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Fundamental Theorem of Calculus Sufficient/Necessary Conditions?
I refer, here, to the Fundamental Theorem of Calculus in the context of measure theory. For increasing functions,
Int_[a,b] f' <= f(b) - f(a)
A perfect example of strict inequality is the Lebesgue-Cantor function, defined on [0,1] (The integral is 0, but f(1) = 1, and f(0) = 0, hence the RHS is 1). The question I have is:
What is a necessary condition for equality above? An equivalent question:
What is a necessary condition for a function to be recovered from its derivative?
I know that a sufficient condition is absolute continuity. Is there a weaker necessary condition that may be used to generalize this theorem further?
ThatDeadDude:
What about functions which are everywhere continuous but nowhere differentiable? They do exist, and equality in this case cannot hold, as the LHS is undefined.
Nor is differentiability almost everywhere enough, as the Cantor-Lebesgue function shows. This is more general than the single-variable calculus presentation of the theorem, but rather a more general treatment.
I know a function whose derivative is almost everywhere continuous is also not guaranteed to give equality, as the Cantor-Lebesgue function again shows. So is there something weaker than absolute continuity?
Thanks anyway.
Steiner:
Thanks for the reply. That was the condition I wasn't certain about. Since I have your attention, might I ask you to help me get started on a proof?
How would I prove the following:
Riemann Integrability of f' does not necessarily imply absolute continuity of f, but absolute continuity of f implies Riemann Integrability of f'.
Thanks again.
Riemann Integrability of f' does not necessarily imply absolute continuity of f, but absolute continuity of f implies Riemann Integrability of f'.
Wait...this is truly trivial...
absolute continuity --> Riemann Integrability:
f is absolutely continuous, hence continuous and bounded on a compact interval, thus integrable.
Riemann Integrability does not imply absolute continuity:
Riemann Integrability doesn't even imply continuous in general.
3 Answers
- SteinerLv 71 decade agoFavorite Answer
A sufficient condition for ∫ (a to b) f'(x) dx = f(b) - f(a) is that f' is Riemann integrable over [a, b]. f' doesn't need to be continuous, but must be Riemann integrable and the fact that it is the derivative of f does not ensure it is Riemann integrable over [a, b].
You can check that if
f(x) = x² cos(1/x²) for x ∈ (0, 1]
f(0) = 0
then
f'(x) = 2x cos(1/x²) + (2/x) sin(1/x²) for x ∈ (0, 1]
f'(0) = 0
f' exists everywhere on [0, 1] but is not Riemann integrable over this interval, because f' is unbounded. So, although f' exists, the fundamental theorem of calculus doesn't hold in this case.
Edit
In general, g is integrable over [a, ] if, and only if, g is bounded and continuous almost everywhere on [a, b] (with respect to thy Lebesgue measure).
- Anonymous1 decade ago
If memory serves continuity over the region in question is both a sufficient and necessary condition. I think the statement of equality is an 'if and only if'. Remember that the derivative can only exist at points in an interval if the function is continuous therein.
- Anonymous5 years ago
It follows trivially from the fact that [m→∞]lim [k=1, m]∑(1/m * (k/m)^(1/n)) =n/(n+1).
