Integrable function, show that its limit is 0

Question

Let $A_n=\{\omega \in \Omega: |f(x)|>n\}$ and $f$ is integrable.

Show that $\lim_{n\rightarrow \infty}$ $\int_{A_{n}} f\,d\mu=0$

Any help would be really appreciated! Thanks guys :)

Answer 1

$\lim_{n\rightarrow \infty}\int_{A_{n}} f\,d\mu=\lim_{n\rightarrow \infty}\int_{X} f1_{A_{n}}\,d\mu$, where $1_{A_{n}}(x)$ is 1 iff $x$ is in $A_{n}$.

Since $f$ is measurable and $1_{A_{n}}$ is too, then so is their product. It also holds that $|f1_{A_{n}}|\leq|f|$, and due to dominated convergence theorem you may conclude that $\lim_{n\rightarrow \infty}\int_{A_{n}} f\,d\mu$ is equal to the integral of the pointwise limit of $\{f1_{A_n}\}_{n=1}^{\infty}$. It now suffices to show that the set on which $f$ is infinite has 0 measure (since $f$ is integrable, then its integral is finite), and then use that to conclude that the integral is 0.