I'm trying to show that assuming $f:[a,b] \rightarrow \mathbb{R}$ is nonnegative, bounded function that is Riemann integrable on [a,b], then $\sqrt{f}$ is Riemann integrable on [a,b].
I know we have to construct an $\epsilon$ proof using the fact that f is Riemann integrable. We know that $\exists P$, a partition, such that $U(f,p) - L(f,p) < \epsilon$.
And we want to show that $| \sqrt{f(x)} - \sqrt{f(y)}|$ is somehow less than $\sum (\sup_{[x_i-1, x_i]} f - \inf_{[x_i-1, x_i]} f) \Delta x_i$, which in turn will be less than $\epsilon$. I can't seem to figure this bit out, so a hint or help would be much appreciated! Here's what I've tried (but I'm not sure if it's correct):
$$| \sqrt{f(x)} - \sqrt{f(y)}| \leq |f(x) - f(y)| \leq \sup_{[x_{i-1}, x_i]} f - \inf_{[x_{i-1}, x_i]} f \text{ } (\forall 1 \leq i \leq N) $$
And so $\exists$ partition P such that:
$$U(\sqrt{f}, p) - L (\sqrt{f}, p) \leq U(f, p) - L (f,p) < \epsilon. $$