The Borel sets for a given topology are defined to be the smallest class of sets that contains all open sets and is closed under countable unions and intersections. Therefore, to prove that every Borel set has a certain property, it is sufficient to prove it for open sets (or even for simpler sets such as intervals), and then to prove that countable unions or countable intersections of sets with the property have the property. It is also possible to express such arguments in terms of transfinite induction.
Basic real analysis, measure theory.
The following problem arises in connection with measure differential inclusions, which are problems of the form
where is a closed convex set-valued function with possibly unbounded values.
Given a vector measure with values in and a non-negative measure (positive for open sets), the problem is to show that if
for any open set where is a closed convex set, then
which is known as the asymptotic or recession cone of .
Since we are assuming the result for open sets, it is sufficient to prove that the property is closed under countable unions and intersections. We can make this task slightly easier by noting that finite unions and intersections of sets with the desired property also have the desired property; this in turn means that we only need to deal with nested countable unions and intersections.
Considering nested unions, suppose with . Then either for all , or there is an where . In the former case, for all , and so taking limits, as is a closed convex cone. In the latter case, we have for all , and so taking limits the result again holds.
Considering nested intersections, suppose with . Then either or . In the former case, for all , and taking limits of gives the desired result. If then either for some or for all . In then for all and taking limits of again gives the desired result. Finally, considering for all we see that
Taking limits and using the definition of gives the desired: .
One can think of the Borel sets as being built up from the open sets by means of successive operations of countable unions and intersections. However, there are transfinitely many stages to this process, so transfinite induction is needed.
More precisely, we build up the Borel sets for a metric space as follows: for each ordinal we have a class of sets , where
is the set of open sets,
is the set of countable unions or countable intersections of sets in ,
is is a limit ordinal.
Note that contains all the closed sets (for a metric space) and so contains all the complements of sets in . Thus, the complement of any set in can be found in for finite ; if is infinite, then complements of sets in are in .
To prove the above result, we could use a transfinite induction on the levels of the Borel hierarchy. We start with the fact that (Target) holds for for all .
To deal with successor ordinals, one has to prove that if (Target) holds for sets then it holds for and . This we showed above.