I am currently learning about the analysis basics of Riemann Integration and integration as a whole. For one of my practice exercises, I was able to prove that if f is uniformly continuous over $[a,b]$, we can in fact show that the the difference between the Darboux Upper Sum and the Darboux Lower Sum is less than or equal to $\epsilon (b-a)$.
Knowing that $U(f,P) - L(f,P) \leq \epsilon(b-a)$, how can I go about using that to prove that the limit of the difference between the Darboux Upper Sum and Darbou Lower sum, as the mesh(P) $\to 0$, in fact equals $0$.
That is, how can I conclude lim $(U(f,P) - L(f,P))$ as mesh(P) $\to 0$ actually equals $0$?
Thank you!
hint
You want to prove that
$$\forall \epsilon>0\;\exists \delta>0\;:$$ $$\;||P||<\delta \implies U(f,P)-L(f,P)<\epsilon$$
You can take $ \delta $ as the $ \eta$ of the uniform continuity :
$$\forall \epsilon\;\exists \eta>0\;:$$ $$\;|x-y|<\eta\implies|f(x)-f(y)|<\frac{\epsilon}{b-a}$$