A student claims that the closed interval [2, 5] is not a Borel set because the Borel sigma-algebra is generated by open sets, and [2, 5] is not open. This reasoning is wrong because:
AThe Borel sigma-algebra contains all closed sets as complements of open sets, and sigma-algebras are closed under complement
BThe interval [2, 5] can actually be rewritten as an open set on ℝ
CClosed sets and open sets are the same thing on the real line
DThe Borel sigma-algebra is simply the power set of ℝ, so it contains everything
A sigma-algebra is closed under complement. The interval (−∞, 2) ∪ (5, +∞) is open (a union of open intervals), so its complement [2, 5] is Borel. The key insight is that the Borel sigma-algebra is not just the open sets themselves but the entire sigma-algebra *generated* by them — which immediately includes all closed sets, and much more.
Question 2 Multiple Choice
Which of the following collections, when used as generators, produces a *different* sigma-algebra from ℬ(ℝ)?
AThe closed intervals {[a, b] : a ≤ b, a, b ∈ ℝ}
BThe left-closed half-open intervals {[a, b) : a < b, a, b ∈ ℝ}
CThe rays {(−∞, x] : x ∈ ℝ}
DThe singletons {{x} : x ∈ ℝ}
Closed intervals, half-open intervals, and rays all generate ℬ(ℝ) — this robustness is a key sign that the Borel sigma-algebra captures something intrinsic to the real line. Singletons generate a strictly smaller sigma-algebra (the countable/co-countable sigma-algebra), which contains only sets that are countable or have countable complements. Since many open intervals are uncountable and their complements are uncountable, this sigma-algebra does not contain all open sets.
Question 3 True / False
The set of all rational numbers ℚ ⊆ ℝ is a Borel set.
TTrue
FFalse
Answer: True
ℚ is countable, so it can be written as a countable union of singletons: ℚ = ⋃_{q ∈ ℚ} {q}. Each singleton {q} is a closed set (hence Borel), and a sigma-algebra is closed under countable unions, so ℚ ∈ ℬ(ℝ). This illustrates how the Borel sigma-algebra includes all sets one can explicitly describe.
Question 4 True / False
Most subset of ℝ is a Borel set.
TTrue
FFalse
Answer: False
Non-Borel subsets of ℝ exist — the standard example is the Vitali set. However, constructing any non-Borel set requires the Axiom of Choice; no non-Borel set can be explicitly defined by a concrete rule or formula. This is why for practical probability theory, every set you would naturally want to assign probability to is Borel.
Question 5 Short Answer
Why is the Borel sigma-algebra defined as the *smallest* sigma-algebra containing all open sets, rather than by explicitly listing all open sets, closed sets, G_δ sets, F_σ sets, and so on?
Think about your answer, then reveal below.
Model answer: The explicit hierarchy of Borel sets (open, closed, G_δ, F_σ, G_δσ, ...) is uncountably deep — the transfinite induction that builds it has ω₁ stages. There is no finite or even countably-enumerable description of all Borel sets. Defining ℬ(ℝ) as the smallest sigma-algebra containing the open sets is both simpler and exact: it characterizes the collection without needing to enumerate it. The 'generated by' construction (intersecting all sigma-algebras containing the generators) always produces the smallest such object and is well-defined because the power set of ℝ is always available as a containing sigma-algebra.
The 'generated sigma-algebra' construction is fundamental to measure theory precisely because it gives a clean, well-defined handle on collections that are too complex to describe by explicit enumeration. The same technique is used to define the sigma-algebra generated by a random variable, making this definition paradigmatic.