If $x$ is a Lebesgue point of $f$, $f \in L^p$ and $f(x)=0$, then it is a Lebesgue point of $f^p$ where $p>1$ (finite)? If $x$ is a Lebesgue point of $f$ and $f \in L^p$, then it is a Lebesgue point of $f^p$ where $p>1$ (finite)?
Here I use the following definition with ball replaced by cube.
And I know if a function is integrable, then almost every point is a Lebesgue point.
Can I derive this?
It seems that around $x$ I may bound $f^p$ by $f$ because there it may be $f\leq 1$.
(Here you may make a simplification that $x=0$ by considering a translation.)
 A: It need not be a Lebesgue point of $f^p$ (or $\lvert f\rvert^p$). Consider for simplicity, the situation in $\mathbb{R}$, with $x = 0$. For $k \in \mathbb{N}\setminus \{0\}$, let
$$A_k = \left(\frac{1}{k!+1}, \frac{1}{k!} \right).$$
Then we want to find a sequence $(c_k)$ of positive real numbers such that
$$\lim_{n\to\infty} n!\sum_{k=n}^\infty c_k\left(\frac{1}{k!} - \frac{1}{k!+1}\right) = 0,$$
but
$$\limsup_{n\to\infty} n!\sum_{k=n}^\infty c_k^p\left(\frac{1}{k!}-\frac{1}{k!+1}\right) > 0.$$
Choosing $c_k = (k!)^{1/p}$ gives us
$$n!\sum_{k=n}^\infty c_k^p\left(\frac{1}{k!}-\frac{1}{k!+1}\right) = n!\sum_{k=n}^\infty k!\cdot\frac{1}{k!(k!+1)} \geqslant \frac{n!}{n!+1} \to 1$$
and for $n \geqslant 4$ we have
$$n!\sum_{k=n}^\infty \frac{c_k}{k!(k!+1)} < n!\sum_{k=n}^\infty \frac{(k!)^{1/p}}{(k!)^2} = n!\sum_{k=n}^\infty \frac{1}{(k!)^{1+1/p}} < \sum_{k=n}^\infty \frac{1}{(k!)^{1/p}} < \sum_{k=n}^\infty \frac{1}{2^{k/p}} = \frac{1}{2^{n/p}(1-2^{-1/p})}.$$
Thus $0$ is a Lebesgue point of
$$f = \sum_{k=1}^\infty c_k \cdot \chi_{A_k},$$
but not a Lebesgue point of $f^p$.
