Originally Posted by

**Prometheus** when you study the Lebesgue measure, then usually only in the end you show that it is a sigma algebra. but in general all measure space must have the set of all its measurable sets, which is a sigma algebra.

about the intersection of sigma algebras, most (or at least from what I have seen so far) of the algebraic structures are closed under intersection, and for most of them, it is very easy to prove it (for example, sets, groups, rings, vector spaces, algebras and so on)

anyway, another example for a sigma algebra is the set of all finite subsets and all the sets that have finite complement.

also, to generalize your last example, if you have $\displaystyle B_i ;\;1\leq i\leq k$ where $\displaystyle i\neq j \rightarrow B_i \cap B_j =\emptyset$ and $\displaystyle \bigcup _1 ^k B_i = \mathbb{R}$then $\displaystyle \{\bigcup _{j\in J} B_j \mid J \subseteq [k]\}$ is a sigma algebra.