If $I_n=\sqrt[n]{\int _a^b f^n(x) dx}$,for $n\ge 1$.Find with proof $\lim _{n \to \infty}I_n$ Let $a,b \in \mathbb R ,a<b$ and let  $f:[a,b] \to [0,\infty) $ a continuous and non constant.

attempt,using Reimann series

$I_n=\sqrt [n]{\int _a^b f^n(x) dx}$
$I_n=\sqrt [n]{\lim _{k \to \infty}\sum_{i=1}^k f^n(x_i^*) dx}$
for each $i=1,2,3...n$
$x_i\in [x_{i-1},x_i]$
so our question is what is 
$I_n=\lim_{n\to \infty}\sqrt [n]{\lim _{k\to \infty}\sum_{i=1}^k f^n(x_i^*) dx}$

Obviously, we have
$$
\begin{align}
\limsup_{n\to\infty}\left(\int_a^bf^n(x)\,\mathrm{d}x\right)^{1/n}
&\le\lim_{n\to\infty}(b-a)^{1/n}\left(\max_{[a,b]}f^n\right)^{1/n}\\
&=\max_{[a,b]}f\tag{1}
\end{align}
$$
Suppose that $f(x)\ge m$ on an interval of length $\delta\gt0$. Then, since $\frac{f(x)}{m}\ge1$ on this interval,
$$
\begin{align}
\liminf_{n\to\infty}\left(\int_a^bf^n(x)\,\mathrm{d}x\right)^{1/n}
&=m\liminf_{n\to\infty}\left(\int_a^b\frac{f^n(x)}{m^n}\,\mathrm{d}x\right)^{1/n}\\
&\ge m\lim_{n\to\infty}\delta^{1/n}\\
&=m\tag{2}
\end{align}
$$
If $f$ is continuous, then for any $\epsilon\gt0$, $f(x)\ge m=\max\limits_{[a,b]}f-\epsilon$ on some interval of positive length. Therefore, for any $\epsilon\gt0$,
$$
\small\max_{[a,b]}f-\epsilon\le\liminf_{n\to\infty}\left(\int_a^bf^n(x)\,\mathrm{d}x\right)^{1/n}\le\limsup_{n\to\infty}\left(\int_a^bf^n(x)\,\mathrm{d}x\right)^{1/n}\le\max_{[a,b]}f\tag{3}
$$
Since $(3)$ is true for any $\epsilon\gt0$, we have
$$
\lim_{n\to\infty}\left(\int_a^bf^n(x)\,\mathrm{d}x\right)^{1/n}=\max_{[a,b]}f\tag{4}
$$
