I've proved the next:
Let $(X,S,\mu)$ be a measurable space and $f\in L_{1}(\mu)$ such that $\mu(E_{f})<\infty$ where $$E_{f}=\{x\in X: f(x)\neq0\}.$$ Then $$\displaystyle\lim_{n\rightarrow\infty}\int|f|^{1/n}d\mu=\mu(E_{f}).$$
It's enough to apply Holder's Inequality to $|f|^{1/n}$ and $\mathcal{X}_{E_{f}}$ and by the other hand splits $|f|^{1/n}$ on the intervals $(0,1)$ and $[1,\infty).$
My question is: the above remains valid for $\mu(E_{f})=+\infty?$
Yes. Let $A=\{|f|\ge 1\}$ and $B=\{0<|f|<1\}$. Then $\int_A|f|^{1/n}\to\mu(A)$ by dominated convergence, while $\int_B|f|^{1/n}\to\mu(B)$ by monotone convergence.