I have to show that
$\lim_{n \to \infty} n\cdot m(\{ x \in A | |f(x)| \geq n\}) = 0$, for $(A, \textit{S}, m)$ a measure space and $f: A \rightarrow \mathbb{R}$ an integrable function.
Let $A_n = \{ x \in A | |f(x)| \geq n\}$
Here's what I've been thinking:
I know that $\forall n >0: n\cdot m(A_n) \leq \int_{A_n}|f(x)| dm(x)$ so I thought that this might mean that I have to show that $f(x) = 0$ so that $\int f = 0$, but this seemed odd to show that for a general function. I also noticed, that the $A_n$ form a decreasing sequence, i.e. $A_{n+1} \subset A_n$, which means that $m(\bigcap_{n\geq 0} A_n) = lim_{n\to \infty} m(A_n)$. However I don't know if this knowledge is useful or how to apply it.
If someone could point me in the right direction I'd really appreciate it. Thanks!
A corretion there, it should be $m\left( \cap_n A_n\right) =\lim_n m(A_n)$, since the sequence decreases, but for this you require $m(A_n)<\infty$ for some $n$ (check this). Now note that your function must be finite almost everywhere, else it'd not be integrable.
Finally, don't forget $\mu(B) = \int_B |f|dm$ is a measure too , what happens when $m(B) =0$ to $\mu(B)$? ;)