This is part of a proof I am trying to solve, but I got stuck. Part of the problem is to prove that, for some fixed $y \in \mathbb{R}^n \backslash \{0\}$ $$\int_{\mathbb{R}^n} \frac{1 - \cos(x \cdot y)}{|x|^{n + q}} > 0$$
I was planning on using the theorem that says that says that, for $f$ measurable and non-negative, if $\int_X f(x) \mathrm{d}\mu = 0$ then $f = 0$ almost everywhere.
Therefore, if $\int_{\mathbb{R}^n} \frac{1 - \cos(x \cdot y)}{|x|^{n + q}} = 0$, then we should have $1 - \cos(x \cdot y) = 0$ which is true iff $x \cdot y = 0$ a.e. iff $x \perp y$ a.e.
Then, I tried to prove that $m_n(\{x \in \mathbb{R}^n:x \perp y\}) = 0$ which would mean that it is not true that $x \perp y$ almost everywhere, but I can't prove this last part.
What I did so far was:
Rotate the coordinate system in a way that leaves $y$ paralell to the $x_1$-axis. Therefore, the set of vectors perpendicular to $x_1$ in $\mathbb{R}^n$, is the set of ordered $n$-tuples $$\{x \in \mathbb{R}^n : x = (0,x_2,...,x_n)\} = \{0\} \times \mathbb{R}^{n-1}$$
Therefore,
$$m_n(\{0\} \times \mathbb{R}^{n-1}) = m_1(\{0\}) m_{n-1}(\mathbb{R}^{n-1}) = 0 \cdot \infty = 0 \text{ (at least in measure theory)}$$
Is this acceptable?