Is there a way of proving that $f(x)=x^n$ where $n \in \mathbb{N}$ is Riemann integrable on an arbitrary interval [a,b] without using the fact that f is continuous?
I have settled for the following cases:
$n$ is odd, so $f$ is increasing on every interval and the proof is relatively straightforward (I followed a general textbook proof for Riemann integrability of monotonic functions)
$n$ is even. I reasoned that this would give us $3$ cases: one where $0 \leq a < b$, thus $f$ is increasing and the proof is already done in case $1$; another where $a < b \leq 0$, which means f is decreasing (I modified the proof for case $1$), and lastly one where $a$ is negative and $b$ is positive, and this is the one that gives me trouble.
If $a<0$ and $b>0$, I thought of splitting the interval in two to get one where f is decreasing ($[a,0]$) and one where $f$ is increasing ($[0,b]$).
Let's say I want to formulate the upper and lower sums with m subintervals total on the interval $[a,b]$, so each of them has a length of $\frac{(b-a)}{m}$. Let $x_k$ where $k=-m,...,m$, be the end point of each subinterval.
And that's where I'm stuck. How can I formulate the Riemann sums so that I could use the Riemann criterion to prove integrability?