According to the Darboux definition, a bounded function f is Riemann integrable on [a,b] when...
Af is continuous at every point of [a,b]
BThere exists some partition for which the upper Darboux sum equals the lower Darboux sum
CThe infimum of all upper Darboux sums equals the supremum of all lower Darboux sums
DThe upper Darboux sum decreases to zero as the partition is refined
Integrability requires inf U* = sup L* — the infimum over all partitions of all upper sums equals the supremum over all partitions of all lower sums. This common value is the integral. Option B is wrong because for most integrable functions, no single partition makes U = L exactly; it is the limiting behavior over all partitions that matters. Option A is sufficient but not necessary: functions with finitely many (or even countably many) discontinuities can still be integrable.
Question 2 Multiple Choice
Consider the Dirichlet function on [0,1]: f(x) = 1 if x is rational, f(x) = 0 if x is irrational. Which statement correctly applies the Darboux criterion?
Af is integrable because it is bounded between 0 and 1
Bf is integrable because its integral should equal 0, since rationals form a 'small' set
Cf is not integrable: on every subinterval, sup f = 1 and inf f = 0, so U = 1 and L = 0 for every partition
Df is integrable for sufficiently fine partitions that separate rationals from irrationals
Every subinterval of [0,1], no matter how small, contains both rationals and irrationals. So on every subinterval, the supremum of f is 1 and the infimum is 0. This means every upper sum U(f,P) = 1 and every lower sum L(f,P) = 0, regardless of how fine the partition is. Therefore inf U* = 1 and sup L* = 0, which are not equal — f is not Riemann integrable. This is why the Darboux criterion is powerful: it immediately identifies non-integrable functions via the gap between upper and lower sums.
Question 3 True / False
Adding more points to a partition (refining it) can cause the upper Darboux sum to increase.
TTrue
FFalse
Answer: False
Refinement can only decrease (or leave unchanged) the upper Darboux sum, and can only increase (or leave unchanged) the lower Darboux sum. When you add a point to a partition, each original subinterval either stays the same or splits into two. Splitting replaces the supremum over a larger interval with suprema over two smaller ones — which can only be ≤ the original. This monotone behavior is what guarantees that inf U* and sup L* are well-defined, and that the upper and lower Darboux sums squeeze toward each other as partitions are refined.
Question 4 True / False
A function with exactly 5 jump discontinuities on [a,b] is Riemann integrable on [a,b].
TTrue
FFalse
Answer: True
The Riemann-Lebesgue criterion states that a bounded function is Riemann integrable if and only if its set of discontinuities has Lebesgue measure zero. A finite set of points (including 5 jump discontinuities) has measure zero. To see this directly from Darboux sums: on a subinterval containing one discontinuity, the gap Mᵢ − mᵢ may be large, but the width Δxᵢ of that subinterval can be made arbitrarily small. With careful partition refinement, the contribution of these finitely many 'bad' subintervals to U − L can be made as small as desired.
Question 5 Short Answer
Why does the Darboux definition use suprema and infima on each subinterval, rather than arbitrary sample points as in the standard Riemann sum?
Think about your answer, then reveal below.
Model answer: Using sup and inf gives the tightest possible overestimate (upper sum) and underestimate (lower sum) for each subinterval — extremes that depend only on the partition, not on how we sample. This makes integrability a property of the function and partition alone, without the arbitrary choice of sample points. If even these worst-case upper and lower sums converge to the same value, then any Riemann sum (which lies between them) must also converge to that value. Darboux sums thus provide a clean, sample-point-free criterion for integrability.
Arbitrary sample points can make a function look integrable or not depending on where you pick them — for example, always picking rationals in the Dirichlet function gives sum 1, always picking irrationals gives sum 0. Darboux sums avoid this arbitrariness by using the extreme values. This is why the Darboux approach is the cleanest foundation for the Riemann integral and connects most naturally to the Lebesgue theory.