Questions: Comparison of Riemann and Lebesgue Integrals

Question 1

The Dirichlet function — defined as 1 for rational x and 0 for irrational x on [0,1] — is bounded and defined everywhere on the interval. Why is it not Riemann-integrable?

• A. It oscillates too rapidly for Darboux sums to converge
• B. Its discontinuity set is the rationals, which have positive Lebesgue measure
• C. Its discontinuity set is all of [0,1] — it is discontinuous everywhere — which has positive measure
• D. Bounded functions on finite intervals are always Riemann-integrable

Answer: (C)

The Dirichlet function is discontinuous at every point of [0,1]: every neighborhood of every point contains both rationals and irrationals, so the function cannot be continuous anywhere. Its discontinuity set is therefore all of [0,1], which has Lebesgue measure 1 (not zero). By the Lebesgue criterion, a bounded function on [a,b] is Riemann-integrable if and only if its discontinuities form a measure-zero set — so the Dirichlet function fails this test. Option D is the common false assumption: boundedness alone is not enough for Riemann integrability.

Question 2

A student claims: 'The Lebesgue integral is more powerful than the Riemann integral because it gives different, more accurate values for functions that Riemann can handle.' What is wrong with this claim?

• A. Nothing — Lebesgue integrals do assign different values to the same functions
• B. The real advantage is that Lebesgue handles functions Riemann cannot, and the real power lies in its convergence theorems — when both apply, they give identical values
• C. The claim is wrong because Riemann integrals are actually more general than Lebesgue integrals
• D. The claim is right, but only for functions with infinitely many discontinuities

Answer: (B)

On any interval where a function is Riemann-integrable, both integrals agree exactly: (R)∫f dx = (L)∫f dλ. The Lebesgue integral is a strict generalization — it does everything Riemann does, plus more — but it does not revise Riemann's values. The deeper advantage is twofold: it handles a wider class of functions (those with large discontinuity sets), and it comes with far more powerful convergence theorems (Dominated Convergence, Monotone Convergence) that allow limit-integral interchange under much weaker conditions than Riemann integration permits.

Question 3

A monotone function on [a,b] may have jump discontinuities, but it is still Riemann-integrable.

• T. True
• F. False

Answer: True

A monotone function can have at most countably many jump discontinuities (since each jump has positive size and the total variation is finite). Any countable set has Lebesgue measure zero. By the Lebesgue criterion for Riemann integrability, a bounded function is Riemann-integrable if and only if its discontinuity set has measure zero — so a monotone function, whose discontinuities form a countable (hence measure-zero) set, is always Riemann-integrable.

Question 4

If a function is Lebesgue-integrable on [0,1], then it is also Riemann-integrable on [0,1].

• T. True
• F. False

Answer: False

The Lebesgue integral strictly extends the Riemann integral — not the other way around. The Dirichlet function is Lebesgue-integrable (its Lebesgue integral equals 0, since the rationals have measure zero) but is not Riemann-integrable. The Lebesgue criterion shows that Riemann integrability requires the discontinuity set to have measure zero, a condition the Dirichlet function fails. Every Riemann-integrable function is Lebesgue-integrable, but not conversely.

Question 5

Why does modern probability theory, functional analysis, and measure theory use the Lebesgue integral as the default framework rather than the Riemann integral, even for functions that Riemann could handle?

Model answer: The Lebesgue integral is preferred not primarily because it handles more functions, but because it supports far more powerful convergence theorems. The Dominated Convergence Theorem and the Monotone Convergence Theorem allow one to interchange limits and integrals under conditions much weaker than those Riemann integration requires. Since virtually all advanced analysis involves limit processes — uniform limits of functions, convergence of probability measures, infinite series of functions — these theorems are indispensable. The Riemann integral's convergence results are too restrictive to carry the weight that modern analysis demands.

The key insight is that the Lebesgue integral's practical superiority is not mainly about handling pathological functions like the Dirichlet function — that would rarely matter in applications. The real reason is the convergence theorem package. In probability theory, for instance, taking expectations of limits of random variables requires exactly the kind of limit-integral interchange that the Dominated Convergence Theorem provides. Building this theory on Riemann integration would require imposing uniform convergence conditions everywhere, making it unworkable. Lebesgue integration is the foundation that makes modern analysis tractable.