Why isn't Volterra's function Riemann integrable? 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).  
 A: 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 $\nu$. 

A: 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 $0.$
A: 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 $0$. 
