The Fundamental Theorem of Calculus in Lebesgue Theory

Question

I am interested to what extent the famous identity $$ \int_a^b f'(x) \ dx=f(b)-f(a) $$ is true for a function $f:[a,b]\to \mathbb C$ 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.

Answer 1

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 $x$.

Answer 2

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.

Answer 3

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).

Answer 4

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.

Answer 5

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 $g(b)=\int_{a}^{b}f'dx+f(a)$.