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.  

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.
