I am reading about the monotone convergence theorem which states:
Let $(X,\Sigma,\mu)$ a complete measurable space.
If $f_n\rightarrow f$ monotone increasing and for almost all $x\in D$ then
$\int_Ef_n dx\rightarrow \int_E f$ for every measurable set $E$.
So, this theorem is also valid for not complete measurable spaces but then we must have $f_n\rightarrow f$ for all $x\in D$. I can not read from the prove why this is true. Can someone explain?
I think that by $D$, you meant $X$.
The issue is a technical one on the way we defined "almost everywhere". If $(X,\Sigma,\mu)$ is not complete, we say that a property P(x) is true "almost everywhere" if there is a set $N \in \Sigma$ such that $\mu(N)=0$ and $\{x\in X\: :\: P(x) \textrm { is false }\} \subseteq N$. Note that the set $\{x\in X\: :\: P(x) \textrm { is false }\}$ may not be measurable.
If $(X,\Sigma,\mu)$ is not complete, we may have a sequence of measurable functions $f_n$ converging monotonically to $f$ almost everywhere, but $f$ not being measurable.
Let us see an example. If $(X,\Sigma,\mu)$ is not complete, there is a set $N \in \Sigma$ such that $\mu(N)=0$ and a set $A \subseteq N$ such that $A \notin \Sigma$. Take, for all $n$, $f_n=0$ and take $f=\chi_A$. Then $f_n\rightarrow f$ monotone increasing almost everywhere. But $f$ is NOT measurable and $\int_X f$ is not defined.