Show that $m(\{x:|f(x)|>1\})=0 \Leftrightarrow \lim_{n \to \infty} \int_{X}|f|^n < \infty$

Question

Let $(X,\mathcal{A}, m)$ be a measurable space with $m(A)< \infty$ and $f \in L^{\infty}(X)$, show that $$m(\{x:|f(x)|>1\})=0 \Leftrightarrow \lim_{n \to \infty} \int_{X}|f|^n < \infty$$

$\Rightarrow)$ If $A= \{x:|f(x)|>1\}$ then is easy to see that $|f(x)|^n >1$ implies $|f(x)|>1$ so then $$\int_X |f|^n=\int_{A^c}|f|^n+\int_{A} |f|^n=\int_{A^c}|f|^n+0 \leq \int_{A^c}1=m(A^c) \leq m(X) < \infty$$ and therefore $$\lim_{n \to \infty} \int_X |f|^n < \infty$$ For the other direction I have had problems in finding a proof; this result reminds me of the geometric series but in the integral version. Any hint or help you can provide I will be very grateful.

Answer 1

Notice that $\phi_n:=\mathbb{1}_{\{|f|>1\}}|f|^n$ defines a monotone increasing sequence and that $\lim_n\phi_n=\infty\mathbb{1}_{\|f|>1\}}$. By monotone convergence $$I:=\lim_n\int_X\phi_n\,dm=\lim_n\int_{\{|f|>1\}}|f|^n\,dm=\infty\, m(|f|>1)$$

If $\lim_n\int_X|f|^n\,dm<\infty$, then $I<\infty$ and so $m(|f|>1)=0$

Answer 2

I suggest to prove it by the contrapositive.

Suppose by that $A$'s $m$-measure is positive, and then there will exist $k\in \mathbb{N}$ such that $\{x : |f(x)|>1+\frac{1}{k}\}$ has positive $m$-measure. Can you finish it from here?

This trick, finding an adequate subset of positive measure where a stronger property holds, is a neat one to keep in mind.