Two functions on [0,1] are defined as f(x) = 1 for all x, and g(x) = 1 for all x except at rational points where g(x) = 0. How are f and g treated in L²([0,1])?
AThey are distinct because they differ at infinitely many points
BThey are distinct because the rationals are dense, so differences are not negligible
CThey are identified as equal because the set of rationals has Lebesgue measure zero
DThey are identified as equal because both are bounded
In L^p spaces, functions that agree almost everywhere are identified — they are treated as the same element. The rational numbers are countable, and any countable set has Lebesgue measure zero. So f and g differ only on a null set, meaning f = g a.e., and they define the same element of L²([0,1]). Option A makes the classic mistake of confusing cardinality of the exceptional set with its measure.
Question 2 Multiple Choice
Which of the following is a null set under Lebesgue measure on ℝ?
AThe interval (0, 0.001)
BThe set of all irrational numbers in [0,1]
CThe Cantor set (an uncountable subset of [0,1])
DAny dense subset of [0,1]
The Cantor set is uncountable yet has Lebesgue measure zero — it is the canonical example that null sets can be cardinally large. Option A has measure 0.001. Option B (irrationals in [0,1]) has measure 1, since the rationals are null and [0,1] has measure 1. Option D is wrong: ℚ is dense yet null, but dense subsets can have any measure.
Question 3 True / False
The rational numbers ℚ form a null set in ℝ even though they are dense in ℝ — there is a rational number arbitrarily close to every real number.
TTrue
FFalse
Answer: True
Density and measure zero are independent properties. ℚ is dense (every open interval contains a rational) yet has Lebesgue measure zero because it is countable: μ(ℚ) ≤ Σ μ({qₙ}) = Σ 0 = 0. This is one of the most important intuition-breakers in measure theory — topological 'size' (density) and measure-theoretic 'size' come apart dramatically.
Question 4 True / False
If a property fails to hold on a set containing more than finitely many points, it can seldom hold almost everywhere.
TTrue
FFalse
Answer: False
Almost everywhere means the set of exceptions has measure zero — not that it is finite or even countable. A property can fail on a countably infinite set (like all rationals) or even on an uncountable null set (like the Cantor set) and still hold almost everywhere. The key is measure zero, not the cardinality of the exceptional set.
Question 5 Short Answer
Why can two L² functions be considered identical (as elements of the function space) even if they differ at infinitely many points?
Think about your answer, then reveal below.
Model answer: L^p spaces identify functions that are equal almost everywhere — that is, functions whose difference is supported on a set of measure zero. Since a set of measure zero contributes nothing to any integral, two such functions produce identical values for all integrals, norms, and inner products. The identification is not just a convenience; it is the correct notion of equality for integration theory, where individual point values are irrelevant.
This identification is what makes L^p spaces well-defined vector spaces with a proper norm: ‖f - g‖ = 0 requires f = g a.e., not pointwise equality everywhere. Without this identification, the 'norm' would fail to be a true norm (it would be zero for distinct functions differing on a null set). The almost-everywhere equivalence relation is the glue that connects the analytic and algebraic structure of L^p spaces.