limit of integral and uniform convergence Assume that $f_n : [0,1] \to \mathbb{R}$ are integrable and that $(f_n)$ uniformly converges to $f$. Does the limit $\lim \limits_{n \to \infty} \int \limits_{0}^{1} f_n(x)^{2018} dx$ exists? Sure because $g(x) = x^{2018}$ is continuous on $[0,1]$ and $f_n$ are integrable and uniformly converge to $f$, so this means that $f$ is integrable and $g(f(x))$ is also integrable so the integral $\int \limits_{0}^{1} f(x)^{2018} dx$ exists and I think that $\int \limits_{0}^{1} f(x)^{2018} dx = \lim \limits_{n \to \infty} \int \limits_{0}^{1} f_n(x)^{2018} dx $ because of the interchanging between limit and integral under uniform convergence, but is there another way to prove this without interchanging limits with integrals?

$g(x)$ has a bounded derivative on $[0,1]$ so by the MVT $g(x)$ is Lipschitz-continuous on $[0,1.$ So $g(f_n)$ converges uniformly to $g(f)$ on $[0,1$. And each $g(f_n)$ is integrable so $g(f)$ is integrable. Hence $$\lim_{n\to \infty}\int_0^1g(f_n(x))dx=\int_0^1g(f(x)dx=\int_0^1\lim_{n\to \infty}g(f_n(x))dx.$$

You cannot prove this without an interchange of the order of the limit and the integral because what you are trying to prove is exactly that the order $can$ be interchanged.