Does the Integral of a Derivative Equal the Function's Net Change?

[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

<br /> \int\limits_{[a,b]}f&#039;(x)dm_1(x) = f(b)-f(a)\quad\quad\quad\quad (1)<br />

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.

 

[micromass]

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 $a=0$.

 

[jostpuur]

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

 

[micromass]

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!

 

[Mandelbroth]

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 $a=0$.

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

 

[jostpuur]

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

<br /> f(x) = \left\{\begin{array}{ll}<br /> 0,\quad &amp;x=0\\<br /> x^2\sin\Big(\frac{1}{x^2}\Big),\quad &amp;x\neq 0\\<br /> \end{array}\right.<br />

Now the function is differentiable everywhere, and the derivative is

<br /> f&#039;(x) = \left\{\begin{array}{ll}<br /> 0,\quad &amp;x=0\\<br /> 2x\sin\Big(\frac{1}{x^2}\Big)-\frac{2}{x}\cos\Big(\frac{1}{x^2}\Big),\quad &amp;x\neq 0\\<br /> \end{array}\right.<br />

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&lt;0&lt;b.

However, the limits

<br /> \lim_{\delta\to 0^-}\int\limits_{a}^{\delta} f&#039;(x)dx = f(0)-f(a)<br />

and

<br /> \lim_{\delta\to 0^+}\int\limits_{\delta}^{b}f&#039;(x)dx = f(b)-f(0)<br />

still hold. Suppose we define the integral over [a,b] (with a&lt;0&lt;b) by

<br /> \int\limits_{a}^{b}f&#039;(x)dx \;:=\; \lim_{\delta\to 0^-}\int\limits_{a}^{\delta}f&#039;(x)dx \;+\; \lim_{\delta\to 0^+}\int\limits_{\delta}^{b}f&#039;(x)dx<br />

Now the formula

<br /> \int\limits_{a}^{b}f&#039;(x)dx = f(b)-f(a)<br />

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...

 

[micromass]

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

 

[micromass]

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

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.

 

[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?

 

[Mandelbroth]

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.

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.

 

[micromass]

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?

I don't know, but this seems pretty relevant: http://math.stackexchange.com/questions/112067/how-discontinuous-can-a-derivative-be

 

[micromass]

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 $\int_0^1 f = 0$.

 

[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

<br /> \int\limits_{[a,b]}|f&#039;(x)|dm_1(x) &lt; \infty<br />

holds. Is it now possible that

<br /> f(x) = f(a) + \int\limits_{[a,x]}f&#039;(u)dm_1(u)<br />

would not hold for some x?

I know that if I define a new function

<br /> g(x) = f(a) + \int\limits_{[a,x]}f&#039;(u)dm_1(u)<br />

then g will be absolutely continuous, and it will have a derivative (at least) almost everywhere. Also g&#039;=f&#039; will hold almost everywhere, according to some theorems. But I couldn't find theorems that would imply g=f.

 

[Mandelbroth]

I know that if I define a new function

<br /> g(x) = f(a) + \int\limits_{[a,x]}f&#039;(u)dm_1(u)<br />

then g will be absolutely continuous, and it will have a derivative (at least) almost everywhere. Also g&#039;=f&#039; will hold almost everywhere, according to some theorems. But I couldn't find theorems that would imply g=f.

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...

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

 

[jostpuur]

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...

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

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

For example, if f&#039; is Riemann-integrable, then we can evaluate like that. If we only assume f&#039; 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.

 

[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}

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

Now \phi is differentiable everywhere, and the derivative is

<br /> \phi&#039;(x) = \left\{\begin{array}{ll}<br /> 0,\quad\quad &amp;x=0\\<br /> \frac{3}{2}\sqrt{|x|}\sin\Big(\frac{1}{|x|}\Big) - \frac{1}{\sqrt{|x|}}\cos\Big(\frac{1}{x}\Big),\quad\quad &amp;x\neq 0\\<br /> \end{array}\right.<br />

Now \phi&#039; is not Riemann-integrable over intervals that contain the origo, but still

<br /> \int\limits_{[-1,1]}|\phi&#039;(x)|dm_1(x) &lt; \infty<br />

and

<br /> \int\limits_{[a,b]}\phi&#039;(x)dm_1(x) = \phi(b)-\phi(a)<br />

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

Next, let's define

<br /> \phi_n(x) = \sum_{j=1}^{2^{n-1}} \phi\Big(x - \frac{2j-1}{2^n}\Big)<br />

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

<br /> \phi_1(x) = \phi\Big(x - \frac{1}{2}\Big)<br />
<br /> \phi_2(x) = \phi\Big(x - \frac{1}{4}\Big) + \phi\Big(x - \frac{3}{4}\Big)<br />
<br /> \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)<br />

and so on...

Then set

<br /> \Phi(x) = \sum_{n=1}^{\infty}\frac{1}{2^n}\phi_n(x)<br />

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

 

[micromass]

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?

 

[Mandelbroth]

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

[...]

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

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

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.

 

[jostpuur]

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?

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&#039;(x)|dm_1(x)&lt;\infty, will f be absolutely continuous?

 

[micromass]

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&#039;(x)|dm_1(x)&lt;\infty, will f be absolutely continuous?

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.

 

[jostpuur]

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 am aware of this result. I don't see how it would hint towards negative answer. But different people have different intuitions...

I'll try to look for a counterexample if you want.

I you look my example, you'll see that the actual formula \phi(x)=|x|^{3/2}\sin(\frac{1}{x}) was not really relevant. If \int\limits_{[a,b]}|\phi&#039;(x)|dm_1(x)&lt;\infty holds, then also

<br /> \lim_{\delta\to 0^+} \int\limits_{[a,b]\backslash [x_0-\delta,x_0+\delta]}\phi&#039;(x)dm_1(x) = \int\limits_{[a,b]}\phi&#039;(x)dm_1(x)<br />

will hold for any x_0\in [a,b]. This means that in the counter example the derivative must have an infinite amount of discontinuities. If the points of discontinuities are finite, the integral can be reduced to Riemann-integrals and limits (concerning the integration intervals).

 

[jostpuur]

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

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

Am I correct to assume that you did not really prove this commutation result?

If you want to justify

<br /> D_x \sum_{n=1}^{\infty} f_n(x) = \sum_{n=1}^{\infty} D_x f_n(x)<br />

the most obvious way is to assume that the derivatives f&#039;_n are continuous and that \sum_{n=1}^{\infty}\|f&#039;_n\|_{\textrm{sup}}&lt;\infty. Then you can use the mean value theorem and dominated convergence.

You must prove that

<br /> \lim_{\Delta x\to 0}\sum_{n=1}^{\infty} \frac{f_n(x+\Delta x)-f_n(x)}{\Delta x}<br /> = \sum_{n=1}^{\infty} \lim_{\Delta x \to 0} \frac{f_n(x+\Delta x)-f_n(x)}{\Delta x}<br />

There exists \xi_{n,x,\Delta x} such that

<br /> \frac{f_n(x+\Delta x)-f_n(x)}{\Delta x} = f&#039;_n(\xi_{n,x,\Delta x})<br />

and therefore

<br /> \Big|\frac{f_n(x+\Delta x)-f_n(x)}{\Delta x}\Big| \leq \|f&#039;_n\|_{\textrm{sup}}<br />

for all x and \Delta x &gt; 0.

Now the g(n):=\|f&#039;_n\|_{\textrm{sup}} is the "dominating function" that justifies the commutation of the derivative operator and the sum.

In my example the \phi_n don't have continuous derivatives, and \|\phi&#039;_n\|_{\textrm{sup}}=\infty holds for all n, so the differentiability of \Phi is a mystery at a first glance.

 

[Mandelbroth]

Am I correct to assume that you did not really prove this commutation result?

If you want to justify

<br /> D_x \sum_{n=1}^{\infty} f_n(x) = \sum_{n=1}^{\infty} D_x f_n(x)<br />

the most obvious way is to assume that the derivatives f&#039;_n are continuous and that \sum_{n=1}^{\infty}\|f&#039;_n\|_{\textrm{sup}}&lt;\infty. Then you can use the mean value theorem and dominated convergence.

Uh...no. I can justify it by saying the differential operator is linear and thus satisfies $D_x \left[f(x)+g(x)\right] = D_x[f(x)]+D_x[g(x)]$ and $D_x [\alpha f(x)] = \alpha D_x[f(x)]$.

I don't understand why we need to take these long, winding methods to reprove that the differential operator is linear.

 

[micromass]

Uh...no. I can justify it by saying the differential operator is linear and thus satisfies $D_x \left[f(x)+g(x)\right] = D_x[f(x)]+D_x[g(x)]$ and $D_x [\alpha f(x)] = \alpha D_x[f(x)]$.

I don't understand why we need to take these long, winding methods to reprove that the differential operator is linear.

Those are finite sums. You can't do the same for an infinite sum.

 

[jostpuur]

Uh...no. I can justify it by saying the differential operator is linear and thus satisfies $D_x \left[f(x)+g(x)\right] = D_x[f(x)]+D_x[g(x)]$ and $D_x [\alpha f(x)] = \alpha D_x[f(x)]$.

I don't understand why we need to take these long, winding methods to reprove that the differential operator is linear.

The reason is the infinite sum.

<br /> D_x\sum_{n=1}^N f_n(x) = \sum_{n=1}^N D_x f_n(x)<br />

is clear, but

<br /> D_x \lim_{N\to\infty} \sum_{n=1}^N f_n(x) = \lim_{N\to\infty}\sum_{n=1}^N D_x f_n(x)<br />

is not. You cannot insist that D_x and \lim_{N\to\infty} would commute merely due to linearity of D_x.

 

[Mandelbroth]

Those are finite sums. You can't do the same for an infinite sum.

Derp. Sorry. Don't listen to me. I'm going to go sit in the corner of shame.

 

[jostpuur]

This means that in the counter example the derivative must have an infinite amount of discontinuities. If the points of discontinuities are finite, the integral can be reduced to Riemann-integrals and limits (concerning the integration intervals).

This result seems to be easy:

Assume that f:[a,b]\to\mathbb{R} is differentiable at all points in [a,b], that \int\limits_{[a,b]}|f&#039;(x)|dm(x)&lt;\infty holds, and denote D\subset [a,b] the set of points where f&#039; is not continuous. If the amount of accumulation points of D is finite, \int\limits_{[a,b]}f&#039;(x)dm(x)=f(b)-f(a) will hold.

So D does not need to be finite, but a finite amount of accumulation points implies the result too.

The proof goes so that we first assume for a lemma that D is finite. We know that if there are no discontinuities of derivative on some interval [e_i,d_i] then \int\limits_{[e_i,d_i]}f&#039;(x)dm(x)=f(d_i)-f(e_i) will hold there. So with fixed \delta &gt; 0 we define sequences d_1,\ldots, d_N and e_1,\ldots, e_N so that d_i=x_i-\delta and e_i=x_i+\delta where x_i are the points of D.

<br /> \int\limits_{[a,b]}f&#039;(x)dm(x) = \lim_{\delta\to 0^+}\Big(\int\limits_{[a,d_1]}f&#039;(x)dm(x) + \int\limits_{[e_1,d_2]}f&#039;(x)dm(x) + \cdots + \int\limits_{[e_N,b]}f&#039;(x)dm(x)\Big) = f(b) - f(a)<br />

Next, assume that D is infinite, but has only finite amount of accumulation points. According to the previous result, if some interval [e_i,d_i] has only finite amount of discontinuities of the derivative, then \int\limits_{[e_i,d_i]}f&#039;(x)dm(x)=f(d_i)-f(e_i) will hold. So we can repeat the previous argument by denoting the accumulation points as x_1,\ldots, x_N.

Then the next question is that what if there are infinitely many accumulation points.

 

[jostpuur]

A new question!

If f:[a,b]\to\mathbb{R} is differentiable at all points in [a,b], will

<br /> |f(b)-f(a)|\leq \int\limits_{[a,b]}|f&#039;(x)|dm(x)<br />

always be true?

This looks simple, but I think this question is very critical. If the answer is positive, I believe I know how to settle the previous question about absolute continuity.

 

[jostpuur]

I proved my previous claim now, and the whole thing is settled (assuming I haven't made mistakes). I'll repeat the claim now, since there has been so much distraction above.

Assume that f:[a,b]\to\mathbb{R} is differentiable at all points in [a,b], and also that \int\limits_{[a,b]}|f&#039;(x)|dm(x)&lt;\infty. Then \int\limits_{[a,b]}f&#039;(x)dm(x) = f(b) - f(a).

I have a proof!

micromass, how's the counter example coming?

 

[jostpuur]

Intermediate result 1: Assume that a function \psi:[a,b]\to\mathbb{R} is differentiable at all points in [a,b]. Then |\psi(b)-\psi(a)|\leq\int\limits_{[a,b]}|\psi&#039;(x)|dm(x).

Intermediate result 2: Assume that a function \varphi:[a,b]\to [0,\infty[ is Lebesgue-measurable and \int\limits_{[a,b]}\varphi(x)dm(x)&lt;\infty. Then, for all \epsilon &gt;0 there exists \delta &gt; 0 such that
<br /> m(A)&lt;\delta\quad\implies\quad \int\limits_{A}\varphi(x)dm(x)&lt;\epsilon<br />
for all measurable A\subset [a,b].

Intermediate result 3: Assume that f:[a,b]\to\mathbb{R} is absolutely continuous. Then the derivative f&#039; exists almost everywhere, and \int\limits_{[a,b]}f&#039;(x)dm(x)=f(b)-f(a).

The proof of the claim (in post #29) using the intermediate results 1, 2 and 3:

If we use the assumptions concerning f, and prove that f is absolutely continuous, the main result becomes proven based on the intermediate result 3. So we seek to prove the absolute continuity.

Let \epsilon &gt; 0 be abitrary. According to the intermediate result 2, there exists a \delta &gt; 0 such that
<br /> m(A)&lt;\delta\quad\implies\quad \int\limits_{A}|f&#039;(x)|dm(x) &lt; \epsilon<br />
for all measurable A\subset [a,b]. So for all collections of intervals [a_1,b_1],\ldots, [a_N,b_N] in [a,b] such that \sum_{n=1}^N(b_n-a_n) &lt; \delta a result
<br /> \sum_{n=1}^N\int\limits_{[a_n,b_n]}|f&#039;(x)|dm(x) &lt; \epsilon<br />
holds. According to the intermediate result 1, an inequality
<br /> |f(b_n)-f(a_n)|\leq \int\limits_{[a_n,b_n]}|f&#039;(x)|dm(x)<br />
holds on each of these intervals, so we get
<br /> \sum_{n=1}^N |f(b_n)-f(a_n)| &lt;\epsilon<br />
proving that f is absolutely continuous.

Now, there are three possibilities. 1: There was mistake in the above proof. 2: There will turn out to be mistakes in the proofs of intermediate results. 3: There will be no mistakes anywhere.

Out of the intermediate results the first one is such that I don't have a reference for it, so it is the most suspicious. The second and third one are supposed to have references, assuming I haven't accidentaly changed something in them.