If we have some function $f:\mathbb{R} \to[0,\infty)$ such that $\int_{\mathbb{R}} f(x)dx =N$ for some finite $N$, it intuitively makes sense that there exists some (compact) interval $S=[-M, M] \subset \mathbb{R}$ with $\int_{S}f(x)dx$ as close to N as we want. i.e.
$\forall \epsilon > 0$, $\exists M>0$ such that $|\int_{-M}^{M}f(x)dx-N|<\epsilon$
But I am not sure how to justify/understand more about this. I don't know much about compactness support yet. Any help would be appreciated.
You can use dominated (or monotonic) convergence. If you call $f_M$ the extension by zero of $f$ outside the interval $[-M,M]$, it is clear that $f_M\to f$ point wise as $M\to\infty$ and $f_M$ is increasing. By the monotonic convergence theorem, $$ \int_{-\infty}^{\infty}f_M(x)\,dx=\int_{-M}^{M}f(x)\,dx\to\int_{-\infty}^{\infty}f(x)\,dx. $$