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Suppose $(E_i, \mathcal E_i)$, $i = 1, \dots, n$, are measurable spaces and let $E := E_1 \times \dots \times E_n$, equipped with the product $\sigma$-algebra, denoted by $\mathcal E$.

Suppose $\psi$ is a measure on $E$, and suppose $\psi(B) > 0$ for some $B \in \mathcal E$.

(i) Are there sets $A_i \in \mathcal E_i$, $i = 1, \dots, n$, such that $A:=A_1 \times \dots \times A_n \subset B$ and $\psi(A) > 0$?

Update

(ii) What if $\psi$ is the product measure $\psi_1 \otimes \dots \otimes \psi_n$ of some measures $\psi_i$ on $(E_i, \mathcal E_i)$?

(iii) Possibly easier, what if all $(E_i,\mathcal E_i)$ and measures $\psi_i$ are identical?

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It's not quite as simple for product measures, but the answer is still no.

Say $m_n$ is Lebesgue measure on $\Bbb R^n$. Let $E\subset\Bbb R$ such that $m_1(E)>0$ but $E$ contains no interval. Define $S\subset \Bbb R^2$ by $$S=\{(x,y):x+y\in E\}.$$Then $m_2(S)=\infty$. But if $A\times B\subset S$ then $A+B\subset E$. Hence $m_1(A)$ and $m_1(B)$ cannot both be positive, because that would imply that $A+B$ contained an interval.

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  • $\begingroup$ thank you, very intriguing. There are some details that I would need to verify, such as (i) there exists such a set $E$ -- this is in fact Rudin, Real and Complex Analysis, exercise 6; (ii) $m_2(S) = \infty$; and (iii) $m_1(A) > 0$ and $m_1(B) > 0$ imply that $A+B$ contains an interval. Would you be able to clarify one or more of these? Thanks a lot in advance. $\endgroup$ – Joris Bierkens Jun 14 '16 at 18:02
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    $\begingroup$ @JorisBierkens (i) Enumerate the rationals $r_1,r_2,\dots$. Say $a_n>0$ and $\sum a_n<\infty$. Let $E=\Bbb R\setminus\bigcup(r_n-a_n,r_n+a_n)$. (ii) Use Fubini (or more properly, Tonelli. (iii) This is also somewhere in Rudin. Replacing $A$ and $B$ by subsets if necessary, wlog $0<m(A)<\infty$ and $0<m(B)<\infty$. Let $f=\chi_A$ and $g=\chi_B$. Let $f*g$ denote the convolution. Then $\int f*g>0$ and $f*g$ is continuous, so $f*g>0$ on some interval; that interval is contained in $A+B$. Details for (iii) are in Rudin somewhere - actually I believe that (iii) itself is an exercise... $\endgroup$ – David C. Ullrich Jun 14 '16 at 18:11
  • $\begingroup$ Thanks a lot, I will try to get my head around this! $\endgroup$ – Joris Bierkens Jun 14 '16 at 19:09
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Consider $E_1 = E_2 = \mathbb{R}$, each with the usual Borel $\sigma$-algebra, and let $\phi$ be some measure on $\mathbb{R}$ absolutely continuous wrt Lebesgue measure. Then the measure $\psi$ on $\mathbb{R}^2$ defined on the $\pi$-system of rectangular sets by $\psi(A \times B) = \phi(A \cap B)$ (where $A$, $B$ are Borel sets in $\mathbb{R}$) is supported on the diagonal $\{ (x,x) | x \in \mathbb{R}\}$ of the product space. So the diagonal itself is a set of positive measure under $\psi$, but the only rectangular sets it contains are singletons $\{(x,x)\}$ for $x \in \mathbb{R}$, and by absolute continuity of $\phi$, these singletons all have measure $0$.

An alternative way to think about the measure $\psi$ in the construction above is as the law of a random variable of the form $(X,X)$ taking values in $\mathbb{R}^2$, so that the two coordinates of the random variable are equal almost surely - $\phi$ then corresponds to the law of $X$ on $\mathbb{R}$. This construction is maybe less cryptic than the purely measure-theoretic one in the first paragraph.

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  • $\begingroup$ thanks, excellent and illuminating answer. I have specialized my question to a more specific case I am interested in. Do you have any suggestions? $\endgroup$ – Joris Bierkens Jun 14 '16 at 17:16
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    $\begingroup$ This would be much better imo if you didn't call $\psi$ a "product" measure. (Another way to define $\psi$ that seems less cryptic to me is $\psi(E)=\phi(t:(t,t)=in E\{\})$.) $\endgroup$ – David C. Ullrich Jun 14 '16 at 17:30
  • $\begingroup$ Thanks for pointing out the typo - have now removed "product". $\endgroup$ – Mark Jun 14 '16 at 17:31

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