Is the upper limit projection Borel Let $M$ space metric compact, $\pi:M\times\mathbb{R}^k\rightarrow M$ projection such that $\pi(x,y)=x$. Let $f_n:M\times\mathbb{R}^k\rightarrow \mathbb{R}$ continuous and $f:M\times\mathbb{R}^k\rightarrow \mathbb{R}$ such that $f=\limsup f_n$ 
then, as I can prove that $\pi(f^{-1}\{(0,1 ]\})$ is Borel?
grateful can give me any suggestions
 A: Let $\mathcal{B}$ denote the Borel sets. For all $t\in \mathbb{R}$
$$(\sup_n f_n)^{-1} ([-\infty,t]) = \{ x\in X: f_n(x) \le t ~\forall~ n \in \mathbb{N} \} = \bigcap_{n\in \mathbb{N}} f_n^{-1} ([-\infty, t]) \in \mathcal{B}$$
And so $\sup_n f_n$ is $\mathcal{B}$-measurable.
Similarily, 
$\lim \sup_n f_n$ and $\lim \inf_n f_n$ are $\mathcal{B}$-measurable since they are (by definition) a $\sup$ followed by an $\inf$ or an $\inf$ followed by a $\sup$.
Finally, the projection map is a continuous mapping which is a general fact from topology and if a function is continuous then the preimage of an open set is open (see Munkres or do this as a simple exercise... or check here, here, or here).  
You should know (or can easily show) that $(0,1] \in \mathcal{B}$.  So as $f$ is $\mathcal{B}$-measurable it follows that $f^{-1}((0,1])\in\mathcal{B}$.  And from this it is an elementary fact that $\pi(B)\in\mathcal{B}$ for any Borel set $B$ because of facts about the product topology.  (Note: I am being a little abusive because the $\mathcal{B}$ at the top is a different type of collection of Borel sets than the one in the last sentence for the obvious reason). 
