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Consider a measurable space $([0,1]\times [0,1], \mathcal{B}([0,1]) \times \mathcal{B}([0,1]))$, and a subset $A:=\{(x,y):x=y\}$ (the diagonal). According to the text book, $A \in \mathcal{B}([0,1]) \times \mathcal{B}([0,1])$. Can any body tell me the reason and in general, how do we check if a given set is a Borel set? Thanks:)

PS: what I cannot see is, since $\mathcal{[0,1]} \times \mathcal{[0,1]}$ by definition is generated by the class $\mathcal{C}=\{I_1 \times I_2\}$, where $I_1, I_2$ are intervals on $[0,1]$, I cannot express $A$ as any countable union/intersection, complement of any $I_1 \times I_2$.

K. Huang
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  • You might use the fact that the product topology is metrizable. So to show $A$ is closed you need only consider sequences; if a sequence from $A$ converges, its limit is clearly in $A$. – Quinn Culver Nov 24 '13 at 19:13
  • Don't Brian's answer and my answer directly address the question raised by your edit? – Trevor Wilson Nov 24 '13 at 19:19
  • @QuinnCulver I think the OP wants to do the problem using the definition of the product $\sigma$-algebra, and not using the fact that it is equal to the Borel $\sigma$-algebra of the space $[0,1] \times [0,1]$. – Trevor Wilson Nov 24 '13 at 19:21
  • @TrevorWilson I think you're right. – Quinn Culver Nov 24 '13 at 19:33

3 Answers3

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By the definition of the product $\sigma$-algebra, $\mathcal{B}([0,1]) \times \mathcal{B}([0,1])$ is generated by sets of the form $X \times Y$ where $X,Y \in \mathcal{B}([0,1])$. If $x$ and $y$ are distinct real numbers in the interval $[0,1]$ then we can find disjoint subintervals $X$ and $Y$ of $[0,1]$ containing $x$ and $y$ respectively, with rational endpoints. We have $X,Y \in \mathcal{B}([0,1])$ and the rectangle $X \times Y$ is disjoint from the diagonal.

Therefore the complement of the diagonal is the union of countably many elements of the product $\sigma$-algebra $\mathcal{B}([0,1]) \times \mathcal{B}([0,1])$, so the diagonal itself is in this $\sigma$-algebra.

Trevor Wilson
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  • Thanks, Trevor. But I don't get that since ${(x,y): x\neq y}$ is uncountable, how one can obtain the complement of the diagonal is the union of Countably elements? – K. Huang Nov 24 '13 at 19:23
  • Oh, perhaps I should have said that more explicitly. It is because we can always take our intervals to have rational endpoints, even if $x$ and $y$ are irrational. So there are only countably many intervals that we need to consider. (Note that the diagonal itself is not the union of countably many rectangles, for the reason you mention. That is why we should consider the complement of the diagonal instead.) – Trevor Wilson Nov 24 '13 at 19:24
  • That makes a lot more sense! Thanks VERY MUCH! – K. Huang Nov 24 '13 at 19:33
  • You're welcome. I'm glad I could help. – Trevor Wilson Nov 24 '13 at 19:37
  • Curious about what you said in the comment, what is the difference of product $\sigma$-algebra and Borel $\sigma$-algebra on $[0,1]\times [0,1]$? How is the second one defined? Thanks. – K. Huang Nov 24 '13 at 20:26
  • Given any set $X$, and given a topology on $X$ one can define the corresponding Borel $\sigma$-algebra on $X$. A topology on $X$ is a collection of subsets of $X$ that we declare to be "open" (obeying some rules that are abstracted from properties of open sets in $\mathbb{R}$.) If we know what the open sets are, we can define the Borel sets to be sets generated from the open sets by complementation and countable unions and intersections.... – Trevor Wilson Nov 24 '13 at 20:56
  • We can define the product topology on $[0,1]\times[0,1]$ and then take the $\sigma$-algebra generated by this, or we can take the $\sigma$-algebra of $[0,1]$ first and then take the product of $\sigma$-algebras. In this case the result turns out to be the same. – Trevor Wilson Nov 24 '13 at 20:58
  • So basically on $R^2$, one construction is $\sigma\langle {[a,b) \times [c,d) : a<b,c<d \in R}\rangle$, the other is $\sigma\langle {A \times B: A,B \in \mathcal{B}(R)}\rangle$? – K. Huang Nov 24 '13 at 22:04
  • Well, I would use open intervals in the first case, because the topology on the product is generated by products of open intervals. (However, as far as the generated $\sigma$-algebra is concerned it works out to be the same.) – Trevor Wilson Nov 24 '13 at 22:30
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$[0,1]^2\setminus A$ is a union of countably many open rectangles, since the open rectangles are a base for the topology. Open rectangles are clearly in the product $\sigma$-algebra, so $[0,1]^2\setminus A$ is, and so is its complement $A$.

For more information see the answers to this question.

Community
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Brian M. Scott
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  • Thanks Brian, I am taking a measure theory course, and nothing about topology is mentioned there. I might need to look into it further. Exactly, what is relation between sigma-algebra and topology? Do you have any good intro source that I can use. Thanks! – K. Huang Nov 24 '13 at 19:30
  • @Tim: The topology isn’t actually necessary here; it’s just that knowing it saves you a little work. Try this instead. Let $\mathscr{I}$ be the family of all intervals of the form $(p,q)$ for rational $p,q\in[0,1]$ with $p<q$ together with all intervals of the forms $[0,p)$ and $(p,q]$ for $p\in(0,1)$; $\mathscr{I}$ is countable. Let $$\mathscr{R}={I\times J:I,J\in\mathscr{I}\text{ and }I\cap J=\varnothing};.$$ Then $\mathscr{R}$ is a countable family of sets that clearly belong to the product $\sigma$-algebra; show that $\bigcup\mathscr{R}=[0,1]^2\setminus A$. You’re welcome! – Brian M. Scott Nov 24 '13 at 19:35
  • @BrianM.Scott. Can we extend $[0,1]$ to $\mathbb{R}$? Will the conclusion still hold? – Mike Brown Sep 22 '15 at 15:32
  • @Mike: Yes, with essentially the same argument. – Brian M. Scott Sep 22 '15 at 18:20
  • @BrianM.Scott. I have difficulty to construct two disjoint rational intervals contain two point. Can you give me a hint? – Mike Brown Sep 22 '15 at 18:58
  • @Mike: I don’t understand what you’re trying to do, I’m afraid. – Brian M. Scott Sep 22 '15 at 19:12
  • @BrianM.Scott. It is easy to see $\bigcup\mathscr{R}\subset\mathbb{R^2}\setminus A$. But for $\mathbb{R^2}\setminus A\subset\bigcup\mathscr{R}$, we need to find a $\mathscr{R}$ contains the point $(x,y)$ and $x\neq y$. – Mike Brown Sep 22 '15 at 19:16
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    @Mike: If $x<y$, let $p,q,r\in\Bbb Q$ be such that $p<x<q<y<r$; then $\langle x,y\rangle\in(p,q)\times(q,r)$, and $\big((p,q)\times(q,r)\big)\cap A=\varnothing$. The argument when $x>y$ is similar. – Brian M. Scott Sep 22 '15 at 19:18
  • @BrianM.Scott. Are you implying you always can find a rational interval for a number? – Mike Brown Sep 22 '15 at 19:28
  • @Mike: The rationals are dense in the reals, so you can always do what I described. – Brian M. Scott Sep 22 '15 at 19:29
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Every closed set is a Borel set, and $A$ is a closed subset of $[0,1]\times [0,1]$. (The map $A\to [0,1]$ given by $(x,y)\mapsto x$ is a homeomorphism, so $A$ is compact.)

bradhd
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  • Thanks Brad, I think you are right. But what I cannot see is, since $\mathcal{[0,1]} \times \mathcal{[0,1]}$ by definition is generated by the class $\mathcal{C}={I_1 \times I_2}$, where $I_1, I_2$ are intervals on $[0,1]$, I cannot express $A$ as any countable union/intersection, complement of any $I_1 \times I_2$. Could you help me with this? – K. Huang Nov 24 '13 at 19:05