A question about Lebesgue integrable function on a measurable set

Question

Show that if $f$ is a Lebesgue integrable function on a measureable set $E$ of finite measure and $E_n=\{x \in E : \lvert f(x)\rvert \geq n\}$, then $\lim\limits_{n \to \infty} (n\cdot m(E_n)) = 0$, where $m(E_n)$ denotes the measure of $E_n$.

In my opinion $\lim_{n \to \infty} m(E_n)=0$ since $f$ is lebesgue integrable on $E$. Otherwise area between $x$ axis and the function will not be finite. However, I could not come up with a complete proof for the question.

Thanks.

Answer 1

You have $|f(x)| \ge n 1_{E_n}(x)$, and so $\int_A n 1_{E_n} \le \int_A|f|$. If we set $A=E_n$, we get $n m(E_n) \le \int_{E_n} |f|$.

Note that the $E_n$ are nested, and since $f$ is integrable, $\lim_n\int_{E_n} |f| = 0$ (to see this, note that $|f(x)|1_{E_n^c}(x) \to |f(x)| $, and $|f(x)|1_{E_n^c}(x) \le |f(x)| $, which shows $\lim_n\int_{E_n^c} |f| =\int |f|$ ).

Answer 2

Hint: What can you say about $\displaystyle\int_{E_n}\lvert f\rvert\,dm?$