TITLE: Uniqueness of product measure (non $\sigma$-finite case)
QUESTION [22 upvotes]: Let $(X,\mathscr{A},\mu), (Y,\mathscr{B},\nu)$ be two measure spaces, then we have the product measurable space $(X\times Y, \mathscr{A}\times\mathscr{B})$ where $\mathscr{A}\times\mathscr{B}$ is the $\sigma$-algebra generated by $\{A\times B, A\in \mathscr{A}, B\in \mathscr{B}\}$. Applying Caratheodory extension we can always construct a measure $m$ on $\mathscr{A}\times\mathscr{B}$ so that $m(A\times B)=\mu(A)\nu(B)$. Further, if $\mu$ and $\nu$ are $\sigma$-finite, by monotone class theorem such a measure $m$ is uniquely determined. Now my question is:

When $\mu$ and $\nu$ are not $\sigma$-finite, is product measure still unique?

REPLY [5 votes]: I give more simple example. Let $\mu =\lambda \times \#$ be usual product of $\lambda$ and $\#$. Let $\mu_1$ be defined as follows: $$\mu_1(X)=\sum_{y \in [0,1]}\lambda_y(([0,1]\times \{y\})\cap X)$$for all Lebesgue measurable subset $X$ of $[0,1]\times [0,1]$, where $\lambda_y$ is Lebesgue measure defined on $[0,1]\times \{y\}$. Then measures $\mu$ and $\mu_1$ are agreeing on measurable rectangles but $\mu(\{(x,x):x \in [0,1]\})=+\infty$ and $\mu_1( (\{(x,x):x \in [0,1]\})=0$.