If $f$ is Riemann integrable but not continuous on $[0,1]$, does $\lim_{n \to \infty} \left( \int_0^1 |f(x)|^n dx \right)^{\frac{1}{n}}$ exists?

A first observation is that you can use Hölder's inequality to show that the $a_n$ are increasing: $$ a_n^n = \int_0^1 |f(x)|^n \,dx$$ $$ \leq \left( \int_0^1 |f(x)|^{(n+1)} \,dx \right)^{n/(n+1)} \left( \int_0^1 1^{n+1 \,dx} \right)^{1/(n+1)} $$ $$ = \ (a_{n+1}^{n+1})^{n/(n+1)}=a_{n+1}^n, $$ so $a_n \leq a_{n+1}$.

Thus $\lim_{n \to \infty} a_n$ does always exist in a sense, but if $f$ is unbounded then the limit can (and will) be infinity. As an example you can look at something like $f(x) = x^{-1/10}$. Then $a_n$ is finite for $1 \leq n \leq 9$ but infinite for $n \geq 10$.