Showing set not in $M\times M$ for Lebesgue measure Let $M$ be the Lebesgue-measurable subsets of $\mathbb{R}$. Suppose $E\subseteq [0,1]$ and $E\not\in M$. Then I want to show that $E\times\{0\}\not\in M\times M$, where $M\times M$ is the smallest $\sigma$-field generated by the sets $A\times B$ with $A,B\in M$.
Well, the $\sigma$-field operations are countable unions, set difference, and complementation. Certainly $E\times\{0\}$ is not of the form $A\times B$ with $A,B\in M$, because $E\not\in M$. But how would I prove $E\times\{0\}\not\in M\times M$? I don't quite see how I could take into account the three operations mentioned above.
 A: Here is one way:
Let $f: \mathbb{R} \to \mathbb{R} \times \mathbb{R}  $ be defined by
$ f(x) = (x,0)$. Then $f$ is  measurable in the context of the $\sigma$-fields $M$, $M \times M$ (see below). If $E'=E\times \{0\}$ was measurable, then $f^{-1} E'$ would be measurable. However, $f^{-1} E' = E$, which is not measurable. Hence $E\times \{0\}$ is not measurable.
To see that $f$ is measurable as a map from the $\sigma$-fields $M$ to $M \times M$:
We have $f^{-1} (A \times B) = \begin{cases} A, &0 \in B \\
\emptyset, & \text{otherwise}  \end{cases}$, hence $f^{-1} (A \times B) \in M$ for all $A,B \in M$. Since the set $\Sigma = \{ C | f^{-1} C \in M \}$ is a $\sigma$-field, we have $M \times M \subset \Sigma$, hence $M$ is measurable.
Note: This shows that the $\sigma$-field $M \times M$ is not complete. If it were, then all subsets of measure zero would be measurable. Since $E \times \{ 0 \} \subset \mathbb{R} \times \{0\}$, and the latter has measure zero, completeness would imply that $E \times \{ 0 \} $ is measurable.
In particular, as Nate observed below, $f$ is not measurable as a map from $M$ to $\overline{M \times M}$.
A: You can show inductively that for every set $C$ in the product $\sigma$-field $M \times M$, the "horizontal slice" $E = \{x \in \mathbb{R} : (x,0) \in C\}$ is in $M$. (The fact that $M$ is Lebesgue measure won't matter to the proof.)
This is clearly true for the basic sets $C = A \times B$ where $A$ and $B$ are in $M$.  Then you can show it is preserved under the operations that generate the $\sigma$-field from these basic sets.
Having shown this, one can then conclude that if the set $C = E \times \{0\}$ is in the product $\sigma$-field $M \times M$, its "horizontal slice" $E$ must have been in $M$ to begin with.
