Let $\phi:\mathbb{R\to R}$, be an integrable function with finite integral on $[0,1]$($\int_{[0,1]}\phi(x)dm<\infty$) and $\phi(x)=\phi(x+1)\forall x\in \mathbb{R}$. Prove that $$f(x)=\sum_{n=1}^\infty\frac{\phi(nx)}{n^2}$$ is a.e finite.
Proving using measure theory techniques is not a problem (mainly we say that by dominant convergence theorem it follows) but I wonder how to solve it using only equi-distribution. I think that since $\phi(x)$ is 1 periodic, $\phi(nx)$ is of peroid n (means $\phi(nx)=\phi(nx+n)$) so on every interval $[0,n]\subset\mathbb{R}$, $\frac{\phi(nx)}{n}$ is equi-distributed hence the sum is convergent to integral, means $$\frac 1 n\sum\frac{\phi(nx)}{n}\to\int_{\mathbb{T}}\frac{\phi(nx)}{n}\frac{dt}{2\pi}<\int_{\mathbb{T}}\phi(x)\frac {dt}{2\pi}<\infty$$ hence $f$ is finite a.e.
But I'm not sure why and how the sequence $x_n=\frac{\phi(nx)}{n}$ is equidistributed and on which interval? Is my way of solution correct?