If $(f_n)_{n=1}^\infty$ is a sequence of Riemann integrable functions defined on a compact interval $I$ which uniformly converges with limit $f$, then $f$ is Riemann integrable and:$$\int_If=\lim_{n\to\infty}\int_If_n$$
Uniform convergence of the sequence means that for any $\varepsilon\gt0$, $\exists N:\forall n\gt N,\forall x,|f_n(x)-f(x)|\lt\varepsilon$.
$$0\le\left|\int_If_n-f\right|\lt\int_I|f_n-f|\lt\varepsilon L$$
Where $n$ is such that $|f_n-f|\lt\varepsilon$ (such $n$ must exist by assumption) and $L$ is the range of the compact interval, which is finite. As for any $\varepsilon\gt0$, $n$ exists (but increases as $\varepsilon$ decreases), we know $n\to\infty\implies\varepsilon\to0$. Therefore:
$$0\le\lim_{n\to\infty}\left|\int_If_n-f\right|\lt\lim_{n\to\infty}\int_I|f_n-f|\lt\lim_{n\to\infty}\varepsilon L=0\\\therefore0\le\lim_{n\to\infty}\left|\int_If_n-f\right|\lt0\\\therefore\lim_{n\to\infty}\left|\int_If_n-f\right|=0\implies\lim_{n\to\infty}\int_If_n=f$$
I strongly suspect that my proof is incorrect, since I never used the fact $I$ was compact, only that it was finite, and I also have never studied real analysis to any great degree so I'm probably lacking some rigour here. Can anyone verify or make corrections to my proof?
P.S I know (EDIT: I thought I knew!) $I$ being compact and $f_n$ being integrable over $I$ implies $f_n$ is uniformly continuous for all $n$, but didn't know how to use that fact.
