Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Dear all,

I am interested to what extent the famous identity

$\int_a^b f'(x) \ dx=f(b)-f(a)$

is true for a function $f:[a,b]-> \mathbb C$ continuous on $[a,b]$ and differentiable on $(a,b)$. One famous easy case of this problem is where $f'$ is continuous. In the above identity, the integral is with respect to Lebesgue measure on $\mathbb R$.

I have proven so far that $f'$ is always measurable on $(a,b)$ and that if $f'$ is bounded on $(a,b)$ then the result holds. The proof was reasonably elementary, making heavy use of the mean value theorem and the so-called bounded convergence theorem.

I felt that my condition was an artifact of the proof, as the bounded convergence theorem is considerably weaker than the dominated convergence theorem and its strengthened forms.

So does anyone know of a strengthened version of this result, or perhaps even a full description of all differentiable functions such that the above identity holds?

Thank you for your time and effort.

share|improve this question
5  
Does this help? en.wikipedia.org/wiki/Absolute_continuity –  babubba Jul 14 '10 at 13:29
4  
Try the gauge integral: en.wikipedia.org/wiki/Henstock%E2%80%93Kurzweil_integral –  Aaron Bergman Jul 14 '10 at 14:01
1  
As Aaron Bergman indicates, the best answer seems to be: the theorem you want is always true if you use a sufficiently general kind of integral. As you write, if $f'$ is bounded, then it is certainly Lebesgue integrable, but there are examples where $f'$ is unbounded and indeed not Lebegue integrable. However, the Henstock-Kurzweil (or "gauge", "generalized Riemann", "Denjoy-Perron"...) integral is stronger than the Lebesgue integral and indeed integrates every derivative, bounded or not. –  Pete L. Clark Jul 14 '10 at 15:20
add comment

5 Answers

up vote 4 down vote accepted

See this Wikipedia article.

Your "famous identity" may not be quite what you want it to be; the usual way of stating the FTC is to let $$F(x)=\int_a^x f ~dx$$ for integrable $f$. Then $F'(x)=f(x)$. This is subtly different from what you wrote.

For this statement, it suffices that $f$ be locally (Lebesgue) integrable and continuous at $x$.

share|improve this answer
7  
@Daniel: Since differentiation and integration are supposed to be essentially mutually inverse operators, there are two parts to the fundamental theorem of calculus, one for $D \circ I$ and the other for $I \circ D$. These two cases turn out to be related but distinct (see e.g. my commentson the Mean Value Theorem question). The OP is asking about $I \circ D$. –  Pete L. Clark Jul 14 '10 at 15:16
1  
@Pete L. Clark: Ah, I see. I was thinking of the usual counterexamples to $I\circ D$ in the case of Riemann integration, but I suppose asking when the "backwards fundamental theorem of calculus" (the CTF?) holds is an interesting question. –  Daniel Litt Jul 14 '10 at 15:31
add comment

As I recall Chapter 7 of Rudin's Real and Complex Analysis has a good presentation of the Fundamental Theorem of Calculus in the context of Lebesgue integration.

share|improve this answer
add comment

For every function $f$ with $f'$ integrable there is a function $g$ equal to $f$ everywhere but a point such that $\int_{a}^{b}g'dx=g(b)-g(a)$. Take g(x)=f(x) for x different from b and g(b)=\int_{a}^{b}f'dx+f(a).

share|improve this answer
add comment

The one-dimensional fundamental theorem (with a Lebesgue integral on the right hand side) holds in greatest generality when the derivative $f'$ (taken in the sense of distributions) is a measure -- such functions are called ``bounded variation''. For example, if $f$ is nondecreasing, it is measurable, locally bounded and therefore a distribution. By taking difference quotients, $f'$ is a non-negative distribution and hence a measure (as one can show with the Riesz Representation theorem). You can prove the formula

$f(b) - f(a) = \int_a^b f'(x) dx$

at points $a$ and $b$ to which the measure $f'(x)$ does not assign positive mass (the case $f'$ being in $L^1 $ is exactly when $f$ is absolutely continuous). One proof is by convolving with a mollifier, quoting the result for smooth functions, then put the dual mollifier on the characteristic function of [a,b]. The mollifier converges to the characteristic function everywhere but the endpoints, and, say, the dominated convergence theorem allows you to take the limit as long as $a$ and $b$ do not have positive mass with respect to $f'$. But if you take the Heaviside function, its derivative is a delta function, and the formula essentially fails if you try to use $0$ as an endpoint. However, even in cases like this one the formula works for any $a, b$ not equal to $0$.

The step to prove it for smooth compactly supported functions may be done by applying the dominated convergence theorem to

$\int (f(x+h) - f(x))/h~ dx$

as $h$ tends to $0$. I think that no matter how you try to prove the fundamental theorem, the mean value theorem will enter in somewhere (here it enters to bound the difference quotients).

share|improve this answer
add comment

N.L. Carothers's Real Analysis has a fairly good bit devoted to this in Chapter 20 "Differentiation" but unfortunately the relevant part of the book is not on Google books.

Carothers's exposition focuses entirely on the real line allowing for more focus on what can be proven in this single case while forgoing questions about abstract measures.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.