For $x \in [-1,1]$ and $n \in \mathbb{N}$, consider
$f_n(x) = \frac{x^{2n}}{(1 + x^{2n})}$
I have found that the function converges pointwise to
$f(x) = \begin{cases} \frac{1}{2} & \mathrm{if} x = 1,-1 \\ 0 & \mathrm{if} x \in (-1,1) \end{cases}$
The convergence is not uniform, since $f$ is discontinuous at the endpoints, but I know that $\lim_{n \to \infty} \int_{-1}^{1} f_n(x) dx = \int_{-1}^{1} f(x) dx$ is true (because my instructor told me).
I know that $\int_{-1}^{1} f(x) dx = 0$ because the function is similar to $g(x) = 0$ except on a set of measure $0$ (the set of discontinuities.) However, I am not sure how to calculate the limit of the integral of the sequence of functions. I've tried doing it with some elementary integration methods from Calculus, but I'm not getting anywhere.
My next thought to prove they are equal was to find the exact condition (as minimal as it maybe) regarding when I can move the limit inside of the integral. However, I have not had much luck finding this.
Due to that, my next thought was possibly to prove that if a function is uniformly convergent everywhere except on a set of measure $0$ (which it is), then the limit can be moved inside of the integral. Would this be the best way to go about this problem? And if so, how would one prove this? I'm not really sure where to start this proof.
Note: My class has not covered Borel sets, sigma algebras, or any Lebesgue convergence theorems. My instructor said this is doable with techniques done before this.
HINT:
Note that $0\le \frac{x^{2n}}{1+x^{2n}}\le x^{2n}$
What is $\int_{-1}^1 x^{2n}\,dx$?