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Given $\{f_n\}\rightarrow f$ uniformly on the compact $I$ and each $f_n$ is continuous, prove $\lim_{n\rightarrow\infty}\int_I|f_n - f|^2 = 0$
My proof attempt
By hypothesis given $\sqrt{\epsilon} > 0 $ there exists $N(\epsilon) \in \mathbb{N}$ s.t. for all $n > N(\epsilon)$ we have for all $x\in I$:
$|f_n - f| < \sqrt{\epsilon}$
$|f_n - f|^2 < \epsilon$
At this point I don't know how to arrive to the integral equation
From $f_n \rightarrow f$ uniformly we have $\sup{|f_n-f|}\rightarrow 0,$ in $\mathbb{R}^n$ compact set $I$ is bounded, so $\int_I|f_n-f|^2\le\sup{|f_n-f|}^2|I|$ where $|I|$ is the Lebesgue measure and $|I|<\infty$
Since $I$ is a compact interval, it has finite measure $\lambda(I)$.
Then, for each $\epsilon>0$ there exists some $N$ such that $n>N$ implies $$\|f_n - f\| < \sqrt{\frac{\epsilon}{\lambda(I)}} \Rightarrow \\ \|f_n - f\|^2 < \frac{\epsilon}{\lambda(I)} \Rightarrow \\ |f_n(x) - f(x)|^2 < \frac{\epsilon}{\lambda(I)} \forall x \in I\Rightarrow \\ \int_I |f_n(x) - f(x)|^2 dx < \int_I \frac{\epsilon}{\lambda(I)} dx =\epsilon$$
