Problem. Let $\{A_{\vec{k}}\}$ be a sequence of real numbers indexed by vectors $\vec{k}=(k_{1},\ldots,k_{n})\in\mathbb{N}$. Let $\{r_{\vec{k}}\}$ be a sequence of positive real numbers such that
$$\sum_{\vec{k}}r_{\vec{k}}|\cos(\vec{k}\cdot x+A_{\vec{k}})|<\infty$$
Shown then that $$\sum_{\vec{k}}r_{\vec{k}}<\infty$$
There was a first part to this qual problem, which was show that
$$\liminf_{|\vec{k}|\rightarrow\infty}\int_{\mathbb{R}^{n}}|\cos(\vec{k}\cdot x+A_{\vec{k}})|f(x)dx>0,$$
where $f:\mathbb{R}^{n}\rightarrow (0,\infty)$ is a positive integrable function. Here, $|\vec{k}|:=\sum_{i=1}^{n}k_{i}$. One can prove this result by approximating $f$ from below by step functions. I do not see how, though, to use it for my problem above; nor do I have an idea for another argument. Any suggestions?
I'll work on $\mathbb R.$ The extension to $\mathbb R^n$ will be clear I hope. We fix a real sequence $a_k$ and a positive sequence $r_k$ and assume $\sum_{k=1}^{\infty}r_k|\cos (kx+a_k)| < \infty$ for $x$ in some set of positive measure.
Lemma: Suppose $E$ is a set of positive finite measure. Then there exists $k_0$ such that
$$\tag 1 \int_E |\cos (kx+a_k)|\, dx > m(E)/4 \text { for } k > k_0.$$
Proof: We have $|\cos (kx+a_k)|\ge \cos^2 (kx+a_k) = [1+\cos (2kx + 2 a_k)]/2.$ Therefore
$$\int_E |\cos (kx+a_k)|\, dx \ge \int_E [1+\cos (2kx + 2 a_k)]/2\, dx =m(E)/2 + \int_E [\cos (2kx + 2 a_k)]/2\, dx.$$
An easy argument using Riemann-Lebesgue lemma shows the last integral $\to 0$ as $k\to \infty.$ This proves $(1).$
Define $f(x) = \sum_{k=1}^{\infty}r_k|\cos (kx+a_k)|.$ Then $f$ is a measurable function on all of $\mathbb R,$ with values in $[0,\infty].$ We are given that $f<\infty$ in a set of positive measure. It follows that $f$ is bounded by some $C$ on a set $E$ of positive finite measure. Referring to the lemma we then have
$$\infty> Cm(E) \ge \int_E f \ge \int_E \sum_{k>k_0}^{\infty}r_k|\cos (kx+a_k)|\, dx$$ $$ = \sum_{k>k_0}^{\infty}r_k\int_E |\cos (kx+a_k)|\, dx \ge \sum_{k>k_0}^{\infty}r_k[m(E)/4].$$
This shows $\sum_{k>k_0}^{\infty}r_k <\infty$ and we're done.