Suppose $g : [0, 1] \to [0, 1]$ is a Lebesgue measurable function. Let $f : [0, 1] \to \mathbb{R}$ be continuous with $f(0) \le f(1)$. Show that the $\lim_{n\to\infty} \int_0^1 f(g(x)^n)dx$ exists and lies in the interval $[f(0), f(1)]$ .
$\textbf{My attempt:}$
Is this attempt correct? (I'm not sure how more rigorously I can write that)
Since $f$ is continuous over a compact set $[0,1]$ it attains its extremes. Let $f_n = f(g(x)^n)$ then since $f$ is continuous $f_n$ is measurable and by the fact $f$ attains its maximum $f_n \le M$ for some $M>0$.
moreover, by continuity of $f$ we'd have that
$(\ast) \lim_{n\to\infty} f(g(x)^n) = f(\lim_{n\to\infty} g(x)^n) =\left\{\begin{matrix} f(1) & \text{if} \ g(x)=1\\ f(0) & \ \text{if} \ 0\le g(x) <1 \end{matrix}\right. $
So, using dominated convergence theorem we'd have ;
\begin{align} \lim_{n\to\infty} \int_0^1 f(g(x)^n)dx & = \int_0^1\lim_{n\to\infty} f(g(x)^n)dx\\ &= \int_0^1f(\lim_{n\to\infty} g(x)^n)dx = \left\{\begin{matrix} f(1) & \text{if} \ g(x)=1\\ f(0) & \ \text{if} \ 0\le g(x) <1 \end{matrix}\right . \\ \end{align}