A function f on [0,1] is discontinuous at every rational number (a dense set), but continuous at every irrational number. Is f Riemann integrable on [0,1]?
ANo — a dense set of discontinuities guarantees the upper and lower Darboux sums can never agree
BYes — the rationals in [0,1] form a countable set, which has measure zero, so f satisfies Lebesgue's criterion
COnly if f is also bounded, in which case the dense discontinuities do not matter at all
DIt depends on whether f is monotone; monotonicity is the deciding factor, not the discontinuity set
Lebesgue's criterion states that f is Riemann integrable if and only if its discontinuity set has measure zero. The rationals in [0,1] are countable, and every countable set has measure zero (they can be covered by intervals of arbitrarily small total length). So a dense discontinuity set is not automatically fatal — what matters is the measure of that set, not its density or cardinality. Option A reflects the common confusion between 'dense' and 'large in the sense of measure.'
Question 2 Multiple Choice
Why is the Dirichlet function (f = 1 on rationals, f = 0 on irrationals) not Riemann integrable on [0,1]?
ABecause it is not monotone on [0,1]
BBecause it has infinitely many discontinuities, and Riemann integrability requires only finitely many
CBecause its discontinuity set is all of [0,1], which has positive measure, so the upper and lower Darboux sums cannot be made to agree
DBecause it is not bounded — it oscillates between 0 and 1 without settling
The Dirichlet function is discontinuous at every point of [0,1]: its discontinuity set has measure 1 (positive), so Lebesgue's criterion fails. On any subinterval of [0,1], the supremum of f is 1 (a rational is always nearby) and the infimum is 0 (an irrational is always nearby). So the upper sum is always 1 and the lower sum is always 0, regardless of how fine the partition is. Option B is the common but wrong answer — infinitely many discontinuities are not inherently fatal; only a positive-measure set of them is.
Question 3 True / False
A function with a countably infinite set of discontinuities on [a,b] cannot be Riemann integrable.
TTrue
FFalse
Answer: False
Countably infinite sets have measure zero — they can be covered by open intervals with total length less than any ε > 0. Therefore a function with only countably many discontinuities satisfies Lebesgue's criterion and is Riemann integrable (provided it is bounded). Monotone functions, for instance, can have countably many jump discontinuities and are always Riemann integrable. The criterion is measure zero, not finiteness.
Question 4 True / False
Every monotone bounded function on [a,b] is Riemann integrable.
TTrue
FFalse
Answer: True
A monotone function on a bounded interval can have at most countably many discontinuities (each discontinuity corresponds to a jump, and the jumps must sum to a finite total variation, so there can only be countably many). A countable set has measure zero, so Lebesgue's criterion is satisfied. This is one of the important corollaries of the measure-zero condition: monotone functions — even highly irregular step-like ones — are always integrable.
Question 5 Short Answer
Why is 'the discontinuity set has measure zero' the right criterion for Riemann integrability, rather than 'the function has only finitely many discontinuities'?
Think about your answer, then reveal below.
Model answer: Measure zero is the precise condition under which discontinuities can be 'hidden' by partitions — their total contribution to the gap between upper and lower Darboux sums can be made arbitrarily small. Finite is sufficient but not necessary: a function with countably or even uncountably many discontinuities (as long as they form a measure-zero set) is still integrable, because those discontinuities occupy negligible total length and don't prevent the upper and lower sums from converging. 'Finitely many' is too restrictive a sufficient condition, and it misses the deeper reason why integrability works.
The Darboux sum gap over a subinterval is bounded by the oscillation of f times the subinterval's length. If the discontinuities are confined to a measure-zero set, we can cover them with intervals of total tiny length. On those tiny intervals, the gap can be large, but the total contribution (oscillation × tiny length) is negligible. On the remaining intervals, f is continuous and we can make the oscillation small. The measure-zero condition is exactly the one that makes this argument go through — and it's why Lebesgue's criterion is the right statement, not the ad hoc 'finitely many' version.