Monotone convergence theorem (measure theory) proof 
I am stuck with the final line on the image, why does $f_n \to f$ (which also increases), mean that $\cup_n A_n = \Omega$?

Choose $x \in \Omega$, then $f_n(x) \uparrow f(x)$. If $f(x) = 0$ then $f_n(x) = 0$ for all $n$ and hence $x \in A_n$ for all $n$, so suppose $f(x)>0$.
You have
$f(x) \ge \sum_k c_k 1_{E_k}(x)$ and hence
$f(x) > \sum_k (c_k-\epsilon) 1_{E_k}(x)$. Since $f_n(x) \uparrow f(x)$, it follows that $x \in A_n$ for some $n$.
As an aside, to avoid such difficulties, some authors define the integral as the $\sup$ over simple, measurable $g$ such that $g \le f$ and if $f(x) >0$, then $g(x) < f(x)$. (Then they show that you get the same value by using the 'usual' set of $g$', of course.)

Answer to comment: (I am assuming the $E_k$ are disjoint here, if not, it is
easy to modify them so they are.)
Note that many proofs/concepts have an underlying idea that is then built upon. The basic idea here are that if $A_n$ are an increasing collection of sets then $\mu(A_n) \to \mu(\cup_k A_k)$. This is the essence of the monotone
convergence theorem (think of the functions $1_{A_n}$).
In this question, the $A_n$ represent the collection of points for which $f_n$ is 'more or less' greater than of equal to $g$. By 'more or less', I mean the $\epsilon$ condition. The value of the $A_n$ is that the are increasing (nested) and $\cup_k A_k = \Omega$, so for any measurable $E$, we have $\mu(A_n \cap E) \to \mu(E)$.
Note that $\int f_n \ge \int_{A_n} f_n \ge \int_{A_n} (g-\epsilon)  \ge \int_{A_n} g - \epsilon \sum_k \mu(E_j)$.
Note that $\int_{A_n} g = \sum c_k \mu(E_k \cap A_n)$, and so
$\int_{A_n} g \to \int g$. Since $f_{n+1} \ge f_n$ we have that $\int f_n$ increasing, then the above gives
$\lim_{n \to \infty} \int f_n \ge \lim_{n \to \infty} \int_{A_n} g - \epsilon \sum_k \mu(E_j) = \int g -\epsilon \sum_k \mu(E_j)$.
Since $g$ is simple, $\sum_k \mu(E_j)$ is finite, and since $\epsilon>0$ is
arbitrary, we have $\lim_{n \to \infty} \int f_n \ge \int g$
