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Let $E$ and $F$ be Borel measurable subsets of $\mathbb R^{d_1}$ and $\mathbb R^{d_2}$, respectively. Then $E \times F$ is also Borel measurable in $\mathbb R^{d_1 + d_2}$.

I suppose it is necessary to show that elements of a generator of Borel $\sigma$-algebra on $\mathbb R^{d_1 + d_2}$ is Borel. That is, show that rectangles, i.e., Cartesian product of intervals, are Borel. It seems that I need to use Fubini' theorem. But I do not know how. Any help, please? Thank you!

user156460
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2 Answers2

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Hint: Let $A = \left\{ E \subseteq \mathbb{R}^{d_1} \mid \text{$E \times \mathbb{R}^{d_2}$ is Borel in $\mathbb{R}^{d_1 + d_2}$} \right\}$. Show that $A$ is a $\sigma$-algebra containing all open sets of $\mathbb{R}^{d_1}$, hence all Borel sets. Then make a similar argument for $F$.

Mike
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I don't see the meaning of constructing the $\sigma$-algebra $A$ in Mike's answer. Well, I think actually we can avoid that redundant construction.

What we need to prove is just: If $E$ is a Borel set in $\mathbb R^{d_1}$, then $E\times \mathbb R^{d_2}$ is a Borel set in $\mathbb R^{d_1+d_2}$.

Proof: This is simple. Because the set in the second position of the Cartesian product is the whole space $\mathbb R^{d_2}$, we can interchange the operations of taking Cartesian product and taking union/intersection. We can prove a similar result for $F$ in the same way. And therefore, if $E$ is a Borel set in $\mathbb R^{d_1}$ and $F$ is a Borel set in $\mathbb R^{d_2}$, then $E\times F(=[E\times \mathbb R^{d_2} ]\cap [\mathbb R^{d_1}\times F]\,)$ is a Borel set in $\mathbb R^{d_1+d_2}$.

$\tag*{$\square$}$

Sam Wong
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  • And how do you know that $E\times \Bbb{R}^{d_2}$ is a Borel set? Btw typo in last line it should be $d_1+d_2$. – peek-a-boo Aug 04 '23 at 12:25
  • @peek-a-boo I have proved it by words that Because the set in the second position of the Cartesian product is the whole space $\mathbb R^{d_2}$, we can interchange the operations of taking Cartesian product and taking union/intersection. $\textbf{Mathematical Proof:}$ Since $E$ is Borel in $\mathbb R^{d_1}$, we may write $E$ as a countable union/intersection of open sets in $\mathbb R^{d_1}$, where there may be finite number of set operations involved but the reasoning is the same. Therefore, $E=\cup_i A_i$ for some $A_i$ open in $\mathbb R^{d_1}$, where the union is a countable union. – Sam Wong Aug 04 '23 at 12:50
  • (cont') Thus, $E\times \mathbb R^{d_2}=(\cup_i A_i)\times \mathbb R^{d_2} = \cup_i [A_i \times \mathbb R^{d_2}]$, where $A_i \times \mathbb R^{d_2}$ is open in $\mathbb R^{d_1} \times \mathbb R^{d_2}$, because $A_i$ and $\mathbb R^{d_2}$ are open in $\mathbb R^{d_1}$ and $\mathbb R^{d_2}$, respectively. So $E\times \mathbb R^{d_2}$ is a Borel set. $\blacksquare$ Definitely, $E$ can be something like $\cup_i \cap_j \cup_k A_{i,j,k}$, but the reasoning is pretty much the same. – Sam Wong Aug 04 '23 at 12:53
  • I doubt that’s true about Borel sets. Im no set theory wiz but a brief search about Borel hierarchy and the like already suggests that even if what you said is true (which I highly doubt) it is very non-trivial. The other approach of simply showing something is a $\sigma$-algebra containing the basic things you want is the much simpler option (and also this idea is used in several proofs). – peek-a-boo Aug 04 '23 at 13:31
  • Sigma algebras are typically very huge beasts, and I don’t think they usually come with “simple explicit descriptions”. Working with the general abstract definition of a generated sigma-algebra often suffices (atleast I’ve gotten by without delving into any further set theory). – peek-a-boo Aug 04 '23 at 13:34
  • @peek-a-boo Borel sigma algebra is generated by open sets by definition. I was working on the generated elements, which is a general approach as well. I actually proved that the Cartesian product of a countable union generated elements (i.e. open sets in $\mathbb R^{d_1}$) and the set $\mathbb R^{d_2}$ equals to the countable union of the Cartesian product of the generated elements and the set $\mathbb R^{d_2}$. We can similarly prove that Cartesian product can be swapped with a countable intersection or set complement. A Borel set must be formed by generated elements with these 3 operations. – Sam Wong Aug 04 '23 at 14:33
  • @peek-a-boo Therefore, we rigorously proved that $E\times \mathbb R^{d_2}$ is Borel. $\blacksquare$ I was just too lazy to write down everything, but every step in my reasoning can be written out with the set-theoretic language such that it is as rigorous as you want. – Sam Wong Aug 04 '23 at 14:35
  • Im not disputing the stuff about cartesian products and interchanging with countable unions/intersections. That’s all trivial stuff. What i am disputing is your very first claim about Borel sets. Yes, by definition the Borel sigma algebra is generated by the open sets, and yes these include all countable intersections and unions of open sets. But how can you be sure these are all? I doubt it. See here. In general I doubt the generated sigma algebra has a nice description. – peek-a-boo Aug 04 '23 at 14:38
  • @peek-a-boo We don’t care what a specific Borel set looks like. It must be formed by countable union, countable intersection, and set complement. You can perform finitely many set operations from these three. All these 3 operations can be swapped with the operation of taking Cartesian product. So, $E \times \mathbb R^{d_2}$ is Borel as well. – Sam Wong Aug 04 '23 at 14:44
  • If you don’t care what a specific Borel set looks like, and you’re only using that unions, intersections comlements play nicely with cartesian products then this proof is essentially the same as the other one. The other answer is pretty direct and tells us to show something is a sigma algebra which contains open sets, and hence contains all Borel sets. – peek-a-boo Aug 04 '23 at 14:56