I was having a conversation with another MSE user today and he mentioned that he had heard that "IVT was sufficient for a function to be integrable". When I asked if he meant it was "if-and-only-if" or just "if", he said "if-and-only-if".
This made no sense to me - I imagined a step-function which was 0 for $x<1$ and was 1 for $x\geq1$. This seemed to me to be integrable (with definite integral 1 between 0 and 2) and yet clearly did not satisfy the IVT.
He countered with "step functions are NOT integrable" and further said that he had found the place in the textbook which mentioned this and had verified that it WAS if-and-only-if... Before I could say anything, a movie we wanted to see started.
Where did I go wrong?
I assume he's correct because:
a) He's usually correct and b) I just started calculus and he's been doing it for years.
However, I thought that my step function was integrable, by the definition of integration that I learnt from Spivak's Calculus a la lower and upper sums.
Answer
You are correct: step functions don't satisfy the intermediate value property (unless there's only one step…), but they are always integrable. On the other hand, there are differentiable functions $f$ such that $f'$ is not Riemann-integrable. But, since it is a derivative, $f'$ has the intermediate value property (by Darboux's theorem).
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