How can I prove that for the function $f:\mathbb{R}^n\to \mathbb{R}$:
$$f(x) = \sin(\log |\log |x||)$$
0 is not a Lebesgue point?
The point $y$ is a Lebesgue point if:
$$ \lim_{r\to 0} \frac{1}{|B_r(y)|}\int_{B_r(y)} |f(x)-f(y)|\, dy =0 $$
In the average integral of the $f$ derivative I tried some approximation but was unsuccessful.
Indeed, the function is not defined at $x=0$ or on the unit sphere. But at all other points, the function is defined. But we can definitely define it at $x=0$. Let's show that no matter what value $a$ we assign to $f$ at $x=0$, the expressions
$$ A_r = \frac{1}{r^n}\int_{B_r (0)} |f(y) - a| dy$$
do not converge to $0$ as $r\to 0+$. WLOG we will assume $r < 1/2$. By radial symmetry, $A_r$ is equal, up to a multiplicative constant, to
$$ r^{-n} \int_0^r |\sin (\ln |\ln s|)-a| s^{n-1} ds.$$
Change variables $s \to u = |\ln s|$ gives $du = -\frac{1}{s} ds$, and so
$$(*)\quad r^{-n} \int_{|\ln r|}^\infty |\sin (\ln u)-a| e^{-n u} du$$
Using the priodicity of $\sin$, there exists a sequence of disjoint intervals $I_k = [\alpha,\beta] + 2\pi k~,k=1,2,\dots$ such that $|\sin (x) -a|>c_0>0$ for all $x$ in the union of the intervals (if $a=0$ choose $c_0=\frac 12$ and if $a\ne 0$ choose $c_0= |a|/2$). Now $x \in I_k$ if and only if $\ln (\alpha +2\pi k) \le \ln x \le \ln (\beta + \pi k)$. Therefore the integral is bounded below by the sum
$$ (**) \quad c_0 \sum_{k=k(r)}^\infty (\beta+ 2\pi k)^{-n} (\ln ( \beta +2\pi k) - \ln (\alpha +2\pi k)),$$
and where $k(r)$ is the smallest $k$ such that satisfying $\alpha + 2\pi k \ge |\ln r|$. That is
$$k(r) \sim |\ln r|/2\pi.$$
Below, $c,c',c''$ are constants not depending on $r$.
By a simple comparison argument, the general term of the series in $(**)$ is bounded below by $c k^{-n-1}$ for all $k\ge k(r)$, and therefore the summation is bounded below by $c' k(r)^{-n}=c''|\ln r|^{-n}$. Now $|\ln r|r \to 0$ as $r\to0$, and therefore, $\liminf_{r\to 0+} A_r = \infty$.