Convergence of double Integral Let's assume that $f \in L^p(\mathbb{R})$ ($1 \leq p < \infty$). Does it then hold that
$$
\lim_{n \rightarrow \infty} \int_0^1 \int_0^1 \left \lvert f\left( \frac{\tilde{r}}{n} \right) -f\left(\frac{r}{n} \right) \right \rvert^p ~\mathrm{d}r \mathrm{d}\tilde{r} = 0 \quad ?
$$
And if yes, how can I prove it? I am quite certain that this is true. I tried to use that $\displaystyle \lim_{h \rightarrow 0} \lVert f(\cdot + h) - f \rVert_{L^p} = 0$ but that did not help me as in this case I have a double Integral. Maybe I could youse a density argument or a transformation but I don't quite know how.
Can anyone help me out? Or does there exist a counterexample?
 A: Here is a counterexample with $p=1$. Take
$$ f(x) = \mathbf 1_{(\frac 1 2, \frac 3 4)} + \mathbf 1_{(\frac 1 4, \frac 3 8)} + \mathbf 1_{(\frac 1 8, \frac 3 {16})} + \mathbf 1_{(\frac 1 {16}, \frac 3 {32})} + \dots $$
Then $f \in L^1(\mathbb R)$, with $\| f \|_{L^1(\mathbb R)} = \frac 1 2 $.
Now observe that
$$ x \in [0, 1] \implies f(x) = f\left( \frac  x n \right) \ \ \ \  $$
for all $n$ of the form $n = 2^k$ where $k \in \mathbb N$.
So for $n$ of the form $n = 2^k$, we have
$$ \int_0^1 \int_0^1 | f\left( \tfrac x n \right) - f\left( \tfrac y n \right) | dx dy  =  \int_0^1 \int_0^1 | f\left(x \right) - f\left( y \right) | dx dy = \frac 1 2 $$
Hence it's impossible for $ \int_0^1 \int_0^1 | f\left( \tfrac x n \right) - f\left( \tfrac y n \right) | dx dy  $ to tend to $0$ as $n \to \infty$.
A: Well, for fixed $\tilde{r} \in [0,1]$ we have
$$\lim_{n\to\infty} \int_0^1 \left|f\left(\frac{\tilde{r}}n\right) - f\left(\frac{r}n\right)\right|^p\,dr =\int_0^1 \lim_{n\to\infty}\left|f\left(\frac{\tilde{r}}n\right) - f\left(\frac{r}n\right)\right|^p\,dr= 0$$
by the Lebesgue dominated convergence theorem since the integrand is dominated by $2^p\|f\|_{L^P}^p$ which is integrable on $[0,1]$.
Therefore the sequence of functions
$$\tilde{r} \mapsto \int_0^1 \left|f\left(\frac{\tilde{r}}n\right) - f\left(\frac{r}n\right)\right|^p\,dr$$
converges pointwise to $0$ when $n\to\infty$ and it is bounded again by $2^p\|f\|_{L^P}^p$ so by the Lebesgue dominated convergence theorem we have
$$\lim_{n\to\infty} \int_0^1 \int_0^1 \left|f\left(\frac{\tilde{r}}n\right) - f\left(\frac{r}n\right)\right|^p\,dr\,d\tilde{r} = \int_0^1 \left(\lim_{n\to\infty}\int_0^1 \left|f\left(\frac{\tilde{r}}n\right) - f\left(\frac{r}n\right)\right|^p\,dr\right)d\tilde{r} = 0$$
