The indicator function of the rationals, f(x) = 1 if x ∈ ℚ, f(x) = 0 otherwise, is not Riemann integrable on [0,1]. What is its Lebesgue integral on [0,1]?
A1 — the rationals are dense in [0,1], so the function is effectively always 1
B1/2 — the rationals and irrationals split [0,1] equally by density
C0 — the rationals have Lebesgue measure zero, so f equals 0 almost everywhere
DUndefined — this function is not Lebesgue measurable
The rationals in [0,1] are countable, and any countable set has Lebesgue measure zero (it can be covered by open intervals of total length ε for any ε > 0). Therefore f = 0 almost everywhere on [0,1], and its Lebesgue integral is 0. The Riemann approach fails because, in any sub-interval, f takes both values 0 and 1, so upper and lower Riemann sums cannot be made to agree. The Lebesgue approach succeeds because it measures where the function takes each value, and the set where f = 1 has measure zero.
Question 2 Multiple Choice
A student argues: 'The rationals cannot have measure zero — they are infinite in number and dense in every interval of [0,1], so they must fill positive measure.' What is the flaw?
AThere is no flaw — the student is correct that the rationals have positive measure
BDensity and measure are different properties; the rationals are countable and can be covered by open intervals of arbitrarily small total length
CThe rationals have measure 1 because they are dense, not some measure between 0 and 1
DMeasure is only defined for uncountable sets; countable sets have undefined measure
Density (being everywhere close) and measure (size in the sense of length) are fundamentally different properties. Enumerate the rationals as q₁, q₂, …. Cover qₙ with an open interval of length ε/2ⁿ. The total length is Σε/2ⁿ = ε, arbitrarily small. So the rationals can be covered by intervals of total length as small as desired — outer measure zero. Denseness is a topological property; measure zero is a size property. A set can be topologically ubiquitous (dense) and measure-theoretically negligible.
Question 3 True / False
Every open subset of ℝ is Lebesgue measurable.
TTrue
FFalse
Answer: True
The Lebesgue measurable sets form a σ-algebra that contains all open sets. Open sets on ℝ are countable unions of open intervals, and open intervals have well-defined measure (their length). Countable unions of measurable sets are measurable by the σ-algebra property, so all open sets are measurable. The non-measurable sets that exist (assuming the axiom of choice) are exotic constructions that cannot be built from open and closed sets by standard set operations.
Question 4 True / False
If two functions f and g on [0,1] differ primarily on the set of rational numbers, their Lebesgue integrals over [0,1] may differ.
TTrue
FFalse
Answer: False
The rationals have Lebesgue measure zero, so f and g agree almost everywhere. In Lebesgue theory, two functions that agree almost everywhere have identical integrals — the integral is blind to what happens on measure-zero sets. This is one of the key advantages over Riemann integration and is what makes 'almost everywhere' the natural equivalence relation for integration theory. It allows the theory to treat equivalence classes of functions rather than insisting on pointwise values everywhere.
Question 5 Short Answer
What does it mean for a property to hold 'almost everywhere' (a.e.), and why is this concept essential to Lebesgue integration theory rather than just a convenient shorthand?
Think about your answer, then reveal below.
Model answer: 'Almost everywhere' means the property holds except possibly on a set of Lebesgue measure zero. It is essential — not merely convenient — because Lebesgue integration inherently cannot distinguish between functions that agree almost everywhere: their integrals are equal. This means the natural objects of Lebesgue theory are equivalence classes of functions under 'equal a.e.,' not individual pointwise-defined functions. Powerful convergence theorems (Dominated Convergence, Monotone Convergence) are stated in a.e. terms because pointwise convergence everywhere is too strong to achieve in most analytic settings, while a.e. convergence is sufficient for all integration purposes.
The practical consequence: modifying a function on a measure-zero set changes nothing about its integral or its L² norm. This is why function spaces like L²[0,1] are spaces of equivalence classes — two functions differing only on a null set are identified as the same element. The 'a.e.' qualification also appears throughout probability theory, where almost sure convergence is the natural mode. Far from being a shorthand, 'almost everywhere' is where Lebesgue measure theory makes its most fundamental departure from the Riemann framework.