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In the book "MEASURE THEORY AND FINE PROPERTIES OF FUNCTIONS", the authors claim: enter image description here where enter image description here

Can someone explain to me how can we use the observation to construct the disjoint union. I used $F_1=A_1\times B_2$, $F_2=A_2 \times B_2 -A_1\times B_1 $, $F_3=A_3 \times B_3 -(A_1\times B_1\cup A_2 \times B_2) $ and so one.

I conclude that $(F_i)$ is a disjoint sequence of element of $(A_i\times B_i)$ which equals its union. But I don't know how to use there observation.

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  • $\begingroup$ I guess, $F_1=A_1\times B_1$. How do you write $F_2$ as a disjoint union of sets from $P_0$? $\endgroup$
    – Berci
    Jan 28, 2020 at 0:03
  • $\begingroup$ I dont understand your remark. I think i did F2 $\endgroup$
    – Mary Maths
    Jan 28, 2020 at 18:26
  • $\begingroup$ You wrote $F_2=A_2\times B_2\,-\,A_1\times B_1$, but this is a set difference and not disjoint union of sets of the form $A\times B$. That's where you should use the framed identity. $\endgroup$
    – Berci
    Jan 28, 2020 at 19:16
  • $\begingroup$ But I considerd $F_3=A_3 \times B_3 -(A_1\times B_1\cup A_2 \times B_2)$ and $F_4$ and so one the all $(F_i)$ have a disjoint union. Still don't understand your remark $\endgroup$
    – Mary Maths
    Jan 28, 2020 at 22:10
  • $\begingroup$ The problem is that $F_2\notin \mathcal P_0$, so this splitting is, though a disjoint union, but not of basic rectangles. But, $F_2, F_3,\dots$ can be split further, using the framed observation, this time to elements of $\mathcal P_0$. $\endgroup$
    – Berci
    Jan 28, 2020 at 23:13

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Your splitting to disjoint union in general does not consist of pure rectangles (elements of $\mathcal P_0$).

But, (repeatedly) using the given splitting of a difference of rectangles to smaller rectangles, you can further split $F_2,F_3,\dots$ so that you end up with a countable disjoint union of rectangles.

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