Problem: if $X$ is an infinite measure subset of $\mathbb R$ and $f:\mathbb R\to\mathbb R$ is a continuous function integrable over $X$ and all its translations and rotations, is $f$ integrable over $\mathbb R$?
Trying to find a counterexample, I came up with this:
Does $\int_Xe^xdx$ converges? ---edit: no it doesn't
Is there any easy counterexample?
As said before, it does not converge since $a_n ≥ 0$ for all $n ∈ ℕ$ and so $\exp|_X ≥ 1$. Therefore $∫_X \mathrm{e}^x dx ≥ ∫_X dx = ∞$.
Okay, so the idea is to use the divergence of $Σ_n \frac 1 n$ vs. the convergence of $Σ_n \frac 1 {n^2}$.
Let $f \colon ℝ → [0..1],\, x ↦ \min\{1,\frac 1 {|x|}\}$ and $X = \bigcup_{k ∈ ℤ} I_k$, where $I_k$ is the interval of reals bewteen $k$ and $k+\tfrac{1}{k}$. Then $∫_ℝ f(x)dx = ∞$ and $∫_X f(x)dx \approx \sum_n \frac{1}{n^2} < ∞$, and this probably won’t change with translations or reflections. Now, I cannot give you a proof of this without investing too much time because I am not good at analysis, but I’m very certain that this works out. I hope that helps.
(If you can work this out, I would like to see the full argument.)
