# FTC for Lebesgue Integral Simple Conjecture

I've been attempting to learn and understand Lebesgue integration over the past few days, and have a question about an observation I have made about the Lebesgue integral with my currently limited understanding of it.

I take the limit of the Riemann sum, $\displaystyle\lim_{n \to \infty} \sum_{i=0}^{n} \mu(F^{-1}([y_{i+1}, y_i])) \cdot \Delta{y}$, the sum of the measures of discrete sections the inverse images of our total area $A$ rendered as a set of points. This roughly follows the format of a normal Riemann integral, merely passed through a different function which removes some limitations from what can be integrated, such as limitations concerning continuity.

I would then propose that any aspects of the Fundamental Theorem of Calculus that are not dependant on the generalizations introduced by passing our original function through a Lebesgue measure before integrating across its range would still hold, as fundamentally the Lebesgue integral is a generalization of the Riemann integral for a greater set of measurable functions. Am I missing anything fundamental about this statement which either alters it or proves it false?

More practically speaking, is there a Fundamental Theory of Calculus for Lebesgue Integrable functions such as indicator functions, which does not depend on continuity? If so, how is it different or similar to the notion of a Fundamental Theory of Calculus without dependence or conclusions based on continuity (assuming that statement isn't complete nonsense, which I think it actually could be).

• The FTC makes assumptions about continuity that force the Riemann integral to exist, in which case the Riemann integral equals the Lebesgue integral, in which case the FTC holds for the Lebesgue integral. – Adrian Keister May 31 '18 at 0:34
• The question contains too much vagueness and undefinedness to answer. It would be good if you can be more precise. – zhw. May 31 '18 at 0:39
• I'm sorry about the vagueness, I'm having trouble grasping the concept in full. Adrian Keister's comment answers my question, and states what I observed in far more concise and accurate language. – Romano Jr. Tio May 31 '18 at 1:05
• I don't understand the comment of @AdrianKeister either. – zhw. May 31 '18 at 1:23
• The Lebesgue integral is a generalization of the Reimann integral. From what I understand the comment states: As a consequence of the FTC, the Reimann integral exists. Therefore, for some Lebesgue integrals which are also Reimann integrals the whole FTC holds. – Romano Jr. Tio May 31 '18 at 2:03

## 1 Answer

There is a Fundamental theorem of calculus for Lebesgue integrable functions. More precisely, if $g$ is a Lesbesgue integrable function, then the function $f(x)=\mu((-\infty,x])$ associated to the measure $$\mu(A)=\int_Ag \ dx\;,$$ with $dx$ the Lebesgue measure, is absolutely continuous. You can read more about absolute continuity from wikipedia https://en.wikipedia.org/wiki/Absolute_continuity. Moreover, any absolutely continuous function has an associated Lebesgue integrable derivative (called the Radon-Nikodym derivative). In this weaker sense, the fundamental theorem of calculus holds.