$Q \in \mathcal{X} \otimes \mathcal{Y}$ implies $\overline{Q } \in \mathcal{Y} \otimes \mathcal{X}$ I am trying to show a seemingly easy statement:
Let $( X, \mathcal{X})$ and $( Y, \mathcal{Y})$ be measurable spaces and let $( X \times Y, \mathcal{X} \otimes \mathcal{Y} )$ be the product measureable space. We know that $\mathcal{X} \otimes \mathcal{Y}$ is the $\sigma$-algebra generated by the sets $A \times B$, where $A \in \mathcal{X}$, $B \in \mathcal{Y}$. For a set $Q \in \mathcal{X} \otimes \mathcal{Y}$, let us denote by
$$
\overline{Q} = \{ ( y, x ) \in Y \times X : ( x, y ) \in Q \}.
$$
Now, how can one formally show that $\overline{Q} \in \mathcal{Y} \otimes \mathcal{X}$?

First check that the collection of sets
$$  \mathcal{A} = \{ \overline{Q} \mid Q \in \mathcal{X} \otimes \mathcal{Y} \}$$
satisfies all the properties of a $\sigma$-algebra (when proving this, one relies on the fact that $\mathcal{X} \otimes \mathcal{Y}$ is a $\sigma$-algebra). 
Secondly observe that $\mathcal{A}$ contains all the elementary sets of the form $B \times A$, for $B \in \mathcal{Y}$ and $A \in \mathcal{X}$. Indeed $B \times A = \overline{A \times B}$, hence $B \times A \in \mathcal{A}$.  
We conclude that $\mathcal{A}$ is a $\sigma$-algebra containing all the rectangles, hence $\mathcal{Y} \otimes \mathcal{X} \subset \mathcal{A}$, since by definition $\mathcal{Y} \otimes \mathcal{X}$ is the smallest such $\sigma$-algebra.  
Now we show that $\mathcal{A} \subset \mathcal{Y} \otimes \mathcal{X}$. Define simmilarly
$$ \mathcal{B} = \{ \overline{Q} \mid Q \in \mathcal{Y} \otimes \mathcal{X} \} $$
and 
$$ \mathcal{A'} = \{ \overline{Q} \mid Q \in \mathcal{B} \}.  $$
The same argument as above implies $\mathcal{X} \otimes \mathcal{Y} \subset \mathcal{B}$. Therefore it is clear that $\mathcal{A} \subset \mathcal{A'}$. And since applying $\overline{\: \cdot \:}$ twice does nothing, $\mathcal{A}' = \mathcal{Y} \otimes \mathcal{X}$. Combining all the inclusions we obtain the following series of inclusions
$$ \mathcal{Y} \otimes \mathcal{X} \subset \mathcal{A} \subset \mathcal{A}' = \mathcal{Y} \otimes \mathcal{X},$$
hence 
$$\mathcal{A} = \mathcal{Y} \otimes \mathcal{X}. $$
