What does the Lebesgue integral of the Dirichlet function (1 on rationals, 0 on irrationals) over [0,1] equal?
A1, because there are infinitely many rationals in [0,1]
B1/2, because rationals and irrationals are each dense in [0,1]
C0, because the rationals have Lebesgue measure zero
DUndefined — the Dirichlet function is not Lebesgue measurable
The Dirichlet function equals 0 almost everywhere (the set of rationals has measure zero; irrationals fill the interval in measure). The Lebesgue integral handles this cleanly: the contribution from the value '1' is 1 × μ({rationals}) = 1 × 0 = 0. The Riemann integral fails here because every subinterval contains both rationals and irrationals, so upper and lower Riemann sums never agree. This is a paradigm case of why measure-theoretic integration is strictly more powerful.
Question 2 True / False
The Dominated Convergence Theorem says you can generally interchange limits and Lebesgue integrals: lim ∫ fₙ = ∫ lim fₙ.
TTrue
FFalse
Answer: False
The DCT requires a crucial hypothesis: there must exist an integrable dominating function g such that |fₙ(x)| ≤ g(x) for all n and almost all x. Without this 'ceiling,' interchange of limits and integrals can fail — even pointwise converging sequences can have integrals that don't converge to the integral of the limit (e.g., functions that shift mass toward infinity). The dominating function prevents the fₙ from escaping to infinity, which is what justifies the interchange.
Question 3 True / False
The fundamental difference between Riemann and Lebesgue integration is that Lebesgue integration uses finer partitions of the domain.
TTrue
FFalse
Answer: False
This is the key misconception. The Riemann integral partitions the *domain* into intervals, regardless of how fine. The Lebesgue integral partitions the *range* — it asks 'for which x-values does f(x) lie in [a,b]?' and measures the preimage. This reversal is not a refinement of Riemann's approach; it is a conceptually different strategy. It is what allows the integral to handle irregular functions like the Dirichlet function and to interact cleanly with measure-theoretic structure.
Question 4 Multiple Choice
A sequence of non-negative measurable functions fₙ increases pointwise to f. According to the Monotone Convergence Theorem, what can you conclude?
B∫ fₙ dμ → ∫ f dμ, with no additional conditions needed beyond monotone pointwise convergence
C∫ fₙ dμ → ∫ f dμ, but only if each fₙ is bounded
D∫ fₙ dμ → ∫ f dμ, provided f is Riemann integrable
The Monotone Convergence Theorem requires only that the fₙ are non-negative measurable functions increasing pointwise to f — no uniformity, no boundedness, no Riemann integrability. This is far weaker than what Riemann theory requires (uniform convergence). Uniform convergence is sufficient but far too strong; the MCT shows you need far less. This is the theorem's power: it licenses interchange of limit and integral under the minimal conditions that make the conclusion well-posed.
Question 5 Short Answer
Why does partitioning the range rather than the domain allow the Lebesgue integral to handle functions that the Riemann integral cannot?
Think about your answer, then reveal below.
Model answer: By partitioning the range, you group x-values by their f-output rather than by their position on the x-axis. The measure of each preimage set replaces interval width. This means irregular functions — ones that spike wildly on complicated sets — are handled as long as their preimage sets are measurable. The Dirichlet function's preimage of {1} is the rationals (measure zero), so it contributes nothing to the integral. The Riemann approach fails because every small x-interval contains both 0s and 1s, so no Riemann sum can resolve the function's behavior.
The range-partition strategy decouples the integral from the geometric regularity of the function's graph. Riemann integration implicitly assumes the function is 'nice enough' that its graph can be approximated by rectangles. Lebesgue integration doesn't need this: it only needs the preimage sets to be measurable, a much weaker condition. This is why the Lebesgue integral extends to a vastly larger class of functions and becomes the foundation for modern probability, functional analysis, and Fourier theory.