# Uniform convergence and Riemann integral; from compact $0<t<T<+\infty$ to $(0,+\infty)$ [Hint]

Let be $g, f_n, \forall n \in \mathbb{N}$ well-defined functions in $(0, \infty)$, and Riemann integrable in every compact subset $[t, T]$ where $0 < t < T < \infty$. In addition, $f_n$ converges uniformly to a function $f$ and $|f_n| \leq g$ in every compact subset of $[0, \infty]$. If we have that:

$\int_0^{\infty}g(x)dx < \infty$

Proove that $\lim_{n \rightarrow \infty} \int_0^{\infty}f_n(x)dx = \int_0^{\infty}f(x)dx$

I am supposed to solve this problem without using directly any theorem about Measure Theory (like the Dominated Convergence Theorem, which has similarities with this exercise), only using facts about convergence of functions.

My problem is that I can't find a way to start. I know that if $f_n$ are Riemann integrable in a certain subset and they converge uniformly to a function $f$ in t his subset then f is Riemann integrable in this subset and I can put the limit inside or outside the integral. The problem is that I know it for every interval $[t,T]$ but they are talking about $(0, \infty)$, and this generalization is non-trivial and here the function $g$ has to help me but I don't know how to use the bound that they give me.

Thanks :)

Consider the difference $\int f_n(x)dx-\int f(x)$. You want to try and make a claim that $|f_n(x)-f(x)|<\epsilon$ for all $x$ when $n$ is big enough but you don't quite have this. Instead, use the fact that $g$ is integrable. This means that $\int_c^\infty g(x)dx$ can be made arbitrarely small as $c$ goes to infinity. So now write:
$$\left|\int_0^\infty f_n(x)dx-\int_0^\infty f(x)dx\right|\leq \int_0^c|f_n(x)-f(x)|dx+2\int_c^\infty g(x)dx$$
where you can now use the fact that $|f_n(x)-f(x)|<\epsilon$ on $[0,c]$ for large enough $n$.