How far can we push the Fundamental Theorem of Calculus for Riemann integral?

Let $$f:[a,b] \rightarrow \mathbb{R}$$ be a continuous function such that it is differentiable everywhere unless in a null set $$S$$. Suppose that there is a function $$g$$, which is bounded and Riemann integrable in $$[a,b]$$, such that $$g(x) = f'(x)$$ for every $$x\in [a,b] -S$$. Then,

$$f(b)-f(a) = \int^b_a g$$

is true? If it is false, provide a counterexample and consider the case in which we switch "null set" by "countable set". In this case, it will be true? If not, provides a counterexample. I know that it would be true if we have finite set instead of null set.

Remark: I am asking this question because I am studying a Brazilian book about Fourier Analysis ("Análise de fourier e equações diferenciais parcias" whose author is Djairo), in which he just uses Riemann integral and uses many times integration by parts. However, he just say "let $$f$$ be continuous in a closed and bounded interval such that $$f'$$ is integrable in the same interval" and then he uses integration by parts (the other function is $$\cos$$ or $$\sin$$ usually). And he doesn't define which he means by the "derivative" of a function, because if $$f$$ is differentiable everywhere, it would redundant to say that it is continuous. Would it be differentiable everywhere unless in a null set, countable set, finite set? Idk. This is why i am asking how far I can push the fundamental theorem of calculus, which is used to prove integration by parts.

Edits: Thanks for the comments, the case in which $$S$$ is a null set is already solved. It's false. The counterexample is the Cantor function. It remains the case in which $$S$$ is countable.

Here: "A Fundamental Theorem of Calculus" there's a similar problem, but I m not sure if it is equivalent. Anyway, I would really appreciate if my problem were solved not using Lebesgue theory, which i haven't studied yet.

I had to fix the statement of the problem because it was wrong, as pointed out in the comments. Originally, i thought that it was enough that $$f$$ was differentiable in $$[a,b]-S$$ and one can extend $$f'$$ in any way, but then $$\int_a^b f'$$ will not exist necessarily, and we won't obviously have $$f(b)-f(a) = \int^b_a f'$$.

• Searching for 'continuous singular function' will give you some basic information. – Kavi Rama Murthy Aug 2 '19 at 11:47
• let $f$ [be] continuous in a closed [and bounded] interval such that $f'$ is integrable in the same interval --- This part is needed because $f$ continuous doesn't imply $f'$ is Riemann integrable. See Volterra's function and Mark McClure's answer to Discontinuous derivative. – Dave L. Renfro Aug 2 '19 at 12:18
• – Dave L. Renfro Aug 2 '19 at 12:26
• Possible duplicate of A Fundamental Theorem of Calculus – ibnAbu Aug 2 '19 at 13:24
• If $S$ is infinite then that "one can define $f'$ arbitrarily in $S$" really doesn't work (see the Note at the bottom of my answer). So if $S$ is infinite it's not clear exactly what you're asking - you want to know whether $f(b)-f(a)$ is equal to $\int_a^b$ of what??? – David C. Ullrich Aug 2 '19 at 16:21

Yes, if $$f'$$ is Riemann integrable then $$\int_a^bf'=f(b)-f(a)$$. I've posted a proof of this before, but it's simple enough that giving the proof again seems easier than trying to find that post:

Say $$a=x_0<\dots is a partition of $$[a,b]$$. The Mean Value Theorem shows that there exists $$\xi_j\in(x_{j-1},x_j)$$ such that $$f(x_j)-f(x_{j-1})=f'(\xi_j)(x_j-x_{j-1}).$$So $$f(b)-f(a)=\sum_j(f(x_j)-f(x_{j-1}))=\sum_jf'(\xi_j)(x_j-x_{j-1}).$$But that last sum is precisely a Riemann sum for $$\int_a^b f'$$, so for any $$\epsilon>0$$ the last sum above is within $$\epsilon$$ of $$\int_a^bf'$$ if $$\max_j(x_j-x_{j-1})$$ is small enough.

So $$\left|f(b)-f(a)-\int_a^b f'\right|<\epsilon$$for every $$\epsilon>0$$.

Now what if $$f$$ is just differentiable on $$[a,b]\setminus S$$? No if we assume just that $$S$$ is a null set. I don't know the answer if $$S$$ is countable, but I suspect it's no. Yes if $$S$$ is finite (and $$f$$ is globally continuous):

Say $$S=(a_j)$$, where $$a_1<\dots. The case proved above shows that $$f(a_{j+1})-f(a_j)=\lim_{\epsilon\to0}(f(a_{j+1}-\epsilon)-f(a_j+\epsilon))=\lim_{\epsilon\to0}\int_{a_j+\epsilon}^{a_{j+1}-\epsilon}f'=\int_{a_j}^{a_{j+1}}f';$$now take the sum over $$j$$.

• How can you apply the MVT if $f$ is not known to be differentiable on the interval $(x_{j-1},x_j)$? – Paul Frost Aug 2 '19 at 15:52
• @PaulFrost I said "if $f'$ is Riemann integrable". A function Riemann integrable on $[a,b]$ has to be defined at every point of $[a,b]$, so $f$ is differentiable on $[a,b]$. – David C. Ullrich Aug 2 '19 at 15:57
• Ah, you are right ;-) In fact, the OP said "Suppose that $f ′$ is bounded and Riemann integrable in $[a,b]$", but certainly it was not his real intention to consider the case $S = \emptyset$. – Paul Frost Aug 2 '19 at 16:02
• @PaulFrost Regardless of his real intention, and in fact regardless of anything he said, all I assert when I assert something is the assertion I'm actually making! Everything I said above is (as far as I know) true. – David C. Ullrich Aug 2 '19 at 16:06
• I agree: Your answer is absolutely correct.. – Paul Frost Aug 2 '19 at 16:10

I don't know the answer assuming that $$S$$ is countable. But yes, FTC holds if $$S$$ is a countable closed set:

Lemma 0. Suppose $$f:(-1,1)\to\Bbb R$$ is continuous and $$f'(t)=0$$ for all $$t\ne0$$. Then ($$f$$ is differentiable at the origin and) $$f'(0)=0$$.

Proof: $$f$$ is constant on $$(-1,0]$$ and constant on $$[0,1)$$; hence $$f$$ is constant.

Lemma 1. Suppose $$S\subset[0,1]$$ is a countable closed set, $$f:[0,1]\to\Bbb R$$ is continuous and $$f'(t)=0$$ for all $$t\in[0,1]\setminus S$$. Then $$f$$ is constant.

Proof: For $$E\subset\Bbb R$$ let $$I(E)$$ be the set of isolated points of $$E$$. Define $$S_\alpha$$ for ordinals $$\alpha$$ by $$S_0=S$$, $$S_{\alpha+1}=S_\alpha\setminus I(S_\alpha)$$and $$S_\alpha=\bigcap_{\beta<\alpha}S_\beta\quad(\alpha\text{ is a limit ordinal)}.$$ Show by induction on $$\alpha$$ that $$S_\alpha$$ is a countable closed set and $$f'=0$$ on $$[0,1]\setminus S_\alpha$$.

There must exist $$\alpha$$ with $$S_{\alpha+1}=S_\alpha$$. A nonempty closed set with no isolated points is uncountable (look up "perfect set" on Wikipedia); hence $$S_\alpha=\emptyset$$.

Prop. Suppose $$S\subset[0,1]$$ is countable closed set, $$f:[0,1]\to\Bbb R$$ is continuous and $$f$$ is differentiable on $$[0,1]\setminus S$$. If there exists a Riemann integrable function $$g$$ such that $$g=f'$$ on $$[0,1]\setminus S$$ then $$f(1)-f(0)=\int_0^1g(t)\,dt.$$

Proof. Define $$F(x)=f(x)-\int_0^xg(t)\,dt.$$(Note that $$F$$ is continuous since $$g$$ is bounded.)

Suppose $$[a,b]\subset[0,1]\setminus S$$. Since FTC holds for differentiable functions with a Riemann integrable derivative, $$F(b)-F(a)=f(b)-f(a)-\int_a^bf'(t)\,dt=0.$$

Since $$S$$ is closed, the previous paragraph shows that $$F'(t)=0$$ for all $$t\in[0,1]\setminus S$$. So Lemma 1 implies $$F$$ is constant, hence $$f(1)-\int_0^1g(t)\,dt=F(1)=F(0)=f(0).$$

Edit: In fact Lemma 1 holds if $$S$$ is any countable set; see here or here. But I don't quite see how to get the Prop from Lemma 1 if $$S$$ is not closed.

• What is an ordinal? Can i just think abt them as natural numbers and your reasoning will work? – Rafael Deiga Aug 5 '19 at 21:58
• What is an ordinal is a long story that you can read about in various places. Natural numbers are ordinals, but there are not enough to make that proof work: There need not be a natuural number $n$ with $S_{n+1}=S_n$. – David C. Ullrich Aug 5 '19 at 22:40