Given $f(x)$ is integrable on $[0, 1]$ and $0 < f(x) < 1$, prove that $\int_{0}^{1} (f(x))^{n} \mathop{dx}$ converges to $0$. 
Given $f(x)$ is integrable on $[0, 1]$ and $0 < f(x) < 1$, prove that
  $\int_{0}^{1} (f(x))^{n} \mathop{dx}$ converges to $0$.

I understand why the statement is true intuitively because as $n \to \infty$, since $f$ lies between $0$ and $1$, it will be like a fractional value, which converges to $0$ since the fractions get smaller and smaller.
However, I am not sure about how to prove this rigorously.  

You may use the following theorem due to Arzelà :---
Let $\{f_n\}$ be a sequence of Riemann integrable Functions on $[a,b]$ and converges point-wise to $f$, also there is a positive number $M$ such that $|f_n(x)|≤M,\forall x\in [a,b],\forall n\in \Bbb N$. Now if $f$ is Riemann integrable over $[a,b]$ then , $$\lim_{n\rightarrow \infty}\int_a^bf_n(x)dx=\int_a^b\lim_{n\rightarrow \infty} f_n(x)dx=\int_a^b f(x) dx.$$

Here $f_n(x)=(f(x))^n\rightarrow 0$ as $n\rightarrow \infty$ $,\forall
 x\in [0,1]$.

