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Here is a bad definition of integrability that my analysis instructor taught me about and explained why it was bad. Consider a real-valued function f with the interval [a, b] as its domain. Partition [a, b] evenly into n intervals of length (b-a)/n. If the upper Darboux sum and lower Darboux sum of f in these partitions converge and equal each other as n approaches infinity, then call f integrable on [a,b]. To see why this definition fails, consider the following function:

f(x) = n if x = 1/n for some natural number n

f(x) = 0 otherwise

On [0,1], the upper Darboux sums of f on partitions from this definition must be greater than or equal to 1 and the lower Darboux sums converge to 0. So f is not integrable with respect to this definition. But f is Riemann integrable.

Edit: Okay, I don't remember the exact example he used.

"Actually, I'm pretty sure the condition you wrote down is equivalent to Darboux integrability. The "bad" definition we spoke about is an exercise in Pugh's book: looking at convergence of Riemann sums with equally-spaced subintervals as you do, but only allowing the midpoints of these subintervals as sample points. Then according to this definition, the characteristic function of the irrationals on [0,1] is integrable (which we know is not, at least if you're talking Riemann/Darboux) since all sample points would be rational so all Riemann sums would be 0, and hence they converge. This shows why we need to consider arbitrary sample points in the definition of Riemann integrability."

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Here is a bad definition of integrability that my analysis instructor taught me about and explained why it was bad. Consider a real-valued function f with the interval [a, b] as its domain. Partition [a, b] evenly into n intervals of length (b-a)/n. If the upper Darboux sum and lower Darboux sum of f in these partitions converge and equal each other as n approaches infinity, then call f integrable on [a,b]. To see why this definition fails, consider the following function:

f(x) = n if x = 1/n for some natural number n

f(x) = 0 otherwise

On [0,1], the upper Darboux sums of f on partitions from this definition must be greater than or equal to 1 and the lower Darboux sums converge to 0. So f is not integrable with respect to this definition. But f is Riemann integrable.

Edit: Okay, I don't remember the exact example he used.

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Here is a bad definition of integrability that my analysis instructor taught me about and explained why it was bad. Consider a real-valued function f with the interval [a, b] as its domain. Partition [a, b] evenly into n intervals of length (b-a)/n. If the upper Darboux sum and lower Darboux sum of f in these partitions converge and equal each other as n approaches infinity, then call f is integrable on [a,b]. To see why this definition fails, consider the following function:

f(x) = n if x = 1/n for some natural number n

f(x) = 0 otherwise

On [0,1], the upper Darboux sums of f on partitions from this definition must be greater than or equal to 1 and the lower Darboux sums converge to 0. So f is not integrable with respect to this definition. But f is Riemann integrable.

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