# Every measurable set in $X \times Y$ is contained in a measurable rectangle

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

• I probably don’t understand the question, but isn’t $X \times Y$ a measurable rectangle containing any set? Nov 24, 2020 at 22:39
• You're right! but we are not certain that $X \times Y$ is a measurable rectangle Nov 24, 2020 at 22:49
• Then what is a measurable rectangle? Nov 24, 2020 at 22:50
• According to Halmos: "A measurable space is a set $X$ and $\sigma$-ring $S$ of subsets of $X$ with the property that $\bigcup S = X.$" Nov 24, 2020 at 22:53
• The question is about measurable spaces, which Halmos defines in terms of sigma rings, as quoted in these comments. Nov 24, 2020 at 23:39

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.