Q:

Lebesgue integrable function and limit: $\lim_{n\to\infty} n\cdot m(A_n)=0$

Show that if $f$ is a Lebesgue integrable function on $A\subset\mathbb R$ and $$A_n=\{x\in A:|f(x)|\geq n\}$$ for $n\in\mathbb N$, then $\lim_{n\to\infty} n\cdot m(A_n)=0$.
My solution which is different from the answer provided is as follows:
Assume to the contrary that $\lim_{n\to\infty} n\cdot m(A_n)\neq0$.
Then $\exists\varepsilon>0$ s.t. $\forall n\in\mathbb N,n\cdot m(A_n)>\varepsilon$.
$A=\bigcup_{n\in\mathbb N}A_n$, so $m(A)=\sum_{n=1}^{\infty}m(A_n)$
So $\int_A|f|d\mu=\sum_{n=1}^{\infty}\int_{A_n}|f|d\mu\geq\sum_{n=1}^{\infty}n\cdot m(A_n)\geq\sum_{n=1}^{\infty}\epsilon\rightarrow\infty$.
So we get a contradiction and hence $\lim_{n\to\infty} n\cdot m(A_n)\neq0$.
Could somebody check if this argument works? Thanks!

A:

It does not work.  For $m>n$, we have $A_m\supset A_n$.  Thus though we do have $A=\bigcup_{n=0}^\infty A_n$, because this union is not disjoint we do not have $m(A)=\sum_{n=0}^\infty m(A_n)$ -- we have $m(A)\leq \sum_{n=0}^\infty m(A_n)$. Thus the first equality in your penultimate line is not valid.

