If $X$ and $Y$ are any two sets (not necessarily subsets of the
same space), the Cartesian product $X \times Y$ is the set of all ordered
pairs $(x,y)$, where $x \in X$ and $y \in Y$. Thus, for instance, if $A \subset X$ and $B \subset Y$, we shall call the set $E = A \times B$ (a subset of $X \times Y$) a rectangle
and we shall refer to the component sets $A$ and $B$ as its sides.
In our context the word "class" should be understood as a set of sets. So the class of measurable rectangles refers to the set of measurable rectangles. So I want to use the fact that a rectangle $A \times B$ in the Cartesian product of two measurable spaces $(X, S)$ and $(Y, T)$, (where $S$ and $T$ are $\sigma$-rings of subsets of $X$ and $Y$ respectively), is measurable if $ A \in S $ and $ B \in T $.
I am reading the Measure Theory book by Paul Halmos, I have found this.
Exercise (6) Section 33.
If $(X,S)$ and $(Y,T)$ are measurable spaces, then every measurable set in $X \times Y$ is contained in a measurable rectangle.
My idea is the following:
the class of all those sets which way be covered by a measurable rectangle is a $\sigma$-Ring.
I think this class contains every measurable set of $ X\times Y$, on the other hand if I test that it is a $ \sigma $-Ring I have finished
Best Answer
Halmos develops the Measure theory based on $\sigma$-rings. Measurable set of $X\times Y$ means sets in product $\sigma$-ring, that is generated by $A \times B$, $A \in S$, $B \in T$ where $S$ and $T$ are $\sigma$-rings. In general, $X\times Y$ may not be a measurable rectangle.
Your idea on how to solve exercise 6 (section 33) is correct. It is exactly the way to go.
Here is a detailed solution to exercise 6 in section 33:
Let $$C=\{ E \subseteq X\times Y : \textrm{ there are } A\in S, B \in T \textrm{ such that } E \subseteq A\times B\}$$
Just checking the definition of $\sigma$-ring, it is immediate that $C$ is $\sigma$-ring.
It is also immediate that for all $A\in S, B \in T$, $A\times B \in C$.
So $C$ is a $\sigma$-ring and all measurable rectangles are in $C$.
Since $S \times T$ is the smallest $\sigma$-ring having all measurable rectangles, we have $S \times T\subseteq C$. It means that every measurable sets of $X \times Y$ is contained in a measurable rectangle.