Let $f:[0,1]\to \mathbb{R}$ be a continuous function, $g:[0,1] \to \mathbb{R}$ be a Lebesgue measure and $0\leq g(x)\leq 1$ for a.e $x \in [0,1]$. Find the limit $$ \lim_{n\to \infty} \int_0^1 f(g(x)^n) dx. $$
Thanks for any help.
proof: Following the hints:
Consider $E_1=\{x: g(x)=1\}$, $E_2=\{x: 0\leq g(x)<1\}$ and $E:=E_1 \cup E_2$. Now, set $$ h_n(x):=f(g(x)^n)\cdot \chi_{E}(x)=f(g(x)^n)\cdot \chi_{E_1}(x)+f(g(x)^n)\cdot \chi_{E_2}(x) \quad E_1 \cap E_2 =\emptyset. $$ Thus, \begin{align*} \lim_{n \to \infty} h_n(x)&=\lim_{n \to \infty}f(g(x)^n)\cdot \chi_{E_1}(x)+\lim_{n \to \infty} f(g(x)^n)\cdot \chi_{E_2}(x) \\ &=f(1)\chi_{E_1}(x)+f(0)\chi_{E_2}(x) \quad \textrm{ since $f$ is continuous.} \end{align*} Moreover, we see $ |f(g(x)^n)|\leq \|f\|_{\infty}$ since $f$ attains a maximum on a compact set, in our case $g([0,1])$ is compact. Hence, we have \begin{align*} \lim_{n \to \infty } \int_0^1f(g(x)^n)dx &= \int_0^1 \lim_{n \to \infty} f(g(x)^n)dx \quad \textrm{ By Dominate Convergence Theorem }\\ &=\int_0^1 f(1)\cdot \chi_{E_1}(x)dx+ \int_0^1 f(0)\cdot \chi_{E_2}(x)dx \\ &=f(1)\int_{E_1}dx+f(0)\int_{E_2}dx\\ &=f(1)m(E_1)+f(0)m(E_2) \\ \end{align*}