TITLE: Does $f^3$ integrable imply $f$ integrable?
QUESTION [14 upvotes]: Let $f$ be a function defined on the close interval $[a,b]$. Does the riemann stieltjes integrability of $f^3$ imply the riemann stieltjes integrability of $f$ ? 
The answer is trivially no in the case of $f^2$, but I am not able to find a counterexample for the case $f^3$.

REPLY [16 votes]: You could try proving the following theorem : If $f:[a,b]\to \mathbb{R}$ is Riemann-integrable, and $\varphi : [c,d]\to \mathbb{R}$ is continuous where $f([a,b])\subset [c,d]$, then $\varphi\circ f$ is Riemann integrable.
You need to begin with the Riemann condition for integrability (difference between Upper and Lower sum is small), and use the uniform continuity of $\varphi$. It takes a little more work than that, but that is the essence of the argument.
Now just take $\varphi(x) :=x^{1/3}$, to see that your statement is true.