# Integral of derivative

1. Jun 11, 2013

### jostpuur

Assume that a function $f:[a,b]\to\mathbb{R}$ is differentiable at all points in $[a,b]$ (we accept left and right sided derivatives at the end points). Will

$$\int\limits_{[a,b]}f'(x)dm_1(x) = f(b)-f(a)\quad\quad\quad\quad (1)$$

hold, where the integral is the Lebesgue integral?

Now, becareful with this thing. I know it looks simple, but I was unable to find an answer after going through my pedagogical material. The question contains different assumptions than the most commonly known theorems.

Last edited: Jun 11, 2013
2. Jun 11, 2013

### micromass

Staff Emeritus
Billingsley gives the following counterexample:

Define $f(x) = x^2 \sin(1/x^2)$ for $0< x\leq 1/2$. Define $f(x) = 0$ for $x\leq 0$ and $x>1$. Define $f$ on $1/2<x<1$ in such a way that $f$ is continuously differentiable on $(0,+\infty)$. Then $f$ is everywhere differentiable. But $f^\prime$ is not even integrable, which makes the equality impossible for $a=0$.

Last edited: Jun 11, 2013
3. Jun 11, 2013

### jostpuur

I had never seen $\frac{1}{x^2}$ being put inside a trigonometric function. Only $\frac{1}{x}$...

4. Jun 11, 2013

### micromass

Staff Emeritus
The issue here is clearly that $f^\prime$ doesn't need to be Lebesgue integral. This has to be seen as a shortcoming of the Lebesgue integral, and not of our function $f^\prime$. Luckily, there is a more general, and better behaved notion of integral of the real line. This is the Henstock-Kurzweil integral. This generalizes Lebesgue integrals and extended Riemann integrals. We have a really nice fundamental theorem in that case:

If $f$ is a differentiable function on $[a,b]$, then $f^\prime$ is Henstock-Kurzweil integrable and

$$\int_a^b f^\prime = f(b) - f(a)$$

See Bartle "A modern theory of integration". There is even an extension of the theorem where we allow countable many points where $f^\prime$ does not exist!

5. Jun 11, 2013

### Mandelbroth

I follow you up to your last sentence, at which point I'm lost. Why is $f^\prime$ not integrable?

6. Jun 11, 2013

### jostpuur

Suppose $f:\mathbb{R}\to\mathbb{R}$ is defined by

$$f(x) = \left\{\begin{array}{ll} 0,\quad &x=0\\ x^2\sin\Big(\frac{1}{x^2}\Big),\quad &x\neq 0\\ \end{array}\right.$$

Now the function is differentiable everywhere, and the derivative is

$$f'(x) = \left\{\begin{array}{ll} 0,\quad &x=0\\ 2x\sin\Big(\frac{1}{x^2}\Big)-\frac{2}{x}\cos\Big(\frac{1}{x^2}\Big),\quad &x\neq 0\\ \end{array}\right.$$

The derivative is not integrable over intervals containing the origo, and also is not integrable overs sets $[a,0]$ or $[0,b]$ with any $a<0<b$.

However, the limits

$$\lim_{\delta\to 0^-}\int\limits_{a}^{\delta} f'(x)dx = f(0)-f(a)$$

and

$$\lim_{\delta\to 0^+}\int\limits_{\delta}^{b}f'(x)dx = f(b)-f(0)$$

still hold. Suppose we define the integral over $[a,b]$ (with $a<0<b$) by

$$\int\limits_{a}^{b}f'(x)dx \;:=\; \lim_{\delta\to 0^-}\int\limits_{a}^{\delta}f'(x)dx \;+\; \lim_{\delta\to 0^+}\int\limits_{\delta}^{b}f'(x)dx$$

Now the formula

$$\int\limits_{a}^{b}f'(x)dx = f(b)-f(a)$$

holds. No ordinary theorem would tell that it holds, but it holds anyway.

--------------

I was just going to complain here that I have a feeling that there is some result missing from my pedagogial materials. But now I see that while writing this message, micromass has already written something about more general integrals.... anyway, I'll submit this example still...

7. Jun 11, 2013

### micromass

Staff Emeritus
jostpuur, what you wrote is exactly the idea and motivation of the Henstock-Kurzweil integral!

8. Jun 11, 2013

### micromass

Staff Emeritus
So we care about the integral

$$\int_0^a \frac{\sin(1/x^2)}{x} dx$$

By substitution $t = 1/x^2$, we transform this into an integral of the form

$$\int_b^{+\infty} \frac{\sin(t)}{t}dt$$

This integral is well known not to be Lebesgue integrable. A proof can be found in several textbooks. The proof consists of

$$\int_{b}^{+\infty} \frac{|\sin(t)|}{t}dt = \sum_{n=c}^{+\infty} \int_{(n-1)\pi}^{n\pi}\frac{|\sin(t)|}{t}dt \geq \sum_{n=c}^{+\infty} \frac{1}{n\pi}\int_0^\pi |\sin(t)|dt$$

But this last series diverges. Again, this is a shortcoming of the Lebesgue integrable.

9. Jun 11, 2013

### jostpuur

I have a vague memory, that Baire's theorem can be used to prove something about how discontinuous derivatives can be, but I cannot find the theorem anywhere anymore. Do you know anything about it?

10. Jun 11, 2013

### Mandelbroth

Sorry. I thought you were saying that $f^\prime$ was not integrable at all. For some reason, I forgot we were talking specifically about Lebesgue integrals. Sorry.

11. Jun 11, 2013

### micromass

Staff Emeritus
12. Jun 11, 2013

### micromass

Staff Emeritus
Bartle deals with this example in his book "A modern theory of integration". The difference between a Riemann integral, is that there you had to make the partition "uniformly small". Now you also have access to a gauge, which means you can make some parts smaller than other parts.

Here is what Barle writes in his book:

Let $\{r_k~|~k\in \mathbb{N}\}$ be an enumeration of the rational numbers in $[0,1]$ and let $\varepsilon >0$. We define the gauge

$$\delta_\varepsilon = \left\{\begin{array}{l}\varepsilon/2^{k+1}~\text{if}~t= r_k\\ 1~\text{if}~t~\text{is irrational}\end{array}\right.$$
Now let $\{(I_i,t_i)\}_{i=1}^n$ be a $\delta_\varepsilon$-fine partition of $[0,1]$. IF the tag $t_i\in I_i$ is rational, then $f(t_i)=1$, but the length $l(I_i)=x_i - x_{i-1}$ is small. More precisely, if $t_i = r_k$, then $I_i \subseteq [r_k -\delta_\varepsilon(r_k), r_k + \delta_\varepsilon (r_k)]$ so that $l(I_i)\leq \varepsilon 2^k$. Further, if $r_k$ is the tag for two consecutive subintervals, the sum of the lengths of these two nonoverlapping subintervals is $\leq \varepsilon /2^k$. We conclude that the rational $r_k$can contribute at most $\varepsilon /2^k$ to the Riemann sum. Since only rational tags make a nonzero contribution to the Riemann sum, we have
$$|\sum_P f| \leq \sum \frac{\varepsilon}{2^k} \leq \varepsilon$$
Since $\varepsilon$ is arbitrary, the function is integrable and $\int_0^1 f = 0$.

13. Jun 14, 2013

### jostpuur

This is still unclear: Assume that a function $f:[a,b]\to\mathbb{R}$ is differentiable at all points in $[a,b]$. Also assume that

$$\int\limits_{[a,b]}|f'(x)|dm_1(x) < \infty$$

holds. Is it now possible that

$$f(x) = f(a) + \int\limits_{[a,x]}f'(u)dm_1(u)$$

would not hold for some $x$?

I know that if I define a new function

$$g(x) = f(a) + \int\limits_{[a,x]}f'(u)dm_1(u)$$

then $g$ will be absolutely continuous, and it will have a derivative (at least) almost everywhere. Also $g'=f'$ will hold almost everywhere, according to some theorems. But I couldn't find theorems that would imply $g=f$.

14. Jun 14, 2013

### Mandelbroth

I don't understand. Maybe I'm just unfamiliar with some of your notation and think it means something else, but can't we just evaluate it? You could say it is a special case of the fundamental theorem of calculus, but...

$$g(x)=f(a)+\int\limits_{[a,x]}f'(u) \, du = f(a)+f(x)-f(a)=f(x).$$

15. Jun 14, 2013

### jostpuur

You need to use some theorem when you "evaluate" the integral like that, and the theorem will involve some assumptions.

For example, if $f'$ is Riemann-integrable, then we can evaluate like that. If we only assume $f'$ to be Lebesgue-integrable, then it's not obvious.

I have also found a theorem that says that if $f$ is absolutely continuous, then the result will follow without Riemann-integrability.

16. Jun 14, 2013

### jostpuur

Here's one example that can be seen as a motivation for these questions.

First let's define $\phi:\mathbb{R}\to\mathbb{R}$

$$\phi(x) = \left\{\begin{array}{ll} 0,\quad\quad &x=0\\ |x|^{3/2}\sin\Big(\frac{1}{x}\Big),\quad\quad &x\neq 0\\ \end{array}\right.$$

Now $\phi$ is differentiable everywhere, and the derivative is

$$\phi'(x) = \left\{\begin{array}{ll} 0,\quad\quad &x=0\\ \frac{3}{2}\sqrt{|x|}\sin\Big(\frac{1}{|x|}\Big) - \frac{1}{\sqrt{|x|}}\cos\Big(\frac{1}{x}\Big),\quad\quad &x\neq 0\\ \end{array}\right.$$

Now $\phi'$ is not Riemann-integrable over intervals that contain the origo, but still

$$\int\limits_{[-1,1]}|\phi'(x)|dm_1(x) < \infty$$

and

$$\int\limits_{[a,b]}\phi'(x)dm_1(x) = \phi(b)-\phi(a)$$

holds for all $a<0<b$. So we see that sometimes Lebesgue-integral works without limits while Riemann-integral doesn't.

Next, let's define

$$\phi_n(x) = \sum_{j=1}^{2^{n-1}} \phi\Big(x - \frac{2j-1}{2^n}\Big)$$

for all $n=1,2,3,\ldots,$ so that

$$\phi_1(x) = \phi\Big(x - \frac{1}{2}\Big)$$
$$\phi_2(x) = \phi\Big(x - \frac{1}{4}\Big) + \phi\Big(x - \frac{3}{4}\Big)$$
$$\phi_3(x) = \phi\Big(x - \frac{1}{8}\Big) + \phi\Big(x - \frac{3}{8}\Big) + \phi\Big(x - \frac{5}{8}\Big) + \phi\Big(x - \frac{7}{8}\Big)$$

and so on...

Then set

$$\Phi(x) = \sum_{n=1}^{\infty}\frac{1}{2^n}\phi_n(x)$$

What kind of differentiability properties will $\Phi$ have? What happens, when you integrate its derivative?

17. Jun 14, 2013

### micromass

Staff Emeritus
We have the following theorem:

If $f,g:[a,b]\rightarrow \mathbb{R}$ are absolutely continuous and if $f^\prime = g^\prime$ a.e., then $f-g$ is constant on $[a,b]$.

This is theorem 20.16 in "Real Analysis" by Carothers. Does that help?

18. Jun 14, 2013

### Mandelbroth

$$\frac{d}{dx}\Phi(x) = \sum_{n=1}^{\infty}\frac{1}{2^n}\frac{d}{dx}\phi_n(x).$$

Then,

$$\int\frac{d}{dx}\Phi(x)\, dx = \int\sum_{n=1}^{\infty}\frac{1}{2^n}\frac{d}{dx}\phi_n(x)\, dx = \sum_{n=1}^{\infty}\frac{1}{2^n}\phi_n(x) + C$$

In fact,

$$\int\limits_{[0,x]}\frac{d}{dx}\Phi(x)\, dx = \int\limits_{[0,x]}\sum_{n=1}^{\infty}\frac{1}{2^n}\frac{d}{dx}\phi_n(x)\, dx = \sum_{n=1}^{\infty}\frac{1}{2^n}\phi_n(x)$$

By Stokes' Theorem, we have that $\displaystyle \int\limits_{[a,b]}\, df = \int\limits_{\partial[a,b]}f = f(b)-f(a)$. I don't see why it has to be Riemann integrable.

19. Jun 14, 2013

### jostpuur

This leads to the following question: If $f:[a,b]\to\mathbb{R}$ is differentiable at all points in $[a,b]$ and $\int\limits_{[a,b]}|f'(x)|dm_1(x)<\infty$, will $f$ be absolutely continuous?

20. Jun 14, 2013

### micromass

Staff Emeritus
The following are equivalent:
(1) $f$ is absolutely continuous
(2) $f^\prime$ exists a.e., $f^\prime \in L^1[a,b]$ and $f(x) - f(a) = \int_a^x f^\prime$.

So I'd say no. I'll try to look for a counterexample if you want.