I want to prove that if we have a sequence of Riemann integrable functions $f_n:[a,b] \rightarrow \mathbb{R}$ that uniformly converge to $f$ then $f$ is also Riemann integrable.
My proof goes (roughly) as follows:
Let $\varepsilon > 0$, there must exist a natural number $N$ such that $|f(x) - f_N(x)| < \varepsilon$ for all $x \in [a,b]$ (by uniform convergence) and there must also be a P of $[a,b]$ such that $U(f_N,P) - L(f_N,P) < \varepsilon$ (by Riemann integrability of $f_N$).
($U(f,P)$ is defined as $\sum (x_{i} - x_{i-1}) \times \sup_{[x_{i-1},x_i]}f$, where the $x_i$'s are the partition points of $P$, and similarly for $L(f,P)$.)
I claim that $U(f,P) - L(f,P) < (b-a) \times \varepsilon \times (\frac{1}{\alpha} + 2)$ (this would imply that $f$ is Riemann integrable).
My reason is that within each interval of $[x_{i-1},x_i]$ of $P$:
$\sup f_N - \inf f_N < \frac{\varepsilon}{\alpha}$ (define $\alpha$ as the length of the smallest interval in the partition), because $U(f_N,P) - L(f_N,P) < \varepsilon$.
$ \sup f $ is at most $\varepsilon$ greater than $\sup f_N$ and $ \inf f $ is at most $\varepsilon$ less than $\inf f_N$ (this is because $|f(x) - f_N(x)| < \varepsilon$)
Therefore $\sup f - \inf f < \frac{\varepsilon}{\alpha} + 2\varepsilon$.
It follows that $U(f,P) - L(f,P) < (b-a) \times \varepsilon \times (\frac{1}{\alpha} + 2)$.
Is my proof correct?