# Sigma algebra Generated by Product of Lebesgue measures

Here's the question: Let $$\Omega_0$$ = (0, 1], $$\lambda$$ = Lebesgue measure on the unit interval and F = all Lebesgue measurable sets of (0, 1] (i.e. outer measurable).

Let $$\Omega$$ = (0, 1] × (0, 1] and let S be a family of subsets of $$\Omega$$ × $$\Omega$$ given by S = {B = A × (0, 1] : A ∈ F}

For any B ∈ S put P(A × (0, 1]) = $$\lambda(A)$$.

(a) Show that S is a $$\sigma$$-algebra.

(b) Show that P is a probability measure on S.

So I think that each S is going to take the form {(a,b],[0,1]} for $$0\leq a \leq b\leq1$$, but I'm not sure how to show this is a $$\sigma$$-algebra (has to be closed under countable unions, intersections, and complementation). Is this the right way to think about this?

Firstly, I think you're confusing $$S$$ with elements of $$S$$: There is only one $$S$$, but $$S$$ has infinitely many elements; these are of the form $$\{ \langle a,b \rangle \, | \ a \in A, 0 < b \leq 1 \}$$ where $$A \in F$$.
To illustrate, for (a) I'd say just use the definition: If $$B_1,B_2 \in S$$, then $$B_1 = A_1 \times (0,1]$$ and $$B_2 = A_2 \times (0,1]$$ for some $$A_1,A_2 \in F$$. Thus, for example to check the intersection property, $$B_1 \cap B_2 = (A_1 \times (0,1]) \cap (A_2 \times (0,1]) = (A_1 \cap A_2) \times (0,1].$$ Clearly, $$A_1 \cap A_2$$ is measurable, and thus $$(A_1 \cap A_2) \times (0,1] \in S$$. The others are similar.
EDIT: For clarification on the probability measure question: suppose the sets $$B_0, B_1, \ldots \in S$$ (where $$B_i = A_i \times (0,1]$$) are pairwise disjoint. Then, by definition of $$\mathbb{P}$$, $$\mathbb{P} \left( \bigcup_{n \in \omega} B_n \right) = \lambda \left( \bigcup_{n \in \omega} A_n \right).$$ By assumption, $$\lambda$$ is an outer measure, and thus countable additivity holds by assumption; i.e. $$\lambda \left( \bigcup_{n \in \omega} A_n \right) = \sum_{n \in \omega} \lambda (A_n)$$ which is exactly $$\sum_{n \in \omega} \mathbb{P} (B_n)$$ (again by definition of $$\mathbb{P}$$), as required. (This uses implicitly that $$(B_i \times (0,1]) \cup (B_j \times (0,1]) = (B_i \cup B_j) \times (0,1]$$ for all $$i,j \in \omega$$, which can be verified easily.)
• Thanks so much for the first answer! So my next question would be how to show exactly it's a probability measure. I know that four properties have to hold, but I'm a little confused as how to show that the Lebesgue measure of A is a probability measure of the product of A X (0,1]. Specifically, I'm not sure how to show that the subtraction and union properties of a probability measure hold (P = 0 and P($A^C$) are relatively straightforward). – psun Nov 8 '18 at 20:07