\infty}}\le c_p \|f\|_{L^{p,\infty}}$$ 
This would answer the others questions it is not difficult to see that $$f\mapsto  |\|f|\|_{L^{p,\infty}}$$ is a norm for $p>1.$
 A: It suffices to show that there exists a constant $c$ such that for each $E$ of positive and finite measure, 
$$
\int_E\left\lvert f\right\rvert^r\leqslant c\left\lVert f \right\rVert_{\mathbb L^{p,\infty }   }^p\mu\left(E\right)^{1-p/r}.                
$$
To this aim, use Fubini's thereom to get 
$$
\int_E\left\lvert f\right\rvert^r=\int_{\mathbb R }\mu\left(\left\{    x\mid \left\lvert f(x)\right\rvert^r\mathbf 1_E (x) \gt t \right\}        \right)  \mathrm dt     
$$
and observing that 
$$
\left\{    x\mid \left\lvert f(x)\right\rvert^r\mathbf 1_E (x) \gt t \right\} 
=\left\{    x\mid \left\lvert f(x)\right\rvert^r\gt t \right\} \cap E $$
we derive that 
$$
\int_E\left\lvert f\right\rvert^r\leqslant \int_{\mathbb R }\min\left\{   \mu\left(\left\{    x\mid \left\lvert f(x)\right\rvert^r \gt t \right\}        \right),\mu\left(E\right)   \right\}    \mathrm dt.
$$
Since 
$$\mu\left(\left\{    x\mid \left\lvert f(x)\right\rvert^r \gt t \right\}        \right)=\mu\left(\left\{    x\mid \left\lvert f(x)\right\rvert \gt t^{1/r}     \right\}        \right)\leqslant t^{-p/r}\left\lVert f \right\rVert_{\mathbb L^{p,\infty }   }^p,$$
we are able to conclude.
