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I read from the Apostol book calculus Vol 1 that the First fundamental theorem of calculus is as follows:

Let $f$ be a function that is integrable on $[a,x]$ for each $x$ in $[a,b]$. let $c$ be such that $a\le c\le b$ and define a new function $A$ as follows:

$$A(x) = \int^x_c f(t)$$

if $a\le x\le b$. Then the derivative $A'(x)$ exists at each point $x$ in the open interval $(a,b)$ where $f$ is continuous, and for such $x$ we have

$$A'(x)=f(x)$$

The book gives a geometric argument as follows:

$$\int^{x+h}_x f(t)dt = \int^{x+h}_c f(t)dt - \int^{x}_c f(t)dt = A(x+h) - A(x)$$

Assuming $f$ is continuous in $[x,x+h]$, then by mean-value theorem for integrals, then we have $A(x+h) -A(x) = hf(z)$ where $x\le z \le x+h$.

The book told us that the above proof is flawed. What's wrong with this argument? Yes it assumes the continuity of interval $[x,x+h]$, but that's what the assumption of the theorem is about, right?

Dachuan Huang
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1 Answers1

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As Apostol points out, this argument depends on the continuity of $f$ in the interval $[x,x+h]$ because one of the hypotheses of the mean value theorem for integrals is continuity on the whole interval of integration. It is flawed in the sense that it is not sufficient if all we assume is continuity of $f$ at $x$. Instead, we must assume continuity in a neighborhood of $x$.

I disagree with the comment above by Rob Arthan. At this stage of the book, Apostol has already developed more than enough theory concerning integrals. Though Apostol calls the argument "geometric" and gives an illustration with areas, the proof itself doesn't rely in any way on the geometric concept of area.

Edit: I will quote more of what Apostol says in that section of the book.

Hence we have $(A(x+h) - A(x))/h = f(z)$, and, since $x \leq z \leq x + h$, we find that $f(z) \to f(x)$ as $h \to 0$ through positive values. A similar argument is valid if $h \to 0$ through negative values. Therefore $A'(x)$ exists and is equal to $f(x)$.

This argument assumes that the function $f$ is continuous in some neighborhood of the point $x$. However, the hypothesis of the theorem refers only to continuity of $f$ at a single point $x$. Therefore, we use a different method to prove the theorem under this weaker hypothesis.

Anonymous
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  • It looks like Apostol should have made it clearer exactly what he meant. – Rob Arthan Aug 20 '21 at 19:43
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    @RobArthan I referred to the book and found the explanation very clear. I believe the OP's misunderstanding of Apostol's intention led him to omit certain parts of the explanation. – Anonymous Aug 20 '21 at 19:46
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    Just to give a simple example of a function that's (Riemann) integrable on $[-1,1]$ and continuous at $x=0$ but not continuous in any neighborhood of $0$ (because it's discontinous at $x=\pm1/n$ for all $n\ge1$), consider $$f(x)=\begin{cases} 0&\text{if }x=0\\ {\displaystyle1\over\displaystyle\left\lfloor{1\over|x|}\right\rfloor}&\text{if }x\not=0\ \end{cases}$$ – Barry Cipra Aug 20 '21 at 20:18
  • Thanks. I misunderstood the theorem's description. It never said $f$ is continuous in the open interval $(a,b)$. Instead, it said whenever $f$ is continuous at some location $x$ where $a\le x\le b$, then $A'(x)$ exists. That's a weaker assumption. – Dachuan Huang Aug 21 '21 at 05:10
  • By the way, I am surprised that $f$ can be continuous at $x=0$ but not any neighborhood of $0$. Mini proof: we can always find an $n$ such that $\frac{1}{n}<\epsilon$ and let $\delta=\frac{1}{n}$. How to prove it is integrable? It looks like a step function, but things get out of my imagination when getting closer to $0$. – Dachuan Huang Aug 21 '21 at 05:14
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    @DachuanHuang It's easy to construct a function $f$ continuous at $0$ but not in any neighborhood of $0$. Pick any function $f$ such that $-|x| \leq f(x) \leq |x| $ and $f$ has discontinuities (for example jump discontinuities) at $a_1, a_2, a_3, \dots$, with $a_n \to 0$. Try to draw a graph like this and you'll see that there are many ways to do it. – Anonymous Aug 21 '21 at 21:52
  • It's interesting to re-discover that the definition of continuity is for a point and does not say anything about its neighborhood. – Dachuan Huang Aug 21 '21 at 23:42