Question: Show that the collection of measurable rectangles form an algebra.

Denote $\Sigma$ by $\sigma$-algebra.

I found the following set operations. Let $A_1, A_2 \in \Sigma_A$, and $B_1, B_2 \in \Sigma_B$. Then, $(A_1 \times B_1) \setminus(A_2 \times B_2) = [(A_1 \cap A_2) \times (B_1 \setminus B_2)] \cup [(A_1 \setminus A_2) \times B_2)]$.

In addition, $(A_1 \times B_1) \cap (A_2 \times B_2) = (A_1 \cap A_2) \times (B_1 \cap B_2)$.

Then, by De Morgan, I can show the finite union as well.

But, two set operations is intuitively understandable, but is there a simple way to prove these operations formally?

Also, how can we show that the product of empty sets is in the collection? Can we still use the caratheodory criterion in this case?

  • $\begingroup$ In the special case when $A_i,B_i$ are intervals in $\mathbb R$ you can draw a picture and see what the difference between two ordinary rectangles is. That should be enough to understand what those identities really mean. $\endgroup$ – Kavi Rama Murthy Nov 1 '18 at 6:07
  • $\begingroup$ You probably meant to say that the collection of unions of measurable rectangles form a $\sigma$-algebra. Because measurable rectangles certainly don't (just consider two disjoint rectangles). $\endgroup$ – Stefan Mesken Nov 1 '18 at 7:39
  • $\begingroup$ @StefanMesken Sorry. You are right. It is the collection of all finite union of measurable rectangles. But how does this imply that $\emptyset$ and $X$ belong to an algebra? $\endgroup$ – Sihyun Kim Nov 1 '18 at 8:18
  • $\begingroup$ @Daeseon The empty set is the union of the empty collection of rectangles (which is a finite union) and $X$, well... What is $X$ in your case and what is your definition of a measurable rectangle precisely? Depending on the precise definition, there might be a small mistake in this exercise and you just have to through in $X$ to get an algebra. $\endgroup$ – Stefan Mesken Nov 1 '18 at 11:04

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