show that $\sum_{n \geq 0}n \mu(|f| > n) < +\infty \implies f \in L^1$

Question

the series being convergent we have $\lim_{n \to \infty }n \mu(|f| > n) = 0$

because $ n \to \infty $ then $\mu(|f| > n)$ must tend to $0$ with an atleast $O(n^{1+\epsilon})$ speed , right ?

so $\mu(|f| > +\infty) = 0$ and therefore $f \in L^1$

is my work correct ?

Answer 1

You need the hypothesis that $\mu(\{|f| > 0\}) < +\infty$.

Let $E_n = \{ n < |f| \leqslant n+1\} $ and $F_n = \{ |f| > n\}$. Since $E_n \subset F_n$ we have $\mu(E_n) \leqslant \mu(F_n)$ and since $\sum_{n \geqslant 0}n\mu(F_n) < +\infty$, by the comparison test it follows that
$$\sum_{n \geqslant 0}\mu(E_n),\,\, \sum_{n \geqslant 0}n\mu(E_n) < +\infty$$

This also implies that

$$\sum_{n \geqslant 0}(n+1)\mu(E_n) < +\infty$$

Thus,

$$\int|f|\,d\mu = \sum_{n \geqslant 0} \int_{E_n}|f|\, d\mu \leqslant \sum_{n \geqslant 0} \int_{E_n}(n+1) \, d\mu = \sum_{n \geqslant 0} (n+1)\mu(E_n) < +\infty, $$

and $f \in L^1$.