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?
 A: 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.
For (b), check the definition of probability measure; the reasoning is similar; I hope this helps for now.
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.)
