Sigma algebra generator set of product space [duplicate]

Solution 1:

Someone just told me an counterexample: Let $\Omega=\Omega_1=\Omega_2=\left\{ 1, 2, 3 \right\}$, $D_1=\left\{ \left\{ 1 \right\} \right\}$, $D_2=\left\{ \left\{ 3 \right\} \right\}$, then \begin{align*} \mathcal{F}_1&=\left\{ \Omega, \emptyset, \left\{ 1 \right\}, \left\{ 2, 3 \right\} \right\}, \\ \mathcal{F}_2&=\left\{ \Omega, \emptyset, \left\{ 3 \right\}, \left\{ 1, 2 \right\} \right\}. \end{align*} On the other hand, we have $D=\left\{ (1, 3) \right\}$, $$\sigma\left( D \right)=\left\{ \Omega_1\times \Omega_2, \emptyset, \left\{ (1,3) \right\}, \left\{ (1, 3) \right\}^C \right\}.$$ Notice that $\mathcal{F}_1\otimes \mathcal{F}_2$ contains $\left\{ (2,3), (3 ,3) \right\}$ which is not in $\sigma(D)$.