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Consider two finite measures $\mu$ and $\nu$ on the measurable space $(X\times Y, \mathcal{B}(X)\otimes \mathcal{B}(Y))$, where $X$ and $Y$ are separable metric spaces and $\mathcal{B}$ denotes the Borel sigma-algebra generated by the natural metric topology.

If I know that $$ \mu(A\times B) \leq \nu(A\times B) \quad \quad \forall A \in \mathcal{B}(X), B\in \mathcal{B}(Y), $$ can I conclude that $\mu(C)\leq \nu(C)$ for all $C\in\mathcal{B}(X)\otimes \mathcal{B}(Y)=\mathcal{B}(X\times Y)$ ?

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    Do you know transfinite induction? – 2017-01-16
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    @tomasz No, I haven't heard of that. Can it be proven using that? – 2017-01-16
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    Yes, but (as usual in this context) it can be avoided using the monotone class theorem. – 2017-01-16

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Hint: Show that the family of all $C\in\mathcal{B}(X)\otimes \mathcal{B}(Y)$ such that $\mu(C)\leq \nu(C)$ is a monotone class.