For the $(\impliedby)$ direction I'm gettinf it ok, but for the $(\implies)$ direction, I'm getting the wrong inequalities. My ideas so far:$$n\mu(E_n)=\int_{E_n}n\,dx\geq\int_{E_n}|f(x)|dx \implies \sum n\mu(E_n)\geq \sum\int_{E_n}|f(x)|dx=\int|f(x)|dx<||f||_1 $$That doesn't work, neither:$$(n-1)\mu(E_n)=\int_{E_n}n-1\,dx\leq\int_{E_n}|f(x)|dx \implies \sum (n-1)\mu(E_n)\leq \sum\int_{E_n}|f(x)|dx=\int|f(x)|dx<||f||_1 $$
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$\begingroup$Notice that$$ \sum^\infty_{n=1}n\mathbb{1}_{\{n-1\leq |f(x)| <n\}}\leq |f(x)|+1\tag{1} $$
and that
$$ |f(x)|\leq \sum^\infty_{n=1}n \mathbb{1}_{\{n-1\leq |f(x)| <n\}}\tag{2}$$
Since the measure is finite, integrability of $|f|$ implies the integrability of $G(x)=\sum_{n\geq1}n\mathbb{1}_{E_n}$. Conversely, the integrability of $G$ implies the integrability of $|f|$.
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