Let ($\Omega,F,\mu$) be a finite measure space and $f: \Omega \rightarrow [0,\infty]$ be a nonnegative measurable function. Prove that :
If$f(\Omega)\subseteq \mathbb{N}$ then $ \int fd\mu= \sum_{n=1}^{\infty}\mu(A_n) \in [0,\infty]$ where $A_n=${$n\leq f$}
I defined $B_k=${$f \in [k,k+1)$}. We see that $A_n=\cup_{k=n}^{\infty}B_k.$ I thought that we first show it for simpleton functions. Let $f=\sum_{i=1}^{n}\alpha_i 1_{A_i}$ for measurable sets $A_i \in F$. Because $f(\Omega)\subseteq \mathbb{N}$ we have that the $\alpha_i \in \mathbb{N}$ and that $f=\sum_{k=1}^{n}k 1_{B_k}$, where we let $n$ be big enough so that we catch all of them. If for some $k$ $f\notin B_k$ it adds nothing to the sum, as the measure of the empty set is $0$. Ultimately:
$\int_\Omega fdy =\sum_{k=1}^{n}k\mu(B_k)=\sum_{k=1}^{\infty}k\mu(B_k)=\mu(B_1)+\mu(B_2)+\mu(B_2)+\mu(B_3)+\mu(B_3)+\mu(B_3)+...=\sum_{k=1}^{\infty}\mu(B_k)+\sum_{k=2}^{\infty}\mu(B_k)+\sum_{k=3}^{\infty}\mu(B_k)+...=\mu(\cup_{k=1}^{\infty}B_k)+\mu(\cup_{k=2}^{\infty}B_k)+\mu(\cup_{k=3}^{\infty}B_k)+...=\mu(A_1)+\mu(A_2)+\mu(A_3)+...=\sum_{n=1}^{\infty}\mu(A_n)$.
If there is no mistake this should be enough for simpleton function. But now we have to do it for a series of simpleton functions $f_n$ with $f_n\uparrow f$ and I do struggle there. I would be thankful for any advices.
Hint: Apply Monotone convergence theorem to $$ \int_\Omega \lim_{N\to\infty}\sum_{n=1}^N 1_{\{n\le f(x)\}}d\mu. $$