Prove that if $f_n$ converges uniformly to $f(x)$, then $lim_{n\rightarrow \infty} \int f_n(x)dx =\int f(x)dx$
I have tried using the definition of uniform convergence to state the fact that if $f_n$ uniformly converges to $f(x)$ then there $\exists N$ such that if $n\geq N \longrightarrow \mid f_n(x) - f(x) \mid < \epsilon$. Hence, $\mid \int (f_n(x)-f(x)) dx \mid \leq \int \mid f_n(x) - f(x) \mid dx < \int \epsilon dx = \epsilon x$
However I do not know how to continue the proof.
This holds in general when we integrate over compact sets, in which your proof is done. If our space was not compact, we can get counter examples. We see that the functions $ \frac{1}{n}\chi_\mathbb{R} $ converge uniformly to $ f\equiv 0 $, but the integrals clearly do not converge.
If you are trying to show this for integrals over all of $ \mathbb{R} $, you need some other conditions on $ f_n $ (see the Dominated Convergence Theorem).