When does interchangibility of limit and Riemann integral imply uniform convergence? Let  $\{f_n\}$ be a sequence of real-valued functions defined on an interval $[a,b]$ such that each $f_n$ is Riemann integrable, $\{f_n\}$ converges point-wise to $f$, $f$ is Riemann integrable and $\lim_{n \to \infty} \int_a^b f_n$ exists and equals $\int_a^b f$; then under what additional condition(s) can we conclude that $\{f_n\} $ converges uniformly to $f$? 
 A: If $g_n = f_n - f$ is continuous on $[a, b]$ then $f_n$ converges uniformly to $f$ on $[a, b]$ if and only if
$$\lim_{n \to \infty} \sup_{a\leq a' < b' \leq b}\frac{1}{b' - a'}\int_{b'}^{a'}\left| f_n(x) - f(x) \right|dx = 0$$
(Note: last one includes $\lim_{n \to \infty}\int_{b}^{a}f_n(x)dx = \int_{b}^{a}f(x)dx$)
In order to prove first part let's assume $f_n$ converges uniformly on $[a, b]$. Then given $\epsilon > 0$ exists $n' \in \mathbb{N}$ such that
$$\sup_{x \in [a, b]}\left| g_n(x) \right|< \epsilon $$
when $n > n'$.
Which implies 
$$\int_{b'}^{a'}\left| g_n(x)\right|dx < \int_{b'}^{a'}\epsilon dx = (b' - a')\epsilon $$
when $a \leq a' < b' \leq b$.
Now 
$$ \sup_{a\leq a' < b' \leq b}\frac{1}{b' - a'}\int_{b'}^{a'}\left| g_n(x) \right|dx < \frac{b' - a'}{b' - a'} \epsilon = \epsilon$$
Because $\epsilon$ was arbitrary we have shown 
$$\lim_{n \to \infty} \sup_{a\leq a' < b' \leq b}\frac{1}{b' - a'}\int_{b'}^{a'}\left| f_n(x) - f(x) \right|dx = 0$$
To prove second part let's assume 
$\lim_{n \to \infty} \sup_{a\leq a' < b' \leq b}\frac{1}{b' - a'}\int_{b'}^{a'}\left| f_n(x) - f(x) \right|dx = 0$.
Now given $\epsilon > 0$ exists $n' \in \mathbb{N}$ so that when $n > n'$
$$ \sup_{a\leq a' < b' \leq b}\frac{1}{b' - a'}\int_{b'}^{a'}\left| g_n(x) \right|dx < \frac{1}{2}\epsilon$$
Because $g_n = f_n - f$ is continuous for any $x \in [a, b]$ exists $[a', b'] \subset [a, b]$ such that  $\left| g_n(t) - g_n(x) \right| < \frac{1}{2}\epsilon$ and  $x \in [a', b']$ when $t \in [a', b']$.
Using mean value theorem we get
$$g_n(x) - \frac{1}{2}\epsilon <\frac{1}{b' - a'} \int_{a'}^{b'}g_n(t) dt < g_n(x) + \frac{1}{2}\epsilon$$
$$\frac{1}{b' - a'} \int_{a'}^{b'}g_n(t) dt  - \frac{1}{2}\epsilon < g_n(x)<\frac{1}{b' - a'} \int_{a'}^{b'}g_n(t) dt + \frac{1}{2}\epsilon$$
$$- \frac{1}{2}\epsilon  - \frac{1}{2}\epsilon < g_n(x)< \frac{1}{2}\epsilon + \frac{1}{2}\epsilon$$
$$-\epsilon < g_n(x)< \epsilon$$
Because this is for any $\epsilon > 0$ or $x \in [a, b]$ we get $\sup_{x \in [a, b]}\left| g_n(x) \right|< \epsilon $ and $\lim_{n \to \infty} \sup_{x \in [a, b]}\left| g_n(x) \right| = 0$. Thus $f_n$ converges uniformly to $f$ on $[a, b]$.
