Show $ \lim_{n \rightarrow \infty} \int_{\Omega} f_n\,d\mu_n = \int_{\Omega} f\,d\mu $ 
Let $(\mu_n)_{n \in \mathbb{N}}$ be a sequence of measures on $A$ satisfying  $\mu_n \nearrow \mu$ and $f,f_n \in M^+$ ( $n \in \mathbb{N}$ ) satisfying $f_n \nearrow f$. Show that $$ \lim_{n \rightarrow \infty} \int_{\Omega} f_n\, d\mu_n =  \int_{\Omega} f\, d\mu. $$

So my idea is to show this claim for a fixed $f_n$. Maybe we could use this then for the claim itself. Moreover I think that monotone convergence theorem could be helpful.
Remark: $M^+ $ := {$ f | f$ is measurable and non-negative }
 A: So $\mu_{n}$ is absolutely continuous with respect to $\mu$, so $\dfrac{d\mu_{n}}{d\mu}$ exists, and we see that 
\begin{align*}
\mu_{n}(A)=\int_{A}\dfrac{d\mu_{n}}{d\mu}d\mu\leq\mu_{n+1}(A)=\int_{A}\dfrac{d\mu_{n+1}}{d\mu}d\mu,
\end{align*}
so
\begin{align*}
\int_{A}\left(\dfrac{d\mu_{n+1}}{d\mu}-\dfrac{d\mu_{n}}{d\mu}\right)d\mu\geq 0
\end{align*}
holds for every $A$, so 
\begin{align*}
\dfrac{d\mu_{n+1}}{d\mu}\geq\dfrac{d\mu_{n}}{d\mu},~~~~\text{a.e.}.
\end{align*}
Since
\begin{align*}
\mu_{n}(A)=\int_{A}\dfrac{d\mu_{n}}{d\mu}d\mu,
\end{align*}
we take $n\rightarrow\infty$ and using Monotone Convergence Theorem to deduce that
\begin{align*}
\mu(A)=\int_{A}\lim_{n\rightarrow\infty}\dfrac{d\mu_{n}}{d\mu}d\mu,
\end{align*}
this shows that 
\begin{align*}
\lim_{n\rightarrow\infty}\dfrac{d\mu_{n}}{d\mu}=1,~~~~\text{a.e}.
\end{align*}
Now we write that
\begin{align*}
\int_{\Omega}f_{n}d\mu_{n}=\int_{\Omega}f_{n}\dfrac{d\mu_{n}}{d\mu}d\mu,
\end{align*}
and use the fact that $f_{n}\dfrac{d\mu_{n}}{d\mu}\uparrow f$ a.e. to get the result.
