Measure, Integration & Real Analysis, 2021a

Section 2B Measurable Spaces and Functions 29
2.30 Example Borel sets
• Every closed subset of R is a Borel set because every closed subset of R is the
complement of an open subset of R.
• Every countable subset of R is a Borel set because if B = {x 1 , x 2 ,...}, then
B = ⋃ ∞
k=1
{x k }, which is a Borel set because each {x k } is a closed set.
• Every half-open interval [a, b) (where a, b ∈ R) is a Borel set because [a, b) =
⋂ ∞k=1
(a − 1 k , b).
• If f : R → R is a function, then the set of points at which f is continuous is the
intersection of a sequence of open sets (see Exercise 12 in this section) and thus
is a Borel set.
The intersection of every sequence of open subsets of R is a Borel set. However,
the set of all such intersections is not the set of Borel sets (because it is not closed
under countable unions). The set of all countable unions of countable intersections
of open subsets of R is also not the set of Borel sets (because it is not closed under
countable intersections). And so on ad infinitum—there is no finite procedure involving
countable unions, countable intersections, and complements for constructing the
collection of Borel sets.
We will see later that there exist subsets of R that are not Borel sets. However, any
subset of R that you can write down in a concrete fashion is a Borel set.
Inverse Images
The next definition is used frequently in the rest of this chapter.
2.31 Definition inverse image; f −1 (A)
If f : X → Y is a function and A ⊂ Y, then the set f −1 (A) is defined by
f −1 (A) ={x ∈ X : f (x) ∈ A}.
2.32 Example inverse images
Suppose f : [0, 4π] → R is defined by f (x) =sin x. Then
f −1( (0, ∞) ) =(0, π) ∪ (2π,3π),
f −1 ([0, 1]) = [0, π] ∪ [2π,3π] ∪{4π},
f −1 ({−1}) ={ 3π 2 , 7π 2 },
f −1( (2, 3) ) = ∅,
as you should verify.
Measure, Integration & Real Analysis, by Sheldon Axler