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My construction of Volterra's function is as follows.

Let $F(x)=\begin{cases} x^2\sin\left(\frac{1}{x}\right) &\text{ if } x \neq 0\\ 0 &\text { if } x =0 \end{cases}$

On the interval $\left[0,\frac{1}{8}\right]$ we find the most extreme value of $F'(x)=0$. Call that value $x_0$. Then we we get $F(x)$ for all $x \in [0, x_0]$ and $F(x_0)$ for all $x \in[x_0, \frac{1}{8}]$. Then we mirror the function from $\left[\frac{1}{8}, \frac{1}{4}\right]$. Outside of $\frac{1}{4}$, we say the function is 0. We call this function $V_1(x)$. Then we translate the graph of $V_1(x)$ to the first deleted interval of the Smith-Cantor-Volterra set $\left[\frac{3}{8},\frac{5}{8}\right]$.

We repeat this for each interval $\left[0,\frac{1}{2\cdot4^n}\right]$, reflect from $\left[\frac{1}{2\cdot4^n},\frac{1}{4^n}\right]$. Call it $V_n(x)$. Then we translate to the next removed pieces of the Smith-Cantor-Volterra set.

I see that the function $V(x)$ is differentiable everywhere and that $V'(x)$ is bounded, but I'm not sure why its not Riemann integrable (without using Lebesgue's criteria).

emka
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3 Answers3

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The basic idea is that the Volterra function $V(x) $ is such that $V'(x) =0$ for all $x$ which lie in the Smith-Volterra-Cantor set. Moreover at these points the oscillation of $V'$ is $2$ (because oscillation of $F'$ at $0$ is $2$). The Smith-Volterra-Cantor set has Jordan outer content equal to $1/2$ and it is precisely because of this reason that $V'$ is not Riemann integrable. Remember the criteria of Riemann integrability in terms of Jordan content:

Theorem: Let $f$ be bounded on $[a, b] $. Let $S_{\sigma} $ denote the set of points in $[a, b] $ at which oscillation of $f$ is greater or equal to $\sigma$. The function $f$ is Riemann integrable on $[a, b] $ if and only if for each $\sigma>0$ the Jordan outer content of $S_{\sigma} $ is $0$.

The above theorem is precursor to Lebesgue's criterion of Riemann integrability which deals with sets of measure zero. The above theorem deals with sets of outer content $0$.

If you don't want to use the concept of content which is somewhat similar to measure then you can go back to the criterion for Riemann integrability given by Riemann himself:

Riemann's Criterion of Riemann Integrability: Let $f$ be a bounded function on $[a, b] $. Then $f$ is Riemann integrable on $[a, b] $ if and only if for any $\sigma>0, \nu>0$ we can find a number $\delta>0$ such that for any partition of $[a, b] $ with subintervals of length less than $\delta$, the subintervals on which the oscillation of $f$ is at least $\sigma$ have a combined total length less than $\nu$.

Paramanand Singh
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Your question made sense until the phrase "without using Lebesgue's criterion." The set of points where $V'(x)$ is discontinuous has positive measure. Just calculate $F'(x)$ near $0,$ it is discontinuous at $0.$

Will Jagy
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    I don't have the machinery of Lebesgue measure in this course. – emka Apr 21 '13 at 07:05
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    You do not need to know measure theory to understand the Lebesgue criterion regarding Riemann integrablity. – AXH Feb 27 '15 at 04:53
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A function is Riemann integrable if and only if the oscillation $w$ goes to $0$ with $|P|$ goes to $0$. I do not think this holds for $x^{2}\sin[1/x]$. Because $\sin[1/x]$ can take values in $1,-1$ while $x$'s value change relatively little, the oscillation will not go to $0$ if you choose a strange enough partititon set near $0$.

Bombyx mori
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  • What do you mean a strange enough partition? Would a partition such as $[-\eta, \eta]$ suffice? For $\eta>0$ – emka Apr 21 '13 at 07:04
  • I mean you need $|P|\rightarrow 0$ at every stage. So try to construct a sequence of partitions $P_{n}$ such that $P_{n}$ further partitions $P_{n-1}$, and $|P_{n}|\rightarrow 0$. The choice of partition points should be smart enough such that $w>\epsilon$ for some fixed $\epsilon$. Using some bounds this should be not so difficult. – Bombyx mori Apr 21 '13 at 07:10
  • Ahh...I see. So let's say I fix $e=1$. On some arbitrary interval $[x_{i-1},x_i]$, I know that $|sup f(x)-inf(x)|<1$...but the biggest value this function could take is 1 and the smallest is -1...so I get 2 < 1. This was my original idea using the idea of oscillation. Would this be the wrong idea? – emka Apr 21 '13 at 08:16
  • You try to bound $x^{2}\sin[1/x]$'s oscillation from below by taking it in an interval $[P^{n}{i},P^{n}{i+1}]$, where $\sin[1/P^{n}_{i}]=1,\sin[1/P^{n}_{i+1}]=-1$. Then you have $w(P^{n})$ on this interval bounded by $2(P^{n}{i})^{2}$ from below. Choose $P^{n}{i}$ wisely such that $|P^{n}|\rightarrow 0$ and $\sin[1/P^{n}_{i}]=(-1)^{i}$, for example. Then the total variation would be bounded by $2\sum (P^{n}_{i})^{2}\ge \epsilon$ for certain $\epsilon$. Now if you require at each stage $|P^{n}|<\frac{1}{n}$, then you can have a rough bound like $\frac{C\pi^{2}}{6}$ where $C$ is constant. – Bombyx mori Apr 21 '13 at 08:33
  • You also do not need to construct $P^{n}{i}$ explicitly in every stage. You can do it inductively. Suppose at stage $n$ you have already found the points, then just try to pick enough points at stage $n+1$ such that the total $|P{n+1}|$ satisfies the desired relationship. This should not be difficult and the proof is completed. – Bombyx mori Apr 21 '13 at 08:43
  • Actually...I do see an issue? Did you take the derivative of $F(x)$? This may be why I'm not understanding what's going on. – emka Apr 21 '13 at 09:09
  • No, there is no derivative at here. All we use is the definition of Riemann integral. – Bombyx mori Apr 21 '13 at 09:11
  • I should say I made a mistake in calculating the "rough bound". But this should not change the essential part of the proof. – Bombyx mori Apr 21 '13 at 09:18
  • What I means is that I'm trying to show that you cannot integrate $F'(x)$. I understand most of your hint, except I still don't understand how you found an explicit bound. It just seems like at every interval the oscillation will be 2. Do you mind fleshing out the details? – emka Apr 21 '13 at 09:23
  • Note that $F$ is integrable. Volterra's function uses countably many copies of $F$. – pancini Mar 15 '17 at 18:51