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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

Ali Caglayan
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helmonio
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  • So, you want official sources, you don't want someone to merely write a proof here.... OK, if you say so. – GEdgar Aug 29 '14 at 01:36
  • It is difficult to give useful suggestions if we know so little about what you know. Do you know what measurable functions are, and have you studied basic properties of measurable functions? Also, you may want to precise what kind of help exactly you need. Do you need help just starting the question, i.e., do you understand what you're trying to prove? Or are you able to start the question but get stuck (in which case indicate what you have done so far and why you are stuck)? – user78270 Sep 01 '14 at 13:20
  • Related question (no duplicate) of the same user: http://math.stackexchange.com/questions/910147/projection-of-a-set-g-delta. This might help in guessing what @helmonio has already tried. – PhoemueX Sep 04 '14 at 17:22

1 Answers1

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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).

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Squirtle
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