Proving that $1_{\limsup A_n}=\limsup1_{A_n}$. Proving that $$1_{\limsup A_n}=\limsup1_{A_n}$$
$$1_{\limsup A_n}= \begin{cases}1, x \in \bigcap_{n=1}^{+ \infty}\bigcup_{k=n}^{+\infty}A_n \\ 0, x \notin \ldots\end{cases}$$
or that $x$ is in infinitely many $A_n$. And also:
$$\limsup1_{A_n}=\inf_{n\geq1}\{\sup_{k \geq n}\{1_{A_n}\}\}$$ this function seems to be $1$ when $x\in A_n$ (just one set). I clearly am not understanding this $\inf\{\sup\{\ldots\}\}$ definition correctly. Can someone enlighten this for me. I have a test very soon.

Let $f_n(x)=\sup_{k\geq n}\boldsymbol 1_{A_k}(x)$. First,
$$f_n(x)\in\{0,1\},$$
for all $x$ and all $n$. For $x$ fixed, the sequence $(f_n(x))_n$ is decreasing, therefore $$\lim_{n\to \infty }f_n(x)=\inf_{n\in\mathbb N} f_n(x)\in \{0,1\}$$
since an infimum is an adherent point. Let fix $x$. Now, $$\inf_{n\in\mathbb N}f_n(x)=0\iff \exists k:f_k(x)=0.$$ In other word, if only if there is a $k$ s.t. for all $m\geq k$, $\boldsymbol 1_{A_m}(x)=0$. Therefore, $$\lim_{n\to \infty }f_n(x)=0\iff \exists k\in\mathbb N: \forall m\geq k,  \boldsymbol 1_{A_m}(x)=0\iff x\in \bigcup_{k}\bigcap_{m\geq k}A_m^c\iff x\notin \bigcap_{k}\bigcup_{m\geq k}A_m\iff x\notin \limsup A_m^c\iff \boldsymbol 1_{\limsup A_n}(x)=0.$$
I let you do the case where $\lim_{n\to \infty }f_n(x)=1$. 
