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

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

1 Answer 1


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.


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