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Let $X=Y=[0,1]$, equipped $X$ by giving Lebesgue measure on Borel sigma-algebra and equipped $Y$ by giving counting measure on the power set of $Y$. Define $D=\{(x,x):0\leq x\leq 1\}$ then how do we show that $D$ is measurable set in product sigma-algebra?

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$D$ is closed and hence $$ D\in \mathcal{B}([0,1]^2)=\mathcal{B}([0,1])\otimes \mathcal{B}([0,1]). $$

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  • $\begingroup$ Sigma-algebra in $Y$ is not Borel sigma-algebra. $\endgroup$ – user54992 May 12 '14 at 7:22
  • $\begingroup$ No, but the product sigma-algebra obtained by equipping $Y$ with the power set is larger than this and hence... $\endgroup$ – Stefan Hansen May 12 '14 at 7:23
  • $\begingroup$ Oho! Yeah, but why $D\in \mathcal{B}([0,1]^2)$? $\endgroup$ – user54992 May 12 '14 at 7:25
  • $\begingroup$ $D$ is closed in $[0,1]^2$ and $\mathcal{B}([0,1]^2)$ is generated by all open (or closed) sets in $[0,1]$. $\endgroup$ – Stefan Hansen May 12 '14 at 7:29
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    $\begingroup$ Yes, the problem is that $(\lambda\times\mu)(D)$ isn't uniquely defined c.f. my previous comments. I think you need to revisit the section in your book/notes about product measure. I'm sure that somewhere in there it says that $\sigma$-finiteness is needed in order for the product measure to uniquely extend. $\endgroup$ – Stefan Hansen May 12 '14 at 8:11
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Hint: $D$ is a graph of a non-negative measurable function mapping from $Y$ to $X$.

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