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 $0$.

Here is a proof which assumes some amount of measure theory (and I think this is unavoidable, but I may be wrong in thinking so).

Let $f_n(x) =(f(x)) ^n $ then each $f_n(x) $ is Riemann integrable on $[0,1]$ and hence the set $D_n$ of its discontinuities is of measure $0$ and thus the set $D=\bigcup\limits_{n=1}^{\infty}D_n$ is of measure $0$. Let $\epsilon>0$ be given. Then there is a sequence of open intervals $\{J_n\}$ such that $D\subseteq \bigcup\limits_{n=1}^{\infty} J_n$ and the length of these intervals $J_n$ combined is less than $\epsilon$.

Next $f_n(x) \to 0$ as $n\to\infty $ for all $x\in[0,1]$. Let $x\in[0,1]\setminus D$. Then we have a positive integer $n_x$ depending on $x$ such that $f_n(x) <\epsilon$ for all $n\geq n_x$. By continuity of $f_{n_x}$ at $x$ it follows that there is a neighborhood $I_x$ such that $f_{n_x} (x) <\epsilon $ for all $x\in I_x$. Since $f_n$ is decreasing it follows that we have $f_n(x) <\epsilon$ for all $x\in I_x$ and all $n\geq n_x$.

Now the set of all neighborhoods $I_x$ as $x$ varies in $[0,1]\setminus D$ together with the intervals $J_n$ forms an open cover for $[0,1]$ and thus by Heine Borel theorem a finite number of these intervals covers $[0,1]$. Thus we have $$[0,1]\subseteq \bigcup\limits_{i=1}^{p}I_{x_i} \cup\bigcup\limits_{i=1}^{q}J_i$$ Let $N$ be the maximum of integers $n_{x_1},n_{x_2},\dots,n_{x_p}$ then we have $$f_n(x) <\epsilon, \forall x\in\bigcup\limits _{i=1}^{p}I_{x_i} , \forall n\geq N$$ The end points of $J_1,J_2,\dots,J_q$ which lie in $[0,1]$ partition it into a finite number of subintervals. Denote the union of all those subintervals which contain points of $J_1,\dots, J_q$ as $A$ and let the union of remaining subintervals be denoted by $B$. Then length of $A$ is less than $\epsilon$ and $f_n(x) <\epsilon$ for all $n\geq N$ and all $x\in B$. Thus we have $$\int_{0}^{1}f_n(x)\,dx=\int_{A}f_n(x)\,dx+\int_{B}f_n(x)\,dx<\epsilon +\epsilon =2\epsilon $$ for all $n\geq N$. Therefore $\int_{0}^{1}f_n(x)\,dx\to 0$ as $n\to \infty $.

Note that the above argument actually proves the following result:

Theorem: Let $\{f_n\} $ be a sequence of functions $f_n:[a, b] \to\mathbb {R} $ such that each $f_n$ is non-negative and Riemann integrable on $[a, b] $ and $f_n(x) \geq f_{n+1}(x),\forall x\in[a, b] $ and $f_n(x) \to 0$ point wise almost everywhere in $[a, b] $ then $\int_{a} ^{b} f_n(x) \, dx\to 0$.

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 $n\rightarrow \infty$ $,\forall x\in [0,1]$.

Since $f$ is integrable, it is measurable. By Lusin's theorem, for any $\varepsilon>0$ there exists a compact set $K\subset [0,1]$ such that $f$ is uniformly continuous on $K$ and $|K|>1-\varepsilon$. Uniform continuity implies that $\sup_{x\in K} f(x) = \lambda<1$. Thus $$\begin{align} \int_{[0,1]} f(x)^n\, dx &= \int_{K} f(x)^n\, dx + \int_{[0,1]\backslash K} f(x)^n\, dx \\ &\le |K|\lambda^n + \varepsilon\cdot1. \end{align}$$ Take limit as $n\to\infty$ yields $$ \limsup_{n\to \infty} \int_{[0,1]} f(x)^n\, dx \le \varepsilon. $$ Since the above hold for any $\varepsilon>0$, we have $\int_{[0,1]} f(x)^n\, dx\to 0$ as wanted.