My first post in this series about Lebesgue integration presents the concepts of -algebras and measurability along with some basic results along these lines.

The parallel Rudin draws between -algebras and topologies is a pretty nice one. Whereas the sets in a topology are closed under arbitrary union and finite intersection, those in a -algebra are closed under complements and countable union (and by implication, countable intersection and subtraction).

Formally, an **algebra** on a set is defined to be a collection of subsets of such that i) , ii) , iii) .

A **-algebra** shares properties i) and ii), but . This condition of closure under not just finite, but countable union, gives -algebras relevance in the study of integration.

In fact, it’s not terribly uncommon to arrive upon a nontrivial -algebra given an arbitrary collection of subsets of :

Result 1: If is a collection of subsets of , there exists a unique smallest -algebra containing this collection. This -algebra is said to be *generated*** **by .

Proof: Just take the intersection of all -algebras that contain (at the very least, the collection of all subsets of is in this family).

Any space where we can define such a structure is called a **measurable space**; it is easy to see that is a measurable space when endowed with the -algebra consisting of the unions of intervals. The sets in such a collection are the **measurable sets, **and functions from a measurable space to a topological space are **measurable functions** if they pull the open sets of back to measurable sets of . Quickly convince yourself that many familiar Riemann-integrable functions are measurable.

The next theorem says we can essentially do any continuous binary operation on two real, measurable functions and still get a measurable function.

Result 2: If and are real, measurable functions on measurable space , and there exists a continuous map from to a topological space , then is measurable.

Proof: It is easy to see that a continuous function composed with a measurable function is measurable, so it suffices to prove that the map from to is measurable. Furthermore, every open set in is the countable union of some rectangles . The map pulls each such rectangle back to , which is measurable as desired.

This tells us that adding or multiplying two measurable functions together gives another measurable function, so complex functions where and are real, measurable functions are measurable as well. The real and imaginary parts and modulus of a complex measurable function are measurable.

Result 3: The characteristic function is measurable iff the set is measurable.

Proof: The characteristic function will pull any open interval in back to the empty set, , , or the whole set.

The last fundamental we present is the measurability of the supremum/infimum and upper/lower limit of a sequence of measurable functions. As a consequence of the latter, the limit of any convergent sequence of measurable functions is measurable.

Result 4: The supremum and upper limit of a sequence of real, measurable functions are measurable.

Proof: First note that a map that pulls all intervals of the form to measurable sets is measurable. For any , pick a sequence that converges to . Then , which certainly pulls back to a measurable set. Because , is measurable as desired. But the preimage of in is the union of the preimages of in for all , which is measurable as desired.