This is an exercise in section 6.3 of Royden & Fitzpatrick's Real Analysis.
Consider the function
$$f(x)= \begin{cases} x^a\sin\left(\frac{1}{x^b}\right) & \text{if }0 < x \leq 1 \\ 0 & \text{if }x=0. \end{cases}$$
show that if $a>b$, $f$ is of bounded variation on $[0,1]$.
MY ATTEMPT:
A theorem of the chapter says
$f$ of bounded variation $\Leftrightarrow f$ is differentiable a.e. on $(a,b)$ and $f'$ is integrable over $[a,b]$.
So I want to somehow show that
$$f'(x)= \begin{cases} -bx^{a-1-b}\cos(x^{-b})+ax^{a-1}\sin(x^{-b}) & \text{if }0 < x \leq 1 \\ 0 &\text{if }x=0.\end{cases}$$
is integrable. Now, by definition, $f$ is integrable if $|f|$ is integrable, and since $|f|$ is nonnegative, it is integrable if its integral is bounded. If this is the correct approach to this, then since the lebesgue integral is equal to the riemann integral on closed bounded intervals, then all I have to do is show the riemann integral of $f'$ is bounded. But the riemann integral of this thing is way too complicated (mathematica gives me a mess). Is this supposed to be this cumbersome?
Any help is much appreciated.
First, the correct assumption should be $a>b>0$ (check if you overlooked something in the statement of the exercise).
Your basic approach is correct. But to show that the integral of $|f'|$ is bounded you don't need to calculate it explicitly. It is enough to give an upper bound. Note that $|\cos(x^{-b})|\le 1$ and $|\sin(x^{-b})|\le 1$. Using that and the triangle inequality,
$$\int_0^1 |f'(x)| dx \le b \int_0^1 x^{a-b-1} dx + a \int_0^1 x^{a-1} dx = \frac{b}{b-a}+1<\infty$$
Note that in the last step we have used that $a-b-1>-1$ (i.e. that $a>b$) and $a-1>-1$ (that $a>0$), because that is the condition for a power $x^\alpha$ to be integrable at $0$.
To be a bit pedantic, since you mentioned the Riemann integral, $|f'|$ is in general not Riemann integrable on $[0,1]$, because it may be unbounded (if $a-b-1<0$ or $a-1<0$). If you want to keep the perspective of Riemann integrals, you need to look at them as improper Riemann integrals.