Can we make any statements about $\int_0^1 f(x)^n dx$ as $n \rightarrow \infty$ assuming only the continuity of $f$? I would like to show, for $f: [0, 1] \rightarrow [0, \infty)$
$$\int_0^1 f(x)^n dx \sim \int_0^1 g_n(x)^n dx$$
as $n \rightarrow \infty$
where
$$g_n(x) = 
\begin{cases}
f(x), & \log(f(x)) \ge -1/n\\
0, &\mathrm{otherwise}\\
\end{cases}$$
Note: the integral could be calculated using Laplace's method if $f$ satisfied a few conditions, but I would like a more general result assuming nothing but the continuity of $f$.
It seems like we should be able to ignore the smaller parts of $f$ as we increase $n$. Is there an obvious proof or counterexample?
 A: It is just a simple application of Lebesgue's dominated convergence theorem:
$$0\le \int_0^1 f(x)^n - g_n(x)^n dx=\int_0^1 f(x)^n \mathbb{1}_{\{f^n<1/e\}}(x) dx=\int_0^1 h_n(x) dx$$
Now it is easy to prove that $h_n\to0$ pointwise and it is bounded by $1/e$, so $\int_0^1 h_n(x)dx\to 0$
You don't even need continuity, integrability is enough.
EDIT: After more thinking, $f(x)^n$ is not guaranteed to be integrable (think about $f(x)=\frac{1}{\sqrt{x}}$), so continuity is still required for the question to be well-defined, or at least integrability of $f(x)^n$ for all $n$ (another sufficient condition is for $f$ to be bounded almost everywhere)
A: Take $B_n=\{x\in[0,1]:\log(f(x))< - 1/n, f(x)\neq 0\}$. This means $f(x)< \exp(-1/n)$.
So then
$$ \int_{B_n} f(x)^n\,dx < \int_{B_n} \exp(-1)\,dx = \exp(-1) \lambda(B_n)$$
We know that if $m>n$ then $B_m\subseteq B_n$. Also as $\bigcap B_n=\emptyset$ we get
$$ \lim_{n\to\infty} \int_{B_n} f(x)^n\,dx = 0$$
Finally we have $f\equiv g_n$ on $B_n^c$. This means
$$ \int_0^1 f(x)^n\,dx - \int_0^1 g_n(x)\,dx = \int_{B_n} f(x)^n\,dx \xrightarrow{n\to\infty} 0$$
Also nota bene that we do not need $f$ continuous but only $f$ measurable.
