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I am reading the Wikipedia page on the fundamental theorem of calculus and have a hard time understanding the logic of the theorem and its generalization (section 9). I browsed over the other questions on the site, Understanding the Fundamental theorem of Calculus in plain english, but it seems irrelevant to mine.

First, I am going to phrase the theorem in my own way so others may spot the problems in my understanding:

Let $f$ be a function defined on $[a,b]$, define $$F(x):=\int^x_{a}f(x)dx$$ the fundamental theorem of caluclus states that:

  1. $f$ is continuous on $[a,b] \Rightarrow F(x)$ is uniformly continuous on $(a,b)$ and $F'(x)=f(x)$ for all $x\in(a,b)$.
  2. $f$ is Riemann integrable $\Rightarrow \int^b_af(x)dx=F(b)-F(a)$.

(Purpose of Generalization) Since the fundamental theorem of calculus primarily serves to link integration and differentiation, and continuity of $f$ on $[a,b]$ is a very strong condition, it makes sense to relax this condition a bit and still make $F'(x)=f(x)$ and $\int^b_af(x)dx=F(b)-F(a)$ work.

Here comes to my understanding of the Generalizations part of Wikipedia:

"Part 1 of the theorem then says... is differentiable for $x=x_0$ with $F'(x_0)=f(x_0)$"

This part means we can relax the assumption of $f$ is continuous on $[a,b]$ to $f$ is Lebesgue integrable on $[a,b]$. To my understanding, Lebesgue integrable function only requires the function to be measurable, and satisfies either $\int_{X} f^{+} d \mu<\infty $ or $ \int_{X} f^{-} d \mu<\infty$. (Reference). So it seems the condition is indeed weaker. However, I am not sure any $x_0\in[a,b]$ satisfies $F'(x_0)=f(x_0)$ make any different to $F'(x)=f(x)$ for all $x\in[a,b]$.

"we can relax the conditions on $f$ still further... the function $F$ is differentiable a.e and $F'(x)=f(x)$"

This part means $f$ need not be Lebesgue integrable on the whole $[a,b]$ as long as it can be decomposed into small Lebesgue integrable sections. ($F'(x)=f(x)$ will still hold a.e).

I hope so far I am not far off the correct understanding, I have bigger problems understanding the generalization of the second part of the theorem.

"Part II of the theorem is true for any Lebesgue integrable function f...., then $F(b)-F(a)=\int_{a}^{b} f(t) d t$".

Does it mean the statement will hold true when we relax $f$ being Riemann integrable to Lebesgue integrable? However, I am really confused when trying to understand the next paragraph, where states:

This result may fail for continuous functions $F$ that admit a derivative $f(x)$ at almost every point $x$, as the example of the Cantor function shows. However, if $F$ is absolutely continuous, it admits a derivative $F^{\prime}(x)$ at almost every point $x,$ and moreover $F^{\prime}$ is integrable, with $F(b)-F(a)$ equal to the integral of $F^{\prime}$ on $[a, b] .$ Conversely, if $f$ is any integrable function, then $F$ as given in the first formula will be absolutely continuous with $F^{\prime}=f$ a.e.

This paragraph seems to suggest there exists function $f$ with continuous antiderivative $F(x)$ but will fail the equation $F(b)-F(a)=\int_{a}^{b} f(t) d t$. What exactly is $f$ is not explicitly stated. And it seems also suggest as longas $f$ is Lebesgue integrable, $F(x)$ will be absolutely continuous, hence $F(b)-F(a)=\int_{a}^{b} f(t) d t$ will still hold.

Is my logic correct?

JoZ
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  • This is a thoughtful question and has already provoked at least one thoughtful response. But as a side note, purely from my own point of view, generalizing the FTC feels like a dead end to me. It is a topic like pointwise convergence of Fourier series - generalizations are easy to state and think about, but are devlishly hard to prove, and the theorems you end up getting are nowhere near as useful or applicable as you might expect them to be. Which is not to discourage you from going there. Only to give you permission not to go there if you decide it seems like a wilderness. – leslie townes Mar 01 '21 at 17:40

1 Answers1

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In general $F$ is only differentiable almost everywhere if $f$ is Lebesgue integrable. Its derivative is given by the "local average" of $f$, if this local average is well defined independent of the path used to shrink the domain of integration. The precise statement is called the Lebesgue differentiation theorem.

Part II, if $F$ is defined by the integral anyway, is rather trivial, it just follows from the additivity property of the integral. What is more interesting and useful is that if you have $F'=f$ everywhere, then $\int_a^b f(x) dx = F(b)-F(a)$. It doesn't matter where $F$ came from. In this regard I would say your original statement of FTC is a bit deceptive.

But this property doesn't work if you only assume $F'=f$ almost everywhere, because there are non-absolutely continuous functions. In particular, if $F$ is the Cantor function then $F'$ exists almost everywhere and is $0$ where it exists, but $F(x) \neq \int_0^x 0 \, dt$.

Ian
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