Simple inequality for measures

Question

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.

Answer 1

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 $\lambda(B) \ge 0$.

Answer 2

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 $\lambda\geq 0$.