Example of $L^p$ function which does not have much Lebesgue points Let $0<p<1$ and $\mu$ be the $n$-dimensional Lebesgue measure.
Let $f\in L^p(\mu)$ and define $P:=\{x\in \mathbb{R}^n: \lim_{r\to 0} \frac{1}{\mu(B(x,r))} \int_{B(x,r)} |f - f(x)|d\mu =0 \}$.
Then is it possible that $\mu(P^c)>0$?
Lebesgue differentiation theorem asserts that $\mu(P^c)=0$ for locally $L^1$ functions. However, $L^p (0<p<1) $ functions are not locally $L^1$ so I'm curious if there is an example that $\mu(P^c)>0$. Thank you in advance.
 A: I'll take $n=1$ for convenience.  Let's construct a nonnegative function $f$ in $L^{1/2}$ so that 
$$ \int_a^b f(x)\; dx = \infty$$
for all $a < b$.  Then $P = \emptyset$.
Enumerate the nonempty open intervals with rational endpoints as
$(a_n, b_n)$, $n \in \mathbb N$.
First construct inductively a sequence $A_n$ of "fat Cantor sets",  nowhere-dense closed sets with nonzero Lebesgue measure,  such that


*

*$A_n$ are disjoint.

*$A_n \subset (a_n, b_n)$.

*$0 < \mu(A_n) < 1/n^3$.


Now define $f$ by
$$ f(x) = \cases{ 1/\mu(A_n) & if $x \in A_n$\cr
                  0 & if $x \notin \bigcup_n A_n$\cr} $$
Note that $f \in L^{1/2}$ since
$$\int_{\mathbb R} f(x)^{1/2} dx = \sum_{n=1}^\infty
\int_{A_n} f(x)^{1/2} \;dx = \sum_{n=1}^\infty \sqrt{\mu(A_n)} 
< \sum_{n=1}^\infty n^{-3/2} < \infty$$
But $\int_{A_n} f(x)\; dx = 1$, and each interval $(a,b)$ contains infinitely many $A_n$, so $\int_a^b f(x)\; dx = \infty$.
And thus $$\int_{B(x,r)} |f - f(x)|\; du \ge \int_{B(x,r)} f\; d\mu - 2 r f(x) = \infty$$ 
