I was reading a proof on Fourier's Integral Theorem, I won't get too much into that because the reason for the question doesn't really have a lot to do with it so I will just mention the steps which made me ask this. At the end of the proof, they change the order of integration "because there is uniform convergence"
The integral whose order of integration is being inversed is the one in the following expression:
$$\lim_{n \rightarrow \infty} \int_{-\infty}^{+\infty} \int_{-n}^{+n} \frac{1}{2}f(x)e^{-i\omega(x-x_0)}d\omega dx=\pi\frac{f(x_o^-)+f(x_0^+)}{2}$$
Where they change the order from $d\omega dx$ to $dxd\omega$. They do prove there is uniform convergence showing that the following integral is bounded:
$$|\int_{-\infty}^{+\infty}f(x)e^{-i\omega(x-x_0)}dx|\leq \int_{-\infty}^{+\infty}|f(x)|dx$$
Where the RHS of the inequality is $<\infty$ from one of the hypothesis of the theorem they were trying to prove in the first place. Then, using Weierstrass Criteria, it must converge uniformally and then the interchanging of the order of integration was indeed valid.
What I do not understand is why does uniform convergence allows us to invert the order of integrarion. I've looking in my notes of this course and on the internet but I could only find information about Uniform Convergence of series of functions, not about an improper integral of a function. I'd appreciate it if someone could provide me some proof linking Uniform Convergence with being allowed to invert the order of integration.
Thanks in advance.