I recently encountered an exercise I can't solve:
Suppose that for $n\in \mathbb{N}$ the functions $fn,f: (0,\infty)\to \mathbb{R}^+_0$ are improperly integrable and for every $x\in(0,\infty)$ the sequence $(f_n(x))$ is monotone increasing and convergent to $f(x)$. Lastly, suppose that $f_n\to f$ converges uniformly on every compact interval $[a,b]\subset (0,\infty)$. Prove that: \begin{equation} \lim_{n\to\infty}\int_0^\infty f_n=\int_0^\infty f \end{equation}
My initial idea is that I have to prove somehow that \begin{equation} \left| \int_0^\infty f-\int_0^\infty f_n\right|<\varepsilon \end{equation} for a large enough $n$. So my attempt would go something like this: \begin{equation} \left| \int_0^\infty f-\int_0^\infty f_n\right|=\left|\int_0^C f-\int_0^C f_n+\int_C^\infty f-\int_C^\infty f_n\right|\leq \int_0^C |f-f_n|+\int_C^\infty |f-f_n| \end{equation} Now we can discard the first integral due to uniform convergence of $f_n$. I have no idea, however, what to do with the second one.
Note: I am aware that there are similar questions to this, however, I did not understand how to apply their problem to my situation. Thank you for your help!
It is not given that $f_n \to f$ uniformly on $(0,C)$, but rather on any compact interval $[a,b] \subset (0,\infty)$. Proceeding similarly to your attempt, with any $0 < c < C$,
$$\left|\int_0^\infty f - \int_0^\infty f_n\,\right| = \left|\int_0^c f + \int_c^C f +\int_C^\infty f - \int_0^c f_n-\int_c^C f_n-\int_C^\infty f_n\,\right| \\ \leqslant \int_c^C |f - f_n| + \int_0^c |f_n|+ \int_C^\infty |f_n| + \int_0^c |f| + \int_C^\infty |f|$$
Since the $f_n$ are nonnegative and increasing to $f$, we have $|f_n| = f_n \leqslant f = |f|$ and, hence,
$$\left|\int_0^\infty f - \int_0^\infty f_n\,\right| \leqslant \int_c^C |f - f_n| + 2\int_0^c |f| + 2 \int_C^\infty|f|$$
Since $f = |f|$ is improperly integrable on $(0,\infty)$, for any $\epsilon > 0$, there exist $c'$ and $C'$ such that
$$\int_0^{c'} |f| \leqslant \frac{\epsilon}{6}, \quad\int_{C'}^\infty |f| \leqslant \frac{\epsilon}{6}$$
Thus,
$$\left|\int_0^\infty f - \int_0^\infty f_n\,\right| \leqslant \int_{c'}^{C'} |f - f_n| + \frac{2\epsilon}{3}$$
Now apply the uniform convergence of $f_n$ to $f$ on the compact interval $[c',C']$ to finish.