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In this errata of Real Analysis by Royden and PM Fitzpatrick, it corrects the definition of measurable rectangles to be

$A\times B$ if $A\in\mathcal{A}, B\in\mathcal{B}$, and $\mu(A)$ and $\mu(B)$ are finite

at the right beginning of chapter 20. This correction makes this measurable rectangle definition different from others which usually don't require the finiteness of $\mu(A)$ and $\mu(B)$. The problem is I don't see the reason why we need this finiteness in its following argument and proof of Fubini theorem.

Update on 20th, Feb, 2017

I guess the reason why the book defines measurable rectangle this way is to avoid $0\cdot\infty$ when defining outer measure by those rectangles.

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    Its been a while since I had a copy of Royden on hand, but normally things like this are because Royden will do a sequence of things to prove a result (like how he needs 3 or 4 sections to construct the lebesgue integral on the real line). The extra assumptions are usually necessary for his approach, but with a little bit of work, you can normally just do the proof in one shot like Folland or Rudin or whatever would. – 2017-02-04
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    @Batman I know there must be a reason for this correction. I guess I need to read the book more carefully to figure out where this additional requirement is needed. – 2017-02-04
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    I guess the reason why the book defines measurable rectangle this way is to avoid $0\cdot\infty$ when defining outer measure by those rectangles. – 2017-02-20

2 Answers 2

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It's not a big limitation. You will construct product measure on sigma finite spaces, so every measurable rectangle can be replaced by countable many with finite measure

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This correction is surprising because the concept of measurable rectangle is independent of any measure.

Halmos, for example, introduces this terminology the following way:

We shall frequently use the concept of measurable rectangle. Two equally obvious and natural definitions of this phrase suggest themselves. According to one, a rectangle in the Cartesian product of two measurable spaces $(X,\mathbb S)$ and $(Y,\mathbb T)$ is measurable if it belongs to $\mathbb S \times \mathbb T$, and, according to the other , $A\times B$ is measurable if $A\in\mathbb S$ and $B\in\mathbb T$. [...] for the time being we adopt the second of our proposed definitios.

(Halmos, Measure Theory, Van Nostrand, 7$^{th}$ edition, p. 140.)

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    I guess the book uses the additional finiteness requirement somewhere but I don't know where it is currently. – 2017-02-04