Let we have $f:D\rightarrow \mathbb{R}$ a measurable function and $f_n(x)$ defined in the following way: $f_n(x)=f(x)$ when $|f(x)|\le n$ and $f_n(x)=0$ when $|f(x)| > n$. I have to prove that $\lim_{n\rightarrow \infty} \int_D f_n d\mu=\int_D fd\mu$ when $f$ is integrable.
I'm not sure but I'm thinking to uste the dominated convergence theorem:
As $f(x)$ is measurable by hypothesis and we know that $0$ is also measurable, then $f_n$ is measurable.
Now, to use the theorem $f_n$ has to be pointwisely convergent to $f$. (Is that true in this case? If it is, how can I prove it?)
By hypothesis I know that $f$ is integrable so using the dominated convergence theorem I know that the following equality is true: $\lim_{n\rightarrow \infty} \int_D f_n d\mu=\int_D fd\mu$
Is that ok?