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I am reading a chapter on probability in my textbook where they give a counterexample to show that Cartesian products of sigma algebras are not necessarily sigma algebras:

It is not true that $\mathcal F_1 \times \mathcal F_2 = \sigma( \mathcal F_1 \times \mathcal F_2)$. Take, for example, $\mathcal F_1 = \mathcal F_2 = 2^{\{1,2\}}$. Then, $|\mathcal F_1 \times \mathcal F_2| = 1+3\times 3 =10$ (because $\emptyset \times X = \emptyset$), while, since $\mathcal F_1 \times \mathcal F_2$ includes the singletons of $2^{\{1,2\} \times \{1,2\}}$, $\sigma( \mathcal F_1 \times \mathcal F_2) = 2^{\{1,2\} \times \{1,2\}}$. Hence, six sets are missing from $\mathcal F_1 \times \mathcal F_2$. For example, $\{(1,1), (2,2)\} \in \sigma( \mathcal F_1 \times \mathcal F_2) \setminus \mathcal F_1 \times \mathcal F_2$.

Can someone please help me parse the notation and why this is not a sigma algebra?

From what I understand, the set $\mathcal{F}_1 \times \mathcal{F}_2 = \{\varnothing, (\{1\}, \{1\}), (\{1\}, \{2\}), (\{1\}, \{1,2\}), (\{2\}, \{1\}), (\{2\}, \{2\}), (\{2\}, \{1,2\}), (\{1, 2\}, \{1\}), (\{1, 2\}, \{2\}), (\{1, 2\}, \{1,2\})\}$.

The ‘missing’ elements in the above set would be those with $\varnothing$ but in the text they say these are set to $\varnothing$ so I don’t understand what elements are missing in order to make this a sigma algebra.

I can ‘see’ why, in general, cartesian products don’t lead to sigma algebras. I’m just having trouble understanding this example.

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1 Answer 1

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The Cartesian Product of sigma-algebras isn't literally the cartesian product (the set of all pairs). Instead, $$\mathcal F_1 \times \mathcal F_2 := \{A\times B \mid A\in \mathcal F_1, B\in \mathcal F_2\},$$ so in the example, for the space $X=\{1,2\}$ and sigma-algebras $\mathcal F_1 = \mathcal F_2 = 2^{\{1,2\}}$, we get $$\mathcal F_1 \times \mathcal F_2 = \{\emptyset, \{1\}, \{2\}, \{1,2\}\}\times \{\emptyset, \{1\}, \{2\}, \{1,2\}\}$$ $$=\left\{ \begin{array}{c} \emptyset \times \emptyset, \emptyset \times \{1\}, \emptyset \times \{2\}, \emptyset \times \{1,2\},\\ \{1\} \times \emptyset, \{1\}\times \{1\}, \{1\}\times \{2\}, \{1\}\times \{1,2\},\\ \{2\} \times \emptyset, \{2\}\times \{1\}, \{2\}\times \{2\}, \{2\}\times \{1,2\},\\ \{1,2\} \times \emptyset, \{1,2\}\times \{1\}, \{1,2\}\times \{2\}, \{1,2\}\times \{1,2\}\\ \end{array} \right\}$$ $$=\Big\{\emptyset, \{(1,1)\}, \{(1,2)\}, \{(1,1),(1,2)\}, \{(2,1)\}, \{(2,2)\}, \{(2,1),(2,2)\}, \{(1,1),(2,1)\}, \{(1,2),(2,2)\}, \{(1,1),(1,2),(2,1),(2,2)\}\Big\}.$$ Then, we can see that this set is missing the subset $\{(1,1),(2,2)\}$, which should be in there for it to be a sigma-algebra, since sigma-algebras are closed under countable unions, and this contains $$\{(1,1)\}, \{(2,2)\} \in\mathcal F_1 \times \mathcal F_2.$$

Pictorially, $\mathcal F_1 \times \mathcal F_2$ is a collection of subsets of $\{1,2\}\times \{1,2\}$, a $2\times 2$ grid, and this missing element is a diagonal on the grid.

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