Let $f: [0,+\infty) \to \mathbb{R}^+$ a continuous and non negative function, $f(x) \ge 0$, so that $\sum_{n=1}^\infty f(na)$ converges $\forall a \ge 0$. I would like to prove that $\sum_{n=1}^\infty f(nx)$ converges uniformally in a compact interval to the function $g(x)= \sum_{n=1}^\infty f(nx)$.
So, I consider the functions $g_m(x)=\sum_{n=1}^m f(nx)$, $g_m(x)$ is a continuous function since it's the sum of a finite series of continuous functions. $F_m^N=\{g_m(x) \le N\}$ is closed, $B_m^N=g^{-1}(F_m^N)$, from continuity of $g$ $B_m^N$ is closed as well. Now, let $A_N=\bigcap_{m \in \mathbb{N}}B_m^N$, $A_N=\{x \ge 0 g(x) \le N\}$ is also closed. Since $[0, +\infty)= \bigcup_N A_N$ by Baire's theorem there must be $N_0$ so that $Int(A_{N_0}$)$\ne \emptyset$. So let $E=[x_0-r, x_0+r] \subset A_{N_0}$,
(now this is the point in which I have more doubts)
$$\begin{split} 0 \le \lim_{k \to \infty} \sup_E | \sum_{n=k}^\infty f(nx)| &= \lim_{k \to \infty} \sup_E \sum_{n=k}^\infty f(nx)\\ &= \lim_{k \to \infty} \sup_E \biggl(\sum_{n=1}^\infty f(nx)- \sum_{n=1}^{k-1} f(nx) \biggr)\\ &\le N- \lim_{k \to \infty} \sum_{n=1}^{k-1} f(nx) =0 \end{split}$$
So $\lim_{k \to \infty} \sup_E | \sum_{n=k}^\infty f(nx)|=0$ and this proves the uniform convergence in $E$.
Is this correct? Thank you!