1

I'm hoping for a bit of clarification. The First FTC in Spivak says that if a function is integrable on $[a,b]$ and is also continuous on $c \in [a,b]$, then $F(x) =\int_a^xf$ is differentiable at c and $F'(c) = f(c)$.

Then the Second FTC says if the integrable function, whether continuous or not, has an antiderivative $g' = f$, then $\int_a^bf=g(b) - g(a)$, which if I'm not mistaken implies the result of the First FTC.

Initially, I thought of the 2nd FTC as a way for the result of the 1st FTC to apply to a greater variety of functions, namely those that may not be continuous. But I can't think of any such functions. The derivative of a function, if I'm correct, can't be a jump or removable discontinuity. If the function is somewhere unbounded or undefined, it won't be integrable. And I'm not sure but I don't think an everywhere discontinuous function is integrable, if there even is such a function who's derivative is everywhere discontinuous.

So either I haven't considered all the possibilities of an integrable, somewhere discontinuous function that has an antiderivative, or I've misunderstood Spivak. Any help would be greatly appreciated.

hiroshin
  • 195
  • 10
  • There are functions with an antiderivative that are not continuous, any function with a derivative which isn't continuous gives rise to one for example. If you use a suitably large notion of the integral, any function you can construct is integrable (though this integral can be infinite) – Jon Hillery May 24 '21 at 23:14
  • @Elliot G Sorry I don't follow, what is f supposed to be an example of? My assumption is based on if f is integrable and has an antiderivative on $[a,b]$, then it's also integrable and has an antiderivative on $[a,x]$ for all $x$ in the interval. Then since the RHS of $F(x) = g(x) - g(a)$ is differentiable, $F$ is also differentiable, and $F'(x) = g'(x) = f(x)$ which is the result of the First FTC. – hiroshin May 24 '21 at 23:47
  • Sorry I just deleted my comment; I think I missed your question initially. – pancini May 24 '21 at 23:48
  • @Jon Hillery Well I am assuming Spivak is making these propositions based on his given notion of an integral. I'm aware we have derivatives that are discontinuous, I'm just not sure if there's one that's both discontinuous AND integrable. – hiroshin May 24 '21 at 23:51
  • 2
    You may have a look at this answer https://math.stackexchange.com/a/1900844/72031 – Paramanand Singh May 25 '21 at 01:46

1 Answers1

4

Using the terminology in Spivak, there are three results:

  • First FTC
  • Corollary of first FTC
  • Second FTC

You write

I thought of the 2nd FTC as a way for the result of the 1st FTC to apply to a greater variety of functions

To be clear, the second FTC is a generalization of the corollary of the first FTC. I guess what you're asking about is a situation when the second FTC can be applied but the corollary cannot. For this, consider the function $g:\Bbb{R}\to\Bbb{R}$ \begin{align} g(x):= \begin{cases} x^2\sin\left(\frac{1}{x}\right)& \text{if $x\neq 0$}\\ 0& \text{if $x=0$} \end{cases} \end{align} Then, $g$ is differentiable everywhere and \begin{align} g'(x)&= \begin{cases} 2x\sin\left(\frac{1}{x}\right)-\cos\left(\frac{1}{x}\right) & \text{if $x\neq 0$}\\ 0 & \text{if $x=0$} \end{cases} \end{align} Let me define $f=g'$. Then, $f$ is continuous everywhere except at the origin, and $f$ is also bounded in a neighborhood of the origin. Thus, for any $[a,b]$, it follows that $f$ is (Riemann) integrable on $[a,b]$, and clearly all the hypotheses for the second FTC are satisfied. On the other hand, $f$ is NOT continuous, so the corollary itself cannot be applied (if $0\in [a,b]$).

peek-a-boo
  • 65,833
  • 1
    I guess this is the opposite of what OP wanted but Volterra's function based on this is pretty cool https://en.wikipedia.org/wiki/Volterra%27s_function – pancini May 25 '21 at 00:26
  • 1
    @ElliotG indeed that's an interesting function to take note of, and this is one of many instances in analysis whereby if we have an example of two "pathological" things, we can usually combine them to make an even more pathological thing (sometimes even "infinitely more pathological":) – peek-a-boo May 25 '21 at 00:29