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Let $(X, \Lambda)$ be a measurable space and $\mu$ be a measure on $(X^2, \Lambda \otimes \Lambda)$. Assume that the diagonal $\Delta = \{(x, y) : x = y\}$ is measurable. Suppose $\mu$ has the property that $$\mu(S \times T) = \mu((S \cap T)^2)$$ for any sets $S, T \in \Lambda$. Is it the case that $\mu$ must be nonzero only on the diagonal, i.e. $\mu(\overline{\Delta}) = 0$?

This seems intuitively obvious but I can't figure out how to prove it. My hope was to use a decreasing sequence of covers of the diagonal using squares $S \times S$, but I don't know how to obtain such a sequence without more assumptions on $X$.

lily
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    I don't think I see why it is intuitively clear. If the measure on $X^2$ has the property that every measurable set can be written as a countable union of sets of the form $S\times T$ then the result follows, however if this is not the case, I doubt there is much you can say – Carlyle Apr 25 '24 at 17:12
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    @Carlyle Well, it seemed intuitive when I was picturing a unit square :) Can you give an example of $(X, \Lambda)$ which doesn't have that property? – lily Apr 25 '24 at 17:28
  • I realized it may have been unclear but I meant $X^2$ should be equipped with the product $\sigma$-algebra $\Lambda \otimes \Lambda$ – lily Apr 25 '24 at 17:35

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Equipping $X^2$ with product algebra,

$$\Lambda \otimes \Lambda = \sigma(\{B\times T | B,T\in \Lambda\})$$

i.e. the $\sigma$-algebra obtained by closing the set

$$\{B\times T| B,T\in\Lambda\}$$

under countable unions, intersection and complements.

We then note that since $\Delta$ is measurable, so is it's complement, $\overline{\Delta}$. hence

$$ \overline{\Delta} \in \Lambda \otimes \Lambda $$

Which means we can write

$$\overline{\Delta} = \bigcup_{n\in\mathbb{N}}B_n\times T_n $$

And since each $B_n\times T_n$ does not intersect the diagonal, we have $B_n\cap T_n=\emptyset$. Then

\begin{align*} \mu(\overline{\Delta}) &= \sum_{n=1}^{\infty}\mu(B_n\times T_n) \\ &= \sum_{n=1}^{\infty}\mu((B_n\cap T_n)^2) \\ &= \sum_{n=1}^{\infty}\mu(\emptyset) \\ &= 0 \end{align*}

Carlyle
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  • Hello. Is it obvious that $\overline \Delta$ can be written as such a countable union of boxes? I believe this is true because of the proof of Proposition 2.1 here (linked here), but I can't see a simpler reason. In general a product-measurable set need not be of this form, right? (Take the diagonal itself in an uncountable space, for instance). – Izaak van Dongen Apr 25 '24 at 22:51
  • @IzaakvanDongen perhaps I am confused, but it seems that the definition of the product $\sigma$-algebra is precisely all those elements obtained by taking countable unions and complements of boxes. – Carlyle Apr 26 '24 at 07:08
  • @IzaakvanDongen but I realise now then that the diagonal being measurable and the question stating that the relevant $\sigma$-algebra is the product algebra, necessarily implies that $X$ is countable for the reason you pointed out – Carlyle Apr 26 '24 at 07:09
  • The definition I know of the product sigma-algebra is the sigma-algebra generated by boxes. What I'm saying is that in general, "$A$ lies in the sigma-algebra generated by $(B_i){i \in I}$" does not imply that $A$ is a countable union of $B_i$. This is because "the sigma-algebra generated by $(B_i){i \in I}$" also contains unions of complements, complements of unions, unions of complements of unions, complements of unions of complements, and so on (transfinitely)... I don't think the diagonal being measurable implies $X$ is countable - take $\Bbb R$ with the Borel algebra, for example. – Izaak van Dongen Apr 26 '24 at 09:45
  • @IzaakvanDongen you are correct. Apologies for my mistake. – Carlyle Apr 26 '24 at 11:39