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I am studying out of Folland and I think I understand the proof of Theorem 2.36, but I am not sure why we can't just look at disjoint unions of rectangles.

Here is the Theorem:

Let $(X,\mathcal{M}, \mu)$, $(Y,\mathcal{N}, v)$ be $\sigma$-finite measure spaces. Then for any $E\in\mathcal{M}\bigotimes\mathcal{N}$, $x\to v(E_x)$ and $y\to \mu(E^y)$ are measurable (on $X$ and $Y$ respectively) and

$$\mu\times v (E)=\int \mu(E^y)\ dv=\int v(E_x)\ dx.$$

It is obvious for rectangles ($\chi_{E_x}=\chi_{A}v(B)$ if $E=A\times B$) and for finite disjoint unions of rectangles, so why can we take an countable disjoint collection of rectangles (Theorem 2.15)? Wouldn't this just be taking the limit of a finite disjoint union of rectangles?

If so, then the claim follows. So why use the Monotone Class Lemma? I know I have to be missing something.

Mark B
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  • What's your definition of sigma algebra on product space? – user160738 Dec 10 '16 at 05:27
  • Neither $\mathcal{M}$ nor $\mathcal{N}$ need be complete. Is that what you are asking? – Mark B Dec 10 '16 at 05:37
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    No, I just don't see how you would be able to draw conclusion without Monotone class theorem (according to my definition of product sigma algebra)/ – user160738 Dec 10 '16 at 05:41
  • Theorem 2.15 in Folland is a consequence of MCT. If ${f_n}$ is a finite sequence or infinite sequence in $\mathcal{L}^+$ (space of measurable functions from $X$ to $[0,\infty]$) and $f=\sum_n f_n$, then $$\int f=\sum_n\int f_n.$$ – Mark B Dec 10 '16 at 05:48
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    And how does that help you getting to conclusion? Measurable sets in product space are not unions of rectangles etc – user160738 Dec 10 '16 at 05:51
  • Then what are they? My understanding is that the algebra is generated by finite disjoint unions of rectangles (sets of the form $A\times B$ with $A\in \mathcal{M}$ and $B\in \mathcal{N}$. Then isn't the product algebra generated by these rectangles countable unions of elements in the algebra. Thanks, for helping me clear this up. – Mark B Dec 10 '16 at 06:01

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Not every element of $\mathcal{M}\otimes\mathcal{N}$ is a countable union of rectangles. So it does not suffice to just prove the theorem when $E$ is a countable union of rectangles.

For instance, if $\mathcal{M}$ and $\mathcal{N}$ are the Borel algebra on $\mathbb{R}$, note that every open subset of $\mathbb{R}^2$ can be written as a countable union of open rectangles, and so every open set is in $\mathcal{M}\otimes\mathcal{N}$. But then so is every closed set (since the complement of any element of $\mathcal{M}\otimes\mathcal{N}$ is in $\mathcal{M}\otimes\mathcal{N}$), including sets like $\{(x,x):x\in\mathbb{R}\}$ which cannot be written as a countable union of rectangles. (As an exercise, try and prove for yourself that this set cannot be written as a countable union of rectangles!)

Eric Wofsey
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  • Of course! If $D={(x,x):x\in\mathbb{R}$ is contained in a countable union of rectangles ${A_i\times B_j}$, Then at least one rectangle $A_k\times B_k$ has uncountably many elements of the form $(x,x)$. If $(x,x),(y,y)\in A_k\times B_k$, then $(x,y), (y,x)\in A_k\times B_k$. Thus every countable union of rectangles that contains $D$ is proper. Is this correct? – Mark B Dec 10 '16 at 06:44
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    That's correct. – Eric Wofsey Dec 10 '16 at 06:52
  • Would you give me another example which cannot be written as a countable union of rectangles and whose measure is not zero? – khashayar Mar 05 '22 at 22:45
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    @khashayar: Well, you could just take a union of the previous example with some set of positive measure. Less obviously, there are examples which do not differ from a countable union of rectangles by a null set. See https://math.stackexchange.com/questions/3845826/does-every-positive-lebesgue-measure-set-in-mathbbr2-contain-a-product-of. – Eric Wofsey Mar 05 '22 at 23:40