Monotone convergence of $\sum\nolimits_{n=1}^\infty n^{- \alpha} f(nx) < \infty $

Question

I'm doing following exercise atm. and need some help.

Let $f:\mathbb{R} \to [0,\infty]$ Lebesgue-integrable. Show that, for all $\alpha > 0$ and for almost all $x \in \mathbb{R}$ holds $$\sum\nolimits_{n=1}^\infty n^{- \alpha} f(nx) < \infty . (1)$$

To show (1) my plan is, to show that $g(x)=\sum\nolimits_{n=1}^\infty n^{- \alpha} f(nx)$ for $x \in \mathbb{R}$ , $g:\mathbb{R} \to [0,\infty]$ is lebesgue-integrable and I have to compute the integral $\int_{-\infty}^\infty g(x) \ \ dx.$

To show that $g(x)=\sum\nolimits_{n=1}^\infty n^{- \alpha} f(nx)$ is lebesgue-integrable,I have to show first, that $g$ is lebesgue-measurable.

For the verification that $g$ is lebesgue-measurable, there are equivalent statements:

• For a function f from a measuring space $(X,\mathcal{A})$ to ($\mathbb{R},\mathcal{B}(\mathbb{R})$) holds that f is measurable if and only if one of the following conditions is fulfilled:
  • ${f \le a} \in \mathcal{A} \ \ \forall a \in \mathbb{R},$
  • ${f < a} \in \mathcal{A} \ \ \forall a \in \mathbb{R},$
  • ${f \ge a} \in \mathcal{A} \ \ \forall a \in \mathbb{R},$
  • ${f > a} \in \mathcal{A} \ \ \forall a \in \mathbb{R}.$

Here {$f \le a$} is an abbreviation for {$x \in X : f(x) \le a$} = $f^{-1}((-\infty,a]).$

For our case it is {$g \le a$} = $g^{-1}((\infty,a])$ = {$x \in \mathbb{R}:\sum\nolimits_{n=1}^\infty n^{- \alpha} f(nx) \le a, \ \ a \in \mathbb{R}$} My problem is to specify a concrete interval that satisfies the property of $g^{-1}((\infty,a]).$ How can I construct such a Intervall?

The second thing I want to do is, to specify a majorant function.

Informations that I collected to (1):

• the series converge for $\alpha > 1$, therefore we only consider the case $\alpha > 1$.
• this one is just an assumption: when $f<\infty$ than $(1) <\infty$ because the multiplication of two finite things are finite.

I would appreciate some help for solving this exercise.

kind regards,

WomBud.

Answer 1

$\int_{-\infty}^\infty g(x) dx = \int_{-\infty}^\infty \sum_{n=1}^\infty n^{-\alpha} f(nx) dx = \sum_{n=1}^\infty \int_{-\infty}^\infty n^{-\alpha} f(nx) = \sum_{n=1}^\infty \frac{1} {n^{\alpha+1}} \int_{-\infty}^\infty f(x)dx. $

We set $g_N(x) = \sum_{n=1}^N f(nx) \to g(x) , for \ \ \ N \to \infty.$

It holds that $0 \le g_N \uparrow g$, because $f(nx) \ge 0.$

With this $\lim \limits_{n \to \infty} \int_{-\infty}^\infty g_N(x)dx = \int_{-\infty}^\infty g(x)dx. $