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Let $(X,\mathscr B_X)$ and $(Y,\mathscr B_Y)$ be two measure spaces and $(Z,\mathscr B_Z)$ be their product space. Consider two finite measures (not necessarily product measures) $\mu,\nu$ on $(Z,\mathscr B_Z)$. Suppose that for any $A\in \mathscr B_X$ and for any $B\in \mathscr B_X$ it holds that $$ \mu(A\times B)\geq \nu(A\times B). $$ Does it mean that $\mu(C)\geq \nu(C)$ for any $C\in \mathscr B_Z$?

Some thoughts: clearly, the question can be equivalently stated as suppose that a measure $\lambda$ on $(Z,\mathscr B_Z)$ is non-negative on rectangles. It it a non-negative measure?

I was going to apply monotone class-like arguments, but I do not know what to do with complements. Clearly, the inequality is preserved under countable disjoint unions, though.

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up vote 3 down vote accepted

Edit: This answer is not quite correct as stated; see the comments for corrections.

The answer is yes. We'll consider your formulation: if $\lambda$ is a signed measure which is nonnegative on all rectangles, then $\lambda$ is a nonnegative measure.

First observe the following consequence of the Carathéodory extension theorem:

Fact. Let $(X, \mathcal{F})$ be a measurable space, and $\mu$ a positive finite measure on $\mathcal{F}$. Suppose $\mathcal{A}$ is an algebra that generates $\mathcal{F}$. Then for any $B \in \mathcal{F}$, we have $$\mu(B) = \inf \{\mu(A) : A \in \mathcal{A}, B \subset A\}.$$

(Consider $\mu$ as a premeasure on $\mathcal{A}$ and let $\mu^*$ be the corresponding outer measure. The right side of the above equation is $\mu^*(B)$, but the uniqueness in the Carathéodory theorem says that $\mu = \mu^*$ on $\mathcal{F}$.)

Now apply the above fact with $\mu = |\lambda|$ the total variation, and $\mathcal{A}$ the collection of all finite unions of rectangles. If $B$ is any measurable subset of the product, then for any $\epsilon > 0$ we may find $A$ a finite union of rectangles with $B \subset A$ and $|\lambda|(A \setminus B) < \epsilon$. But then $|\lambda(A) - \lambda(B)| = |\lambda(A \setminus B)| \le |\lambda|(A \setminus B) < \epsilon$. We have $\lambda(A) \ge 0$ by assumption so $\lambda(B) \ge -\epsilon$. $\epsilon$ was arbitrary so $\lambda(B) \ge 0$.

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Thanks, Nate. Your proof is almost clear to me - the only question concerns the Fact. Namely, in the definition of the outer measure generated by the premeasure that I have in hand (Folland,Bogachev) the infimum is taken over countable sums: $$ \mu^*(B):= \inf\left\{\sum\limits_{j=1}^\infty \mu(A_j):A_j\in \mathcal A,B\subset \bigcup_{j=1}^\infty A_j\right\}. $$ Since the countable union does not have to be an element of $\mathcal A$, in general it only follows from the definition (directly) that $\mu^*(B)\leq \inf\{\mu(A):A\in \mathcal A,B\subset A\}$ [...] – Ilya Aug 1 '12 at 7:13
[...] I guess, it is ok since in the second part of your answer (where you talk about $\lambda$) we can take $A$ to be the countable union of rectangles - we know already that $\lambda$ is non-negative on such sets. However, I wonder if we the following definition of the outer measure is also correct: $$ \mu^*(B):= \inf\left\{\sum\limits_{j=1}^n \mu(A_j):A_j\in \mathcal A,B\subset \bigcup_{j=1}^n A_j,n\in \mathbb N\right\}. $$ I tried to prove it by myself, but wasn't succesful in it. – Ilya Aug 1 '12 at 7:18
@Ilya I think, this definition of outer measure it nor the same. Because for this definition $\mu^*(\mathbb{Q}\cap[0,1])\geq 1$ – Norbert Aug 1 '12 at 7:33
@Norbert: here we are talking about $\mathcal A$ consisting of finite unions of $(a,b)$ - right? Then would it mean that $$ \mu(\mathbb Q\cap [0,1])\neq \inf\{\mu(A):A\in \mathcal A,B\subset A\} $$ since the RHS is non-zero ($\mu$ being Lebesgue measure)? – Ilya Aug 1 '12 at 7:42
In the comment above, $B = \mathbb Q\cap[0,1]$ - cannot fix it directly there. – Ilya Aug 1 '12 at 8:02

For any $M\in \mathscr{B}_Z$ and $\sigma$-additive measure $\lambda$ there exist

$$\{A_{i,j}:i,j\in\mathbb{N}\}\subset\mathscr{R}_Z \qquad\text{ and }\qquad A_0\in\mathscr{B}_Z$$

such that

$$ M=\left(\bigcap\limits_{i=1}^\infty\bigcup\limits_{j=1}^\infty A_{i,j}\right)\setminus A_0 $$ $$ A_{i,j}\subseteq A_{i,j+1} \qquad\text{ and }\qquad \bigcup\limits_{j=1}^\infty A_{i+1,j}\subseteq \bigcup\limits_{j=1}^\infty A_{i,j} \qquad\text{ and }\qquad \lambda(A_0)=0 $$ I think these properties are enough to show that $\lambda\geq 0$.

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Thanks - but why do such sets $A_{ij}$ exist? – Ilya Jul 31 '12 at 20:57
Oh, this is a long story. I've found this result in Действительный анализ в задачах Ульянов П.Л., Бахвалов А.Н. problem 7.45. You can download it here – Norbert Jul 31 '12 at 21:00
Sure, I'll take a look. – Ilya Jul 31 '12 at 21:03
Privet, @SeyhmusGüngören! I'm from Russia, you can realize this by looking into my profile :) – Norbert Jul 31 '12 at 22:56

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