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My confusion I think stems from lack of understanding of Borel sets. I saw this exact question basically but I didn't understand the portion that's confusing me. Let me state the relevant part of the proof briefly:

Let $E\in\mathcal{B}(\mathbb{R}^n)$, $F\in\mathcal{B}(\mathbb{R}^m)$. Then $E\times F \in\mathcal{B}(\mathbb{R}^{n+m})$.

Step 1: Pick an open set $D\subset \mathbb{R}^n$ and consider $\{F\subset \mathbb{R}^m:D\times F\in\mathcal{B}(\mathbb{R}^{n+m})\}$. This is a $\sigma$-algebra.

Is there a certain way to go about showing that $D^c\times F$ is also a Borel set, or should that just be obvious?

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letslearnmath
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    The Borel sets are the $\sigma$-algebra generated by the open sets. The Borel $\sigma$-algebra contains the open sets and the closed sets, but also a whole lot more sets. Honestly, it seems like you don't understand what a $\sigma$-algebra is, so I'd suggest looking that up, perhaps on wikipedia or something. – jgon Nov 08 '17 at 01:58
  • -_- yeah idk why but my friends and I just weren't thinking about the $\sigma$-algebras part. We reviewed them a couple of weeks ago, then came back to measure theory recently, and we got stuck on thinking Borel was either open sets or closed. Our text says Borel sets are "the $\sigma$-algebra generated by the family of all open sets in $\mathbb{R}^n$" which we confused as open sets or closed sets, but you're right - $\sigma$-algebras also have finite and infinite unions and intersections of sets and complements so of course we'll have a mix. Thanks! – letslearnmath Nov 08 '17 at 02:12
  • Okay I just updated my question, should it also be obvious that $D^c\times F$ is also a Borel set? Or should I try to prove this by showing that it's created by some union/intersection of open sets/their complements? – letslearnmath Nov 08 '17 at 02:20
  • I think I need more context, why are you trying to show $D^c\times F$ is a Borel set? – jgon Nov 08 '17 at 02:24
  • Ah, sorry, I'm trying to show that ${F\subset \mathbb{R}^m:D\times F\in\mathcal{B}(\mathbb{R}^{n+m})}$ is a $\sigma$-algebra. So actually I need to show that $F^c$ is also in the set, and so $D\times F^c$ is Borel. Thanks for the responses by the way. – letslearnmath Nov 08 '17 at 02:28
  • Ok hang on I'll post an answer in a bit then. – jgon Nov 08 '17 at 02:30

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$\newcommand{\BB}{\mathcal{B}}\newcommand{\RR}{\mathbb{R}}\newcommand{\AA}{\mathcal{A}}$ So you want to show that $\AA_D=\{F\subset\RR^m : D\times F \in\BB(\RR^{n+m})\}$ is a $\sigma$-algebra. So if $F\in\AA_D$, you want to show $F^c\in\AA_D$. So we have that $D\times F$ is Borel, and $D\times \RR^m$ is open, and hence Borel.

Now we have that $$(D\times F)^c \cap (D\times \RR^m)= (D\times \RR^m)\setminus (D\times F) = D\times (\RR^m \setminus F) = D\times F^c.$$ Therefore $D\times F^c$ is Borel, so $F^c \in \AA_D$.

jgon
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