If $f$ is measurable and $A>0$ then the truncation $f_{A}$ defined by:
$$f_{A}(x)=\begin{cases} f(x)&\text{if $\left | f(x) \right |\leq A$}\\ A&\text{if $ f(x)> A $}\\ A&\text{if $f(x)< A $} \end{cases}$$
Let $f$ be an $X$ measurable function on $X$ to $\mathbb{R}$. For $n\in \mathbb{N}$, let $(f_{n})$ be a sequence of truncates of $f$. If $f$ es integrable with respect to $\mu$, then $\int fd\mu =\lim\int f_{n}d\mu $.
Conversely, if $\sup\int\left | f_{n} \right |d\mu<{+}\infty $ then $f$ is integrable.
I think I´ve already prove the first part, I applied Lebesgue Dominated Convergence Theorem, because que can see that $f_{n+1}\leq f_{n} \leq f, n\in\mathbb{N}$ so $ (f_{n})\leq \left |f \right | $ and $f$ is integrable, so I have to prove that $(f_{n})$ is a sequence of integrable functions which converges almost everywhere to a real-valuated function and I can applied the theorem, right???
The other part is the one which I have problems to prove it, does anybody can help me a little with that?? Thanks!
I think that the problem wants us to define $f_N(x) = f(x)$ if $0 \leq f(x) \leq N$, $f_N(x) = f(x)$ if $-N \leq f(x) \leq 0$ and $f_N(x) = 0$ otherwise.
Then we have that $f_n \to f$ monotonically and so we can apply the monotone convergence theorem to conclude that $\lim_n \int f_n = \int \lim_n f_n = \int f$. Now if $\sup_n \int |f_n| < \infty$, then clearly $\int |f| < \infty$ since for any bounded monotonic sequence we have that $\sup_n a_n = \lim_n a_n$.