I read that given $\Omega = \Omega_1 \times \Omega_2$, where $\Omega_1$ and $\Omega_2$ are sets, the smallest $\sigma$-field generated by rectangles $A = A_1 \times A_2$, where $A_1$ is an interval over $\Omega_1$ and $A_2$ is an interval over $\Omega_2$ can also be generated by 'cylinder sets', i.e.
$$C_1 = \{ A_1 \times \Omega_2\}$$
$$C_2 = \{ \Omega_1 \times A_2 \}$$
such that the $\sigma$-field over $\Omega$ generated by $C_1 \cup C_2$ is the same $\sigma$-field generated by $A$. My question is why couldn't we have the following set as a basis:
$$C^* = \{ \Omega_1 \times \Omega_2 \}$$
How does the $\sigma$-field generated by the one above differ from that of $A$ ?... Is the $\sigma$-field generated by $C^*$ still Lebesgue-measurable?