Using the Lemma of Fatou, I've deduced a statement together with a counter-example, so I am now looking where my mistake lies.
The statement:
Let $(\Omega,\mathcal{A},\mu)$ be a measurable space and $A\in\mathbb R$. For all $n\in\mathbb N$ let $f_n:\Omega\to \mathbb R$ be measurable such that $\ \mu\left(\bigcup_{n\geq1}{f_n^{-1}\left(A\right)}\right)<\infty$.
Then we have: $$ \limsup_{n\rightarrow\infty}\mu\left(f_n^{-1}\left(A\right)\right)\le \mu\left(\left({\limsup_{n\rightarrow\infty}{f_n}}\right)^{-1}\left(A\right)\right) $$
The proof:
We have in general, that $$ \mu\left(\{w\in\Omega\mid\left({\limsup_{n\rightarrow\infty}{f_n\left(w\right)}}\right)\in A\}\right)=\int_{\Omega}\mathbf{1}_{\{w\in\Omega\mid{\limsup_{n\rightarrow\infty}{f_n\left(w\right)}}\in A\}}d\mu =\int_{\Omega}\limsup_{n\rightarrow\infty}{\mathbf{1}_{\{w\in\Omega\mid f_n\left(w\right)\in A\}}}d\mu $$
Furthermore, if we have $\int_{\Omega}\sup_{n\geq1}\mathbb{1}_{\{w\in\Omega\mid f_n\left(w\right)\in A\}}d\mu<\infty$, we can apply Fatous Lemma to pull out the $\limsup$: $$ \geq\limsup_{n\rightarrow\infty}\int_{\Omega}\mathbb{1}_{\{w\in\Omega\mid f_n\left(w\right)\in A\}}d\mu =\limsup_{n\rightarrow\infty}\mu\left(\{w\in\Omega\mid f_n\left(w\right)\in A\}\right) $$
Finally, we have that $$ \int_{\Omega}\sup_{n\geq1}\mathbf{1}_{\{w\in\Omega\mid f_n\left(w\right)\in A\}}d\mu =\int_{\Omega}\mathbf{1}_{\bigcup_{n\geq1}{\{w}\in\Omega\mid f_n\left(w\right)\in A\}}d\mu=\mu\left(\bigcup_{n\geq1}{\{w}\in\Omega\mid f_n\left(w\right)\in A\}\right)=\mu\left(\bigcup_{n\geq1}{f_n^{-1}\left(A\right)}\right) $$ And therefore have shown the statement.
The counter-example: Let $ f_n:\mathbb{R}\rightarrow\mathbb{R},x\mapsto nx$, and set as measure space the dirac-measure on $\mathbb R$, that ist $\Omega=\mathbb R, \mathcal A = P(\Omega)$ and $\mu(A)=\begin{cases}1, &1\in A\\\ 0, &\text{else} \end{cases}$.
Then we have: $$ \limsup_{n\rightarrow\infty}\mu\left(f_n^{-1}\left(0,\infty\right)\right)=\limsup_{n\rightarrow\infty}\mu\left(\left(0,\infty\right)\right)=\limsup_{n\rightarrow\infty}\ 1 $$
But we have $$\lim\sup_{n\rightarrow\infty}{f_n}=\begin{cases}0 &\text{ for } x=0\\\ \infty &\text{ for } x>0 \end{cases}$$ , and therefore
$$ \mu\left(\left(\lim{\sup_{n\rightarrow\infty}{f_n}}\right)^{-1}\left(0,\infty\right)\right)=\mu\left(\emptyset\right)=0 $$