Weak $L^q$-spaces and Lebesgue measure Let $q>1$ and $f: \mathbb R^n \rightarrow \mathbb R$ Lebesgue measurable. Let $||f||_{w, q} = \sup \lambda_n (A)^{-1/q'} \int_A |f| d \lambda_n$, where the supremum is taken over all $A \in \mathcal M_{\lambda^{*}_n} $ with $\lambda_n(A) < \infty$, and $1/q+1/q' =1$.
Why does it then hold true that $$ < f >_{w,q} \  \leq \ ||f||_{w,q} \ \leq \ \dfrac{q}{q-1} < f >_{w,q}$$
where $ < f >_{w,q} := \sup \alpha \lambda_n ($ { $ x \in \mathbb R^n: |f(x)| > \alpha $ } $)^{1/q}  < \infty$.
 A: The following exercises come from the book of Loukas Grafakos, Classical Fourier Analysis, and your question is a special of them.

Let $(X,\mu)$ be a measure space and let $E$ be a subset of $X$ with $\mu(E)<\infty$. Assume that $f$ is in $L^{p,\infty}(X,\mu)$ for some $0<p<\infty$. Show that for $0<q<p$ we have 
  \begin{align*}
\int_{E}|f(x)|^{q}d\mu(x)\leq\dfrac{p}{p-q}\mu(E)^{1-q/p}\|f\|_{L^{p,\infty}}^{q}.
\end{align*}

Here $L^{p,\infty}(X,\mu)$ is the set of all measurable functions such that $\sup_{\alpha>0}\alpha\mu(x\in X:|f(x)|>\alpha)^{1/p}<\infty$.
Here is the proof.
First of all, we note that 
\begin{align*}
\min(\mu(E),\alpha^{-p}\|f\|_{L^{p,\infty}}^{p})=\mu(E)
\end{align*}
if and only if
\begin{align*}
\mu(E)&\leq\alpha^{-p}\|f\|_{L^{p,\infty}}^{p}\\
\alpha&\leq\mu(E)^{-1/p}\|f\|_{L^{p,\infty}}.
\end{align*}
Likewise,
\begin{align*}
\min(\mu(E),\alpha^{-p}\|f\|_{L^{p,\infty}}^{p})=\alpha^{-p}\|f\|_{L^{p,\infty}}^{p}
\end{align*}
if and only if
\begin{align*}
\alpha\geq\mu(E)^{-1/p}\|f\|_{L^{p,\infty}}.
\end{align*}
Now, we write that
\begin{align*}
\int_{E}|f(x)|^{q}d\mu(x)&=q\int_{0}^{\infty}\alpha^{q-1}\mu(E\cap(|f|>\alpha))d\alpha\\
&=q\left(\int_{0}^{\mu(E)^{-1/p}\|f\|_{L^{p,\infty}}}+\int_{\mu(E)^{-1/p}\|f\|_{L^{p,\infty}}}^{\infty}\right)\alpha^{q-1}\mu(E\cap(|f|>\alpha))d\alpha\\
&=I_{1}+I_{2}.
\end{align*}
We have $\mu(E\cap(|f|>\alpha))\leq\min(\mu(E),\alpha^{-p}\|f\|_{L^{p,\infty}}^{p})$ and hence
\begin{align*}
I_{1}=q\int_{0}^{\mu(E)^{-1/p}\|f\|_{L^{p,\infty}}}\alpha^{q-1}\mu(E)d\alpha=\mu(E)\left(\mu(E)^{-1/p}\|f\|_{L^{p,\infty}}\right)^{q}=\mu(E)^{1-q/p}\|f\|_{L^{p,\infty}}^{q}.
\end{align*}
Also, 
\begin{align*}
I_{2}&=q\int_{\mu(E)^{-1/p}\|f\|_{L^{p,\infty}}}^{\infty}\alpha^{q-1}\alpha^{-p}\|f\|_{L^{p,\infty}}^{p}d\alpha\\
&=-\dfrac{q}{q-p}\|f\|_{L^{p,\infty}}^{p}\left(\mu(E)^{-1/p}\|f\|_{L^{p,\infty}}\right)^{q-p}\\
&=-\dfrac{q}{q-p}\mu(E)^{1-q/p}\|f\|_{L^{p,\infty}}^{q}.
\end{align*}
As a result,
\begin{align*}
I_{1}+I_{2}\leq\left(1-\dfrac{q}{q-p}\right)\mu(E)^{1-q/p}\|f\|_{L^{p,\infty}}^{q}=\dfrac{p}{p-q}\mu(E)^{1-q/p}\|f\|_{L^{p,\infty}}^{q},
\end{align*}
we are done.
Here is the second exercise.

Let $(X,\mu)$ be a $\sigma$-finite measure space and let $0<p<\infty$. Pick $0<r<p$ and define
  \begin{align*}
|||f|||_{L^{p,\infty}}=\sup_{0<\mu(E)<\infty}\mu(E)^{-1/r+1/p}\left(\int_{E}|f|^{r}d\mu\right)^{1/r},
\end{align*}
  where the supremum is taken over all measurable subsets $E$ of $X$ of finite measure. Prove that for all $f\in L^{p,\infty}(X,\mu)$ we have 
  \begin{align*}
\|f\|_{L^{p,\infty}}\leq|||f|||_{L^{p,\infty}}.
\end{align*}

Here is the proof.
We write 
\begin{align*}
X=\bigcup_{k}X_{k},~~~~\mu(X_{k})<\infty,
\end{align*}
where $(X_{k})$ is an increasing sequence of sets.
We have
\begin{align*}
\|f\|_{L^{p,\infty}}=\sup_{\alpha>0}\alpha\mu(x\in X:|f(x)|>\alpha)^{1/p}=\sup_{\alpha>0}\lim_{k\rightarrow\infty}\alpha\mu(X_{k}\cap(|f|>\alpha))^{1/p}.
\end{align*}
Let $\alpha>0$ and $k\in\mathbb{N}$ be given. We can assume without loss of generality that $\mu(X_{k}\cap(|f|>\alpha))>0$ and hence
\begin{align*}
&\alpha^{p}\mu(X_{k}\cap(|f|>\alpha))\\
&=\alpha^{p-r}\int_{X_{k}\cap(|f|>\alpha)}\alpha^{r}\chi_{(|f|>\alpha)}(x)d\mu(x)\\
&\leq\alpha^{p-r}\int_{X_{k}\cap(|f|>\alpha)}|f(x)|^{r}d\mu(x)\\
&=\alpha^{p-r}\mu(X_{k}\cap(|f|>\alpha))^{1-r/p}\mu(X_{k}\cap(|f|>\alpha))^{-1+r/p}\int_{X_{k}\cap(|f|>\alpha)}|f(x)|^{r}d\mu(x)\\
&\leq\alpha^{p-r}\mu(X_{k}\cap(|f|>\alpha))^{1-r/p}|||f|||_{L^{p,\infty}}^{r}\\
&=\left(\alpha\mu(X_{k}\cap(|f|>\alpha))^{1/p}\right)^{p-r}|||f|||_{L^{p,\infty}}^{r}\\
&\leq\|f\|_{L^{p,\infty}}^{p-r}|||f|||_{L^{p,\infty}}^{r}.
\end{align*}
We obtain that
\begin{align*}
\|f\|_{L^{p,\infty}}^{p}&\leq\|f\|_{L^{p,\infty}}^{p-r}|||f|||_{L^{p,\infty}}^{r},
\end{align*}
and hence
\begin{align*}
\|f\|_{L^{p,\infty}}^{r}&\leq|||f|||_{L^{p,\infty}}^{r}\\
\|f\|_{L^{p,\infty}}&\leq|||f|||_{L^{p,\infty}},
\end{align*}
we are done.
The $\sigma$-finitenss of the second exercise cannot be relaxed. For $X=\{1,2\}$ and $\mu(\{1\})$=1 and $\mu(\{2\})=\infty$, so $(X,\mu)$ is not $\sigma$-finite. If we let $f=1$, then
\begin{align*}
|||f|||_{L^{p,\infty}}=\left(\int_{\{1\}}1d\mu\right)^{1/r}=\mu(\{1\})^{1/r}=1.
\end{align*}
But 
\begin{align*}
\|f\|_{L^{p,\infty}}=\sup_{\alpha>0}\alpha\mu(1>\alpha)^{1/p}\geq\dfrac{1}{2}\mu\left(x\in X:1>\dfrac{1}{2}\right)^{1/p}=\dfrac{1}{2}\mu(X)^{1/p}=\infty.
\end{align*}
