The Fundamental Theorem of Calculus in Lebesgue Theory - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T04:00:55Z http://mathoverflow.net/feeds/question/31849 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/31849/the-fundamental-theorem-of-calculus-in-lebesgue-theory The Fundamental Theorem of Calculus in Lebesgue Theory mencius 2010-07-14T13:10:45Z 2011-03-07T00:09:27Z <p>Dear all,</p> <p>I am interested to what extent the famous identity</p> <p>$\int_a^b f'(x) \ dx=f(b)-f(a)$</p> <p>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 $\mathbb R$.</p> <p>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.</p> <p>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.</p> <p>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?</p> <p>Thank you for your time and effort.</p> http://mathoverflow.net/questions/31849/the-fundamental-theorem-of-calculus-in-lebesgue-theory/31850#31850 Answer by Daniel Litt for The Fundamental Theorem of Calculus in Lebesgue Theory Daniel Litt 2010-07-14T13:20:31Z 2010-07-14T13:20:31Z <p>See <a href="http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Generalizations" rel="nofollow">this Wikipedia article</a>.</p> <p>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. </p> <p>For this statement, it suffices that $f$ be locally (Lebesgue) integrable and continuous at $x$.</p> http://mathoverflow.net/questions/31849/the-fundamental-theorem-of-calculus-in-lebesgue-theory/31858#31858 Answer by Noah Stein for The Fundamental Theorem of Calculus in Lebesgue Theory Noah Stein 2010-07-14T14:01:25Z 2010-07-14T14:01:25Z <p>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.</p> http://mathoverflow.net/questions/31849/the-fundamental-theorem-of-calculus-in-lebesgue-theory/31860#31860 Answer by Franklin for The Fundamental Theorem of Calculus in Lebesgue Theory Franklin 2010-07-14T14:05:20Z 2010-07-14T14:05:20Z <p>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). </p> http://mathoverflow.net/questions/31849/the-fundamental-theorem-of-calculus-in-lebesgue-theory/57404#57404 Answer by Phil Isett for The Fundamental Theorem of Calculus in Lebesgue Theory Phil Isett 2011-03-04T21:58:44Z 2011-03-07T00:09:27Z <p>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</p> <p>$f(b) - f(a) = \int_a^b f'(x) dx$</p> <p>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 $0$.</p> <p>The step to prove it for smooth compactly supported functions may be done by applying the dominated convergence theorem to </p> <p>$\int (f(x+h) - f(x))/h~ dx$</p> <p>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).</p> http://mathoverflow.net/questions/31849/the-fundamental-theorem-of-calculus-in-lebesgue-theory/57435#57435 Answer by K. Henriksen for The Fundamental Theorem of Calculus in Lebesgue Theory K. Henriksen 2011-03-05T06:16:14Z 2011-03-05T06:25:27Z <p>N.L. Carothers's <em>Real Analysis</em> 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.</p> <p>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. </p>