Consider the following enumeration of the rational numbers in $[\,0,1)$: $$0, \frac{1}{2},\frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{3}{4}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5}, \frac{1}{6}, \frac{5}{6}, \cdots$$
Where the list is first group sorted by ascending denominator, then each group in ascending numerator, and deleting all reducible elements. Obviously $0$ is an exception to this sorting.
This is an arbitrary choice, but the simplest to read (in my opinion)
Also, the list begins at $0$ i.e. $r_0 = 0, r_1 = \frac{1}{2}, \cdots$
Define the following sequence of functions $\phi_n(x)$
$$\phi_0(x) = x- \pi \cot{\pi x}, \phi_1(x) = x - \pi \left(\cot{\pi x} + \cot{\pi\left(x-\frac{1}{2}\right)} \right) \\\\ \phi_2(x) = x - \pi \left(\cot{\pi x} + \cot{\pi\left(x-\frac{1}{2}\right)} + \cot{\pi\left(x-\frac{1}{3}\right)}\right) \\\\ \phi_3(x) = x - \pi \left(\cot{\pi x} + \cot{\pi\left(x-\frac{1}{2}\right)} + \cot{\pi\left(x-\frac{1}{3}\right)} + \cot{\left(x-\frac{2}{3}\right)}\right) \\\\ \phi_n(x) = x - \pi \left( \sum_{k=0}^n \cot{\pi \left(x-r_k\right)}\right)$$ where $r_k$ is the $k^{\text{th}}$ element in the enumeration described above.
Assuming $\displaystyle A = \text{PV} \left( \int_{-\infty}^\infty f(x) \, \text{d}x \right)$ is finite and is Lesbegue measurable,
consider the following sequence of integrals
$$I_n = \text{PV} \left( \int_{\infty}^\infty f(\phi_n(x)) \, \text{d}x \right)$$
It can be shown with Glasser's Master Theorem, that for any finite $n$, $I_n = A$
What happens in the limit, for $n \to \infty$ ?
Is the limit of $I_n$ still equal to $A$?
If so, how can I rigorously prove it?
Edit:
The motivation of this question was out of simple curiosity.
I wanted to see what would happen to the integral if I densely filled the function's argument with singularities, and whether any infinite weirdness happens, as is often the case.
An internet person argued that if one chooses to accept the Axiom of Countable Choice, then the limit exists and is equal to $A$.
Is this the case?